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1 + 2 + 3 + 4 + ⋯

From Wikipedia, the free encyclopedia

A graph depicting the series with layered boxes and a parabola that dips just below the y-axis
The first four partial sums of the series 1 + 2 + 3 + 4 + ⋯. The parabola is their smoothed asymptote; its y-intercept is  1/12.[1]

The infinite series whose terms are the natural numbers 1 + 2 + 3 + 4 + ⋯ is a divergent series. The nth partial sum of the series is the triangular number

which increases without bound as n goes to infinity. Because the sequence of partial sums fails to converge to a finite limit, the series does not have a sum.

Although the series seems at first sight not to have any meaningful value at all, it can be manipulated to yield a number of mathematically interesting results. For example, many summation methods are used in mathematics to assign numerical values even to a divergent series. In particular, the methods of zeta function regularization and Ramanujan summation assign the series a value of  1/12, which is expressed by a famous formula,[2]

where the left-hand side has to be interpreted as being the value obtained by using one of the aforementioned summation methods and not as the sum of an infinite series in its usual meaning. These methods have applications in other fields such as complex analysis, quantum field theory, and string theory.[3]

In a monograph on moonshine theory, Terry Gannon calls this equation "one of the most remarkable formulae in science".[4]

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  • ✪ Numberphile v. Math: the truth about 1+2+3+...=-1/12
  • ✪ The sum of all counting numbers equals WHAT?
  • ✪ L'incroyable addition 1+2+3+4+...=-1/12 - Micmaths
  • ✪ WTF: 1 + 2 + 3 + 4 + 5 ... = -1/12 ! ● Gehe auf SIMPLECLUB.DE/GO & werde #EinserSchüler
  • ✪ Ramanujan: Making sense of 1+2+3+... = -1/12 and Co.

Transcription

Welcome to the first Mathologer video of the year. Today it's about something very serious and so I'm wearing a totally black t-shirt. You all like Numberphile right? Me too, except for this one video over there in which they prove the infamous identity 1+2+3+...=-1/12 using some simple algebra that even kids in primary school should be able to follow. Since this video was published in 2014 over six million people have watched it and more than 65,000 have liked it. Unfortunately, pretty much every single statement made in this video is wrong. And by wrong I mean wrong in capital letters. In particular, as anybody who knows any mathematics will confirm 1+2+3+... sums to exactly what common sense suggests it should namely plus infinity. And this video was not published on the 1st of April. Also, as we all know, the Numberphile videos are presented by smart guys, in this case university physics professors, who do know their maths and who are definitely not out to mislead us. So how did they get it so horribly wrong and what did they really want to say. Well, they started out with some genuinely deep an amazing connection between 1 + 2 + 3 etc and the number -1/12 but in the effort to explain this connection in really, really simple terms they just went overboard and ended up with an explanation that is not just really simple but also really wrong. Well 6 million views later and the comment sections of all maths YouTubers are being inundated by confused one plus two plus three comments that are a direct consequence of this video. For mathematical public relations it's a disaster. (Marty) It's THE disaster. (Mathologer) Yeah, it's THE disaster. And so I think it's a good idea to have another really close look at the Numberphile calculation step by step, state clearly what's wrong with it, how to fix it, and how to reconnect it to the genuine maths that the Numberphile professors had in mind originally. Lots of amazing maths look forward: non-standard summation methods for divergent series, the eta function, a very well-behaved sister of the Zeta function, the gist of analytic continuation in simple words, some more of Euler's mathemagical tricks etc. Now, I've tried to make this whole thing self-contained. So you don't have to have watched my very different other video on one plus two plus three from over a year ago or anything else to understand this one. Okay, let's get going. So that we are all on the same page, here real quick is the whole Numberphile calculation. They call the unknown value of the infinite series 1 + 2 + 3+... S. As stepping-stones for the calculation they first calculate the sums of these other two infinite series. So 1-1+1-1+... and 1-2+3-4+... Adding up the terms of the first series, we get the partial sums 1. Ok 1 minus 1 is 0, 1 minus 1 plus 1 is 1, 1 minus 1 plus 1 minus 1 is 0 and so on. These partial sums alternate between 0 and 1 and so the Numberphile guys declare that the sum of this infinite series is CLEARLY the average of 1 and 0 which is 1/2 (Marty) That's not all that clear to me. (Mathologer) Alright we'll get to that. They also mention that there are other ways to justify this. We'll also get to that. Ok, second sum. Here they start by considering what happens when you double this sum. So 2 times S2 is equal to the infinite series added to itself but now before adding the two infinite series they shift the bottom series one term to the right. Now 1 plus nothing is 1, minus 2 plus 1 is minus 1, 3 minus 2 is 1, minus 4 plus 3 is minus 1, etc. But that bottom series is the one we already looked at which, remember, is equal to 1/2 and so .... Second sum done, great. Now for last sum, that's the one we're really after. Here the Numberphile guys start by subtracting S2, the sum they just figured out, from S. Now 1 minus 1 is 0, 2 minus minus 2 is plus 4, 3 minus 3 is 0, 4 minus minus 4 is 8, etc. The zeros don't matter so let's get rid of them. Take out the common factor 4. Ah the yellow that's our 1+2+3+... sum S again. Now solve for S, and my usual magic here, and we get -1/12. And here the Numberphile guys take a bow. But, not so fast! !ll this is really nonsense the way it was presented. In particular these three identities are false. This means that if on any maths exam at any university on Earth you're asked to evaluate the sums of these infinite series and you give the Numberphile identities as your answer you will receive exactly 0 marks for your answers. It's critical to realise that in mathematics we have a precise definition which underpins the sum of an infinite series. Wherever you see infinite series this definition and only this definition applies unless there are some huge disclaimers to the contrary in flashing neon lights. Alright, now the Numberphile guys did not include any such disclaimers and so they too should get 0 marks for their effort. (Marty) Or maybe give them -1/12 marks. (Mathologer) Yeah I think I can agree with that. OK, what's this definition and what are the answers that will get you full marks on your maths exam. To evaluate the sum of an infinite series you calculate the sequence of partial sums just like the Numberphile guys did at the very beginning. Now if the sequence of partial sums levels off to a finite number, that is, if the sequence converges, or if it explodes to plus infinity, or if it explodes to minus infinity, then this limit is the sum of the infinite series. If no such limit exists, then the infinite series does not have a sum. That's it, that's the definition. So for the first Numberphile series the sequence of partial sums alternates between 0 and 1 and therefore does not have a limit. This means that this infinite series does not have a sum, neither 1/2 nor anything else. This is the correct answer for your maths exam. Alright, what about the other two infinite series? Hmm, well, in the case of 1 plus 2 plus 3 the partial sums explode to plus infinity and so the sum of the series is infinity. For the infinite series in the middle the partial sums explode in size, but neither just to infinity or just to minus infinity, and so this series also does not have a sum. So these are the answers that get you full marks. In many ways the most important infinite series are those with a finite sum which have not featured here yet. So, to give some perspective, here's a standard example, an infinite geometric series: 1/2+1/4+1/8 and so on. Now here the partial sums exhibit a really nice pattern and clearly they converge to 1. (Marty) Yeah, I think this one is clear. (Mathologer) This one is clear and so the sum of this infinite series is 1. Oh, before I forget, those finite sum series are usually called convergent series and all the other infinite series are called divergent series. Keeping this in mind let's have another look at the Numberphile calculation. Here's the whole thing at a glance. It's just a transcript of the writing on the brown paper in the Numberphile video. Again, as it was presented by Numberphile all this is nonsense and worth 0 marks. (Marty) Or less! (Mathologer) Or less :) THIS. IS. NOT. MATHEMATICS. Don't use it, otherwise you'll burn in mathematical hell. Having said that there should be some method to this madness, right? Those guys are smart! But if there is, then it's clear that the sums you see here cannot possibly represent the usual sums, as about six million people have been misled to believe by this video. Ok let's start by doing something that may also seem a little bit crazy. At first glance, just for fun, and in denial of reality, let's assume for a second that those three Numberphile series were actually convergent, that is, all had a finite sum. Then all, ALL highlighted arguments would be valid. This includes the termwise adding and subtracting of series that was performed here, ... and here, ... and even the shifting to the right before the addition that a lot of people view with suspicion. Why would all these operations be ok if we were dealing with convergent series? Because summation of convergent series is consistent with termwise addition and subtraction, and shifting. Let me explain that too. There are differences between finite and infinite sums. For example, infinite series sometimes don't have a sum whereas finite sums always exist, rearrangement of the terms can change the sum of convergent infinite series, etc. On the other hand, the sums of convergent infinite series do share a lot of the properties of finite sums and it's exactly these properties that make them so useful. Here the most important three such properties. Let's say you have two convergent infinite series, okay< with sums A and B. Then, by adding these two series termwise you get a new infinite series. And now it's quite easy to prove that this new infinite series is also convergent and that its sum is equal to A plus B, of course. And the same stays true if you replace all the pluses with minuses. So termwise addition and subtraction is consistent with summing convergent series. That's property one. Property two. Multiplying the terms of a convergent series with sum A by a number, say five, gives a new infinite series. Again it's really easy to see that the new series is convergent and that its sum is five times A. So termwise multiplication by numbers is also consistent with summing convergent infinite series. That's our second property. Finally, property three. Shifting the terms of a convergent series with sum A by one term is the same as adding a zero as the first term to our series, like that. Obviously, the new series is still convergent and its sum is the same as that of the original series. This also works the opposite way. Removing a zero term at the front does not change the sum. Okay, so that's property three. Now we can use these three properties to build valid arguments, very much like in the Numberphile video. Here's an example. This is the convergent series that I showed you earlier. Remember its sum is 1. Now let's say we did not know its sum or even whether this series is convergent or not. Then we could argue in a legit way like this: Ok, well we don't know whether it's convergent or not, but if it's convergent and it's sum is M (M for mystery number :) then because of the number multiplication property we get that half M is equal to 1/2 times 1/2 which equals to 1/4 plus 1/4 times 1/2 which is equal to 1/8 and so on. But because of the shift property 1/2 M is also equal to this guy here zero plus whatever. Now because of the addition and subtraction property we are justified to subtract as in the Numberphile video. So, on the left side we get M minus 1/2 M, that's 1/2 M, and on the right side we've got 1/2 minus zero that's 1/2, and then everything else kind of goes away. In total we get M is equal to 1. So our assumption that our mystery series is convergent with sum M lets us conclude in a valid way that the only value that M can possibly be is 1. Question: Does this prove that M is 1? No, and this is very important. Because this argument starts with an assumption, to be able to conclude that the sum of the series really is 1, you still have to somehow show that the series we started with is actually convergent. Hmm so this sort of argument gives you an idea of what to expect but it does not get you all the way. Anyway let's see what happens when we unleash this sort of reasoning on the first Numberphile series which we already know is NOT convergent. Well, so just for kicks, let's assume that the 1-1+1 series was actually convergent with the unknown sum S1. Then, because of the shift property, S1 would also be equal to 0 plus all the other junk. Again, because of the addition property, we can add on both sides S1 plus S1 that's 2 S1 and then we add termwise on the right to get, well, 1 there and everything else cancels out as you can see. And so we get this which is exactly what Numberphile said S1 should be. So the plot thickens here, right? In fact, in some ways this line of reasoning would have fitted in better with the rest of their argument than just plucking the number 1/2 out of thin air. So let's have another look at the Numberphile argument and let's fit in what we just did as the first step to justify why S1 should be 1/2. So here we go. This is the Numberphile argument. Let's round it off by inserting the argument for S1 from just a moment ago. Here we go. And so here is one way to rephrase the whole thing to make it into a valid argument. IF, and this is a big, huge monstrous IF, these three infinite series were convergent, THEN this whole argument would be valid and the sums of these three infinite series would be exactly the numbers given by Numberphile. Great, but of course we know that the assumption of this valid argument is false, that the three infinite series are not convergent. So, yes, this argument is valid in itself, but what good is it if the assumption is starts with is false? Well here's an idea. Since no divergent series has a finite value attached to it, let's dream big. What if it was possible to extend the notion of summing a convergent series, what if it was possible to define a super sum. This super sum should have three key properties: first of all, we don't want to lose anything so the super sum of a convergent series should be the same as its normal sum, right? Then all divergent infinite series should be super summable with finite values. And then the last thing we want is that super summing, just like normal summing is consistent with adding, shifting, etc. alright. Now such a super sum extension of the standard infinite sum would be as fantastic as the extension of the real numbers to the complex numbers, with all sorts of cool properties for a smaller world remaining true in the bigger world and at the same time all sorts of new magic appearing in the bigger world. In particular, because consistency made the argument over there valid for convergent series, we would expect the argument to still be valid if we replaced ordinary equality by equality with respect to super summing. Even better, since every infinite series would have a super sum, we could get rid of the if and so the whole Numberphile video could be saved by just saying that we are super summing instead of just boring old summing. What a lovely dream :) (Marty) It's time to wake up! (Mathologer) Yes, sadly. Well, anyway, those of you who watched my last video on this topic know that there is a super sum. However it only assigns a sum to some :) divergent series but not to all. In particular, it sums the first two guys over there. To show how super summing works let's apply it to our first annoying divergent series. Basically super summing builds upon normal summing and then averaging out any bouncing around of the partial sums. We start by calculating the sequence of partial sums. If this sequence converges, then our super sum equals our normal sum and no tricks are needed. However if the sequence of partial sums does not converge, as in the case of this infinite series, then we start with the trickery, building a second sequence out of the first. The Nth term of this new sequence is just the average of the first N terms of the first sequence. So, for our particular series for the first term of the new sequence there's nothing to average, well the average of 1 is just 1. Ok, then the average of 1 is 0 is 1/2, the average of 1 and 0 and 1 is 2/3, etc. Now, every second number here is 1/2 and the remaining numbers also converge to 1/2 and this means that overall the second series converges to 1/2. And that means that our infinite series super sums to 1/2, which is also the number that Numberphile gets. Now, for other infinite series even the second sequence may not converge in which case we generate a third sequence, again by averaging the second sequence. If that doesn't work, then we generate a fourth, then a fifth, etc. As long as any of those sequences converges, the infinite series under discussion has the corresponding limit as super sum. For example, in the case of the second Numberphile series, the first and second sequences diverge but the third sequence converges to 1/4 which then is our super sum. This is also what Numberphile gets and so everything is looking good. Until now, now we have to confront the sad truth that for most infinite series none of the associative averaging sequences of numbers converge and so these series don't have a super sum. In particular, for 1 plus 2 plus 3 and so on all the partial sums are positive and obviously averaging over positive numbers will always only result in positive numbers. In fact, all the associated sequences of numbers will explore to positive infinity and so 1 + 2 + 3 etc definitely does not super sum to anything finite, let alone anything finite and negative like -1/12. Returning to the Numberphile calculation, here's the part that can be totally justified using super sums instead of regular sums. Because not every divergent series has a super sum we still need the big IF to make this part of the argument valid. In itself not too bad, though, since the assumption is actually true, right? And since super summing is really the most natural extension of normal summing, 1/2 and 1/4 are the only reasonable numbers to associate with the first two series. Really nice stuff I think. To recap, we now know finite sums, convergen infinite sums and our new super sums. Oh, by the way, I should mention that in the literature our super sum would be called generalized Cesaro summation or generalized Hölder summation. Anyway, these summation methods are proper extensions of each other and are therefore able to assign meaningful values to larger and larger classes of series. However, being able to do more also comes at a price. The more powerful a summation method, the less well behaved it is. What works for finite sums cannot necessarily be taken for granted for infinite sums. I already mentioned problems with rearranging convergent series, for example. Of course, super sums also lack all the nice summy properties that normal infinite sums lack but they are even summy things that still work for infinite sums that no longer work for our super sums. Yes, yes the three basic properties I've been hammering are fine but to assume that any familiar summy property will also work for super sums in general is risking zero marks (Marty): Or less! (Mathologer) Or less :) For example, inserting or deleting infinitely zeros has no effect on convergent series. However, doing the same to super sums can change things dramatically. For example, if we insert infinitely many zeros into our first series, like this, the super sum of the new series will be different from 1. Little puzzle for you: What's the new super sum? As usual, give your answers in the comments. And this zero problem is important. I glossed over this because it won't have any bearing on our discussion, but at some point in the Numberphile calculation they simply zap infinitely many zeros and this cannot be justified with our three properties. Bad... The effect of losing more and more properties as you go more and more general is actually something that you've all encountered before when you got introduced to larger and larger number systems: fFrst the positive numbers, then to the integers, the rational numbers, the reals and to the complex numbers and even beyond to the quarternions and the octonions. Each time you broaden your world, you lose some nice properties. Second puzzle for you: Can you think of some properties that get lost along the way as we build larger and larger number systems? And another puzzle: Suppose we assume that the 1 + 2 + 3 etc series actually super sums to a finite number, can you manipulate this identity into a couple of contradictory statements using our favourite three properties? What can you then conclude from the fact that you can arrive at contradictory statements? It will be interesting to see what you can come up with in this respect? Anyway, it's time! (drum roll) We have to get serious about the connection between our 1 + 2 + 3 series and - 1/12. So press the pause button, go get your popcorn and your hot chocolate and let me know when you're back :) (jeopardy music) Ready? Here we go. Even at the level of super sums we are pretty far removed from what most people think of as a sum. After all for divergent series, given all the averaging that is going on, the super summing is really more like finding a super average than a real summy sum, don't you think? Well it will get more extreme not only in this respect but also in terms of the maths that is required to understand what is going on with the -1/12 connection. I'm sure that a lot of you will already have heard of this connection, so let me just state it first and then really explain it using the Numberphile calculation as a template. This is the mega famous Riemann zeta function. It is a function of the complex variable z. Written as the infinite sum there it makes sense if the real part of z is greater than 1. However, there is one unique way to extend the Riemann zeta function to an analytic function for all complex numbers z excepting 1. Formally, if you substitute z = -1 you get ... Well, of course, minus 1 is not greater than 1, and so we really don't have equality here, and so let's quickly get rid of their equal sign. Ok, at the same time the right-hand side is our master villain 1 + 2 + 3 etc and the value of zeta at minus 1 is equal to you guessed it - 1/12. And this, in a nutshell, is the genuine, real, actual connection between 1 + 2 + 3 + and - 1/12. But why would anybody describe this connection as a sum, and what has all this to do wit the last part of the Numberphile calculation. Well there's more to explain. First, here's a mini introduction to analytic functions and analytic continuation. This will be a rough and ready intro which is all we need. You all know what a polynomial is, right? One of these guys: a constant function or a linear function or a quadratic function or a cubic, etc. Now let's play a game. Here is a chunk of a mystery continuous function that is defined for all real numbers. So I'm just not showing you the part to the left of the y-axis. Here's the question. Just by looking at this chunk, can you continue the graph and tell me what my function is? Now you might be tempted to say 'Yes' but the answer is 'No'. There are infinitely many ways we could continue to the left and here are a couple of examples. Here's one and there's another one, there's a third one, there's infinitely many different ones. Were you tricked? (Marty) No. (Mathologer) Sure you were not , but you know the game, right? Of course our initial chunk is part of a line and it's natural to think of continuing the function as this same line. But we don't have to. However, if I tell you the mystery function is linear, then your initial chunk tells you exactly how to continue the function, there's only one way to continue so that the whole function is linear. In fact, the same is true if I only told you the function was a polynomial. Just by looking at the chunk you could be absolutely sure that my polynomial is linear and exactly which linear function, right? We can now generalize this simple observation a couple of steps, in a pretty dramatic way actually. Here we go. First, suppose that our initial chunk is part of a parabola, or if you like a cubic, or any polynomial. If I then tell you that my mystery function is a polynomial there's always only going to be exactly one polynomial that continues our beginning chunk. In other words, a polynomial is completely determined by any part of it. Second, all we've said stays true if we think of polynomials as functions of a complex variable and if you begin with a chunk of the polynomial corresponding to a region in the complex plane. So on the left, you see the complex number plane where each point stands for a complex number and I've also colored a small region in the plane. And so in terms of this picture a polynomial is completely pinned down by the values it takes on over a region like this. No other polynomial will take on all the same values there. Again, just relax if all this seems a little bit too much. Now, the polynomials are the simplest and most nicely behaved functions but there is a much larger world of functions that shares a lot of the nicest properties with polynomials. Those are the so-called analytic functions. These are the complex functions that can be expressed locally as either regular finite polynomials or as infinite polynomials, so called power series. For example, the exponential function is an analytic function because it can be written as an infinite polynomial like this. In fact, pretty much all our favorite functions such as the trig functions, rational functions, etc. are analytic. Important for us is that just like a polynomial, an analytic function is completely pinned down by any initial chunk. So if I give you a beginning chunk of an analytic function like the exponential function, then no other analytic function can continue this chunk. This is usually expressed by saying that analytic continuation of an analytic function is uniquely determined. In summary, though there may be many nice ad hoc ways to continue an analytic function there's just one distinguished, most reasonable, absolutely fantastic never to be improved way to do this, leading to a larger analytic function. Of course, as I said, this is all very sketchy and you guys in the know will probably nitpick me to death in the comments. (Marty) Looking forward to it. Ok, in particular I didn't tell you why we need to drag complex functions into the discussion, but please just run with it for today and I promise I'll fill in the details soon. In the meantime you can also read up on things by following the links in the description. All you really have to remember is this: an initial chunk of an analytic function nails down the whole analytic function. Now we can join the dots. We have two completely different notions of best extension. First, for extending sums to super sums of divergent series and second for extending a chunk of an analytic function to the whole analytic function. Combining these two extension ideas, we can finally explain what's going on with 1 + 2 + 3 + and -1/12. Okay, have a look at this infinite series. Notice that it's the same as the zeta function except that it includes minuses. It's also obviously different from the Numberphile series in that it includes a variable z. So it is actually an infinite family of infinite series, one for each complex number z. Let's just make a little list of such series corresponding to a few prominent integer values. For z = 0 we our 1 minus 1 plus 1 series. Ok for x = - 1 we get 1 minus 2 plus 3 and Mathologer regulars know that for z = 1 and 2 the series are convergent. Now, in general, these series are convergent for all complex numbers z in the positive brownish half plane. The infinite series are divergent for all other z including 1 minus 1 plus 1 etc at 0 and the other guy. However, just like the two Numberphile series can be super summed, the same is possible for every z, for all the divergent series in this family. And this allows us to define a close relative of the zeta function, the so-called Dirichlet eta function. And this function turns out to be an analytic function. So to start with, standard summation only gives us part of this analytic function for which two infinite series converge, this part here. Now, most mathematicians will simply discount a divergent series that pop up here as useless artifacts. Instead they will construct the analytic continuation of the eta function by completely different methods. These methods are very slick and ingenious. However, they provide very little intuition and insight into what's really going on here. On the other hand, seeing that the most reasonable extension of an analytic function that is defined on part of the complex plane is actually given by the most reasonable way to assign generalised sums to these supposedly useless divergent series just feels right to me and leads the way to a more intuitive understanding of analytic continuation, at least in this case. But now here's a great thought: we just used a generalised sum to construct an analytic continuation, right? Let's turn the idea around, let's use analytic continuation to identify candidates for a generalised summation method. And this is exactly what happens in the case of 1 + 2 + 3 etc and - 1/12 and the zeta function. You get the zeta function when you replace all the minuses in the eta function by pluses. Well you get part of it, the brownish part. which is the part of the complex plane for which the infinite series on the right converge. For all other z the resulting infinite series are divergent and even super summing doesn't help for the white part on the left. So the super summing trick for eta just doesn't work for zeta. The trick to use is encoded in the finale of the Numberphile pseudo proof. That's the brownish bit down there. Remember this part of the argument takes the sum S2 of the 1 - 2 + 3 series and spits out a sum for the 1 + 2 + 3 series formally these two sums are just what you get when you let z equal to -1 in the infinite series of the eta and zeta functions and actually the Numberphile calculation is just a special instance of a calculation involving eta and zeta. Let's be brave. Ignoring questions of legality, let's unleash exactly the same calculation on eta and zeta. So instead of subtracting S2 from S, let's subtract eta from zeta. Ready? ... Right, take out the common factor down there. That's zeta again in the brackets. Now let's solve for zeta. There my magic again okay that's really exactly the same as the last part of the Numberphile calculation using zetas and eta instead of the Numberphile series. Just as a check after substituting z = -1 this identity turns into this, and with S2 being 1/4 we get this. Okay more magic and we're back to -1/12. But didn't we say that the Numberphile computation was nonsense? (Marty) Yes, we did. (Mathologer) We definitely did. And it is, but some magic happens with zetas and eta which saves our zetas eta calculation from being nonsense and that is the magic of analytic continuation. Both the series for zetas and eta are convergent for every z in this brownish region. This means that for these values of z our calculation and the resulting equation above are pure, correct, 100% approved bona fide mathematics. But, as well, eta is defined and analytic for all z and the same is true for the denominator. But then the right side, as a quotient of two functions that are analytic everywhere, is itself defined and analytic everywhere except possibly at the zeroes of the denominator. In fact, a closer look reveals that the whole right-hand side is analytic everywhere except at z=1. Here comes the punchline and this punchline hinges on the chunks-pin-down-analytic-functions business that I've been going on about. You should really try to understand this. Okay, so both the left side and the right side are analytic in this part of the complex plane here. But since the right side is analytic everywhere, because of our chunks-pin- down-analytic-functions property, the right side has to be equal to the elusive analytic continuation of the zeta function on the left that everybody is really interested, the analytic continuation of the zetas function. So this identity is a real jewel as it gives an explicit way to calculate any value of the Riemann zeta function, analytic continuation and all via the eta function which, remember, we defined via super summing. In other words, it actually makes sense to use this identity as a definition for the zeta function which works for all z. For example, setting z=0 we get this one here and eta of zero was just the super sum of 1 minus 1 plus 1 etc which is equal to 1/2 and so zeta of 0 is equal to -1/2. Here just a couple more interesting values for zeta. Nice. The zeroes are particularly interesting. In fact, it turns out that zetas has zeros at all negative even integers, -2, -4, -6, and so on. These are the so-called trivial zeroes of the zeta function. I'm sure if you made it this far you've also heard of the Riemann hypothesis which is all about other zeros of the zeta function. It says that all other zeros are situated on this blue line here and what's also really interesting is that on the right hand side we actually don't have to super sum to calculate the values of zetas on the blue line because just with ordinary summing eta can be evaluated everywhere in this part of the complex plane, this part here which includes this critical line. Lots and lots of other interesting things one could say here but, well, we are here to wrap up this whole 1 plus 2 plus 3 is equal to -1/12 business. Alright, ok, let's do it. In the first instance it is really our identity up there that the Numberphile video is attempting to capture and it's definitely tempting to express his identity as a new generalized sum maybe like this. I've decorated the equal sign with an R in honour of Ramanujan who seems to have been the first to think this way (NOT Euler as many people think). In fact, in Ramanujan's notebook we can find a calculation very similar to the one that's in the Numberphile video. Of course, there's a huge difference between a monumental genius quickly abbreviating all his complicated stuff in a personal note and a YouTube video addressed to a general audience, right? What you see up there is part of a general method called Ramanujan summation that assigns values to all sorts of divergent series including the three Numberphile series. An important aspect of infinite series which is often overlooked is the order in which the terms are summed. We are not just adding an infinite set we are doing so in a certain order and this has all sorts of important implications. So, for example, no matter how we arrange the natural numbers into an infinite series this infinite series will always diverge to plus infinity using standard summation. However, how exactly, how fast, slow, regular, erratic it diverges to infinity depends very much on the order of the terms in the series. The Ramanujan sum lacks pretty much all the nice summy properties that we encountered today. However, it manages to capture aspects of naturally ordered series and it pops up in many other branches of the theory of divergent series in addition to the one we talked about today. Check out some of the links in the description, especially if you know a lot of maths the article by Terry Tao. Also I'm planning another video on the so called Euler-Maclaurin summation formula which establishes a powerful connection between sums and integrals and which is the starting point for Ramanujan's sum. Just to whet your appetite here's one closely related -1/12 fact. The nth partial sum of our infinite series is N(N+1)/2 So, if we plug in 1, 2, 3, 4 etc. for N the formula will spit out those partial sums 1, 3, 6, 10. Now let's replace N by x and graph the resulting function. That's a quadratic with zeros at 0 and - 1. Now the remarkable thing about this graph is that the signed area here is equal to -1/12. And this is definitely no coincidence. Really amazing stuff, don't you agree? And that's it, finally, for today (and now I go and kill myself :) and I promise the next video will be a LOT shorter.

Contents

Partial sums

The first six triangular numbers
The first six triangular numbers

The partial sums of the series 1 + 2 + 3 + 4 + 5 + 6 + ⋯ are 1, 3, 6, 10, 15, etc. The nth partial sum is given by a simple formula:

This equation was known to the Pythagoreans as early as the sixth century BCE.[5] Numbers of this form are called triangular numbers, because they can be arranged as an equilateral triangle.

The infinite sequence of triangular numbers diverges to +∞, so by definition, the infinite series 1 + 2 + 3 + 4 + ⋯ also diverges to +∞. The divergence is a simple consequence of the form of the series: the terms do not approach zero, so the series diverges by the term test.

Summability

Among the classical divergent series, 1 + 2 + 3 + 4 + ⋯ is relatively difficult to manipulate into a finite value. Many summation methods are used to assign numerical values to divergent series, some more powerful than others. For example, Cesàro summation is a well-known method that sums Grandi's series, the mildly divergent series 1 − 1 + 1 − 1 + ⋯, to 1/2. Abel summation is a more powerful method that not only sums Grandi's series to 1/2, but also sums the trickier series 1 − 2 + 3 − 4 + ⋯ to 1/4.

Unlike the above series, 1 + 2 + 3 + 4 + ⋯ is not Cesàro summable nor Abel summable. Those methods work on oscillating divergent series, but they cannot produce a finite answer for a series that diverges to +∞.[6] Most of the more elementary definitions of the sum of a divergent series are stable and linear, and any method that is both stable and linear cannot sum 1 + 2 + 3 + ⋯ to a finite value; see below. More advanced methods are required, such as zeta function regularization or Ramanujan summation. It is also possible to argue for the value of  1/12 using some rough heuristics related to these methods.

Heuristics

Passage from Ramanujan's first notebook describing the "constant" of the series
Passage from Ramanujan's first notebook describing the "constant" of the series

Srinivasa Ramanujan presented two derivations of "1 + 2 + 3 + 4 + ⋯ =  1/12" in chapter 8 of his first notebook.[7][8][9] The simpler, less rigorous derivation proceeds in two steps, as follows.

The first key insight is that the series of positive numbers 1 + 2 + 3 + 4 + ⋯ closely resembles the alternating series 1 − 2 + 3 − 4 + ⋯. The latter series is also divergent, but it is much easier to work with; there are several classical methods that assign it a value, which have been explored since the 18th century.[10]

In order to transform the series 1 + 2 + 3 + 4 + ⋯ into 1 − 2 + 3 − 4 + ⋯, one can subtract 4 from the second term, 8 from the fourth term, 12 from the sixth term, and so on. The total amount to be subtracted is 4 + 8 + 12 + 16 + ⋯, which is 4 times the original series. These relationships can be expressed using algebra. Whatever the "sum" of the series might be, call it c = 1 + 2 + 3 + 4 + ⋯. Then multiply this equation by 4 and subtract the second equation from the first:

The second key insight is that the alternating series 1 − 2 + 3 − 4 + ⋯ is the formal power series expansion of the function 1/(1 + x)2 but with x defined as 1. Accordingly, Ramanujan writes:

Dividing both sides by −3, one gets c =  1/12.

Generally speaking, it is incorrect to manipulate infinite series as if they were finite sums. For example, if zeroes are inserted into arbitrary positions of a divergent series, it is possible to arrive at results that are not self-consistent, let alone consistent with other methods. In particular, the step 4c = 0 + 4 + 0 + 8 + ⋯ is not justified by the additive identity law alone. For an extreme example, appending a single zero to the front of the series can lead to inconsistent results.[1]

One way to remedy this situation, and to constrain the places where zeroes may be inserted, is to keep track of each term in the series by attaching a dependence on some function.[11] In the series 1 + 2 + 3 + 4 + ⋯, each term n is just a number. If the term n is promoted to a function n−s, where s is a complex variable, then one can ensure that only like terms are added. The resulting series may be manipulated in a more rigorous fashion, and the variable s can be set to −1 later. The implementation of this strategy is called zeta function regularization.

Zeta function regularization

Plot of ζ(s). For s > 1, the series converges and ζ(s) > 1. Analytic continuation around the pole at s = 1 leads to a region of negative values, including ζ(−1) = − 1/12
Plot of ζ(s). For s > 1, the series converges and ζ(s) > 1. Analytic continuation around the pole at s = 1 leads to a region of negative values, including ζ(−1) =  1/12

In zeta function regularization, the series is replaced by the series . The latter series is an example of a Dirichlet series. When the real part of s is greater than 1, the Dirichlet series converges, and its sum is the Riemann zeta function ζ(s). On the other hand, the Dirichlet series diverges when the real part of s is less than or equal to 1, so, in particular, the series 1 + 2 + 3 + 4 + ⋯ that results from setting s = –1 does not converge. The benefit of introducing the Riemann zeta function is that it can be defined for other values of s by analytic continuation. One can then define the zeta-regularized sum of 1 + 2 + 3 + 4 + ⋯ to be ζ(−1).

From this point, there are a few ways to prove that ζ(−1) =  1/12. One method, along the lines of Euler's reasoning,[12] uses the relationship between the Riemann zeta function and the Dirichlet eta function η(s). The eta function is defined by an alternating Dirichlet series, so this method parallels the earlier heuristics. Where both Dirichlet series converge, one has the identities:

The identity continues to hold when both functions are extended by analytic continuation to include values of s for which the above series diverge. Substituting s = −1, one gets −3ζ(−1) = η(−1). Now, computing η(−1) is an easier task, as the eta function is equal to the Abel sum of its defining series,[13] which is a one-sided limit:

Dividing both sides by −3, one gets ζ(−1) =  1/12.

Cutoff regularization

The series 1 + 2 + 3 + 4 + ⋯
After smoothing
A graph showing a parabola that dips just below the y-axis
Asymptotic behavior of the smoothing. The y-intercept of the parabola is  1/12.[1]

The method of regularization using a cutoff function can "smooth" the series to arrive at  1/12. Smoothing is a conceptual bridge between zeta function regularization, with its reliance on complex analysis, and Ramanujan summation, with its shortcut to the Euler–Maclaurin formula. Instead, the method operates directly on conservative transformations of the series, using methods from real analysis.

The idea is to replace the ill-behaved discrete series with a smoothed version

,

where f is a cutoff function with appropriate properties. The cutoff function must be normalized to f(0) = 1; this is a different normalization from the one used in differential equations. The cutoff function should have enough bounded derivatives to smooth out the wrinkles in the series, and it should decay to 0 faster than the series grows. For convenience, one may require that f is smooth, bounded, and compactly supported. One can then prove that this smoothed sum is asymptotic to  1/12 + CN2, where C is a constant that depends on f. The constant term of the asymptotic expansion does not depend on f: it is necessarily the same value given by analytic continuation,  1/12.[1]

Ramanujan summation

The Ramanujan sum of 1 + 2 + 3 + 4 + ⋯ is also  1/12. Ramanujan wrote in his second letter to G. H. Hardy, dated 27 February 1913:

"Dear Sir, I am very much gratified on perusing your letter of the 8th February 1913. I was expecting a reply from you similar to the one which a Mathematics Professor at London wrote asking me to study carefully Bromwich's Infinite Series and not fall into the pitfalls of divergent series. … I told him that the sum of an infinite number of terms of the series: 1 + 2 + 3 + 4 + ⋯ =  1/12 under my theory. If I tell you this you will at once point out to me the lunatic asylum as my goal. I dilate on this simply to convince you that you will not be able to follow my methods of proof if I indicate the lines on which I proceed in a single letter. …"[14]

Ramanujan summation is a method to isolate the constant term in the Euler–Maclaurin formula for the partial sums of a series. For a function f, the classical Ramanujan sum of the series is defined as

where f(2k−1) is the (2k − 1)-th derivative of f and B2k is the 2kth Bernoulli number: B2 = 1/6, B4 =  1/30, and so on. Setting f(x) = x, the first derivative of f is 1, and every other term vanishes, so:[15]

To avoid inconsistencies, the modern theory of Ramanujan summation requires that f is "regular" in the sense that the higher-order derivatives of f decay quickly enough for the remainder terms in the Euler–Maclaurin formula to tend to 0. Ramanujan tacitly assumed this property.[15] The regularity requirement prevents the use of Ramanujan summation upon spaced-out series like 0 + 2 + 0 + 4 + ⋯, because no regular function takes those values. Instead, such a series must be interpreted by zeta function regularization. For this reason, Hardy recommends "great caution" when applying the Ramanujan sums of known series to find the sums of related series.[16]

Failure of stable linear summation methods

A summation method that is linear and stable cannot sum the series 1 + 2 + 3 + ⋯ to any finite value. (Stable means that adding a term to the beginning of the series increases the sum by the same amount.) This can be seen as follows. If

1 + 2 + 3 + ⋯ = x

then adding 0 to both sides gives

0 + 1 + 2 + ⋯ = 0 + x = x by stability.

By linearity, one may subtract the second equation from the first (subtracting each component of the second line from the first line in columns) to give

1 + 1 + 1 + ⋯ = xx = 0.

Adding 0 to both sides again gives

0 + 1 + 1 + 1 + ⋯ = 0,

and subtracting the last two series gives

1 + 0 + 0 + ⋯ = 0

contradicting stability.

Therefore, every method that gives a finite value to the sum 1 + 2 + 3 + ⋯ is not stable or not linear.[17]

Physics

In bosonic string theory, the attempt is to compute the possible energy levels of a string, in particular the lowest energy level. Speaking informally, each harmonic of the string can be viewed as a collection of D − 2 independent quantum harmonic oscillators, one for each transverse direction, where D is the dimension of spacetime. If the fundamental oscillation frequency is ω then the energy in an oscillator contributing to the nth harmonic is nħω/2. So using the divergent series, the sum over all harmonics is  ħω(D − 2)/24. Ultimately it is this fact, combined with the Goddard–Thorn theorem, which leads to bosonic string theory failing to be consistent in dimensions other than 26.[18]

The regularization of 1 + 2 + 3 + 4 + ⋯ is also involved in computing the Casimir force for a scalar field in one dimension.[19] An exponential cutoff function suffices to smooth the series, representing the fact that arbitrarily high-energy modes are not blocked by the conducting plates. The spatial symmetry of the problem is responsible for canceling the quadratic term of the expansion. All that is left is the constant term  1/12, and the negative sign of this result reflects the fact that the Casimir force is attractive.[20]

A similar calculation is involved in three dimensions, using the Epstein zeta-function in place of the Riemann zeta function.[21]

History

It is unclear whether Leonhard Euler summed the series to  1/12. According to Morris Kline, Euler's early work on divergent series relied on function expansions, from which he concluded 1 + 2 + 3 + 4 + ⋯ = ∞.[22] According to Raymond Ayoub, the fact that the divergent zeta series is not Abel summable prevented Euler from using the zeta function as freely as the eta function, and he "could not have attached a meaning" to the series.[23] Other authors have credited Euler with the sum, suggesting that Euler would have extended the relationship between the zeta and eta functions to negative integers.[24][25][26] In the primary literature, the series 1 + 2 + 3 + 4 + ⋯ is mentioned in Euler's 1760 publication De seriebus divergentibus alongside the divergent geometric series 1 + 2 + 4 + 8 + ⋯. Euler hints that series of this type have finite, negative sums, and he explains what this means for geometric series, but he does not return to discuss 1 + 2 + 3 + 4 + ⋯. In the same publication, Euler writes that the sum of 1 + 1 + 1 + 1 + ⋯ is infinite.[27]

In popular media

David Leavitt's 2007 novel The Indian Clerk includes a scene where Hardy and Littlewood discuss the meaning of this series. They conclude that Ramanujan has rediscovered ζ(−1), and they take the "lunatic asylum" line in his second letter as a sign that Ramanujan is toying with them.[28]

Simon McBurney's 2007 play A Disappearing Number focuses on the series in the opening scene. The main character, Ruth, walks into a lecture hall and introduces the idea of a divergent series before proclaiming, "I'm going to show you something really thrilling," namely 1 + 2 + 3 + 4 + ⋯ =  1/12. As Ruth launches into a derivation of the functional equation of the zeta function, another actor addresses the audience, admitting that they are actors: "But the mathematics is real. It's terrifying, but it's real."[29][30]

In January 2014, Numberphile produced a YouTube video on the series, which gathered over 1.5 million views in its first month.[31] The 8-minute video is narrated by Tony Padilla, a physicist at the University of Nottingham. Padilla begins with 1 − 1 + 1 − 1 + ⋯ and 1 − 2 + 3 − 4 + ⋯ and relates the latter to 1 + 2 + 3 + 4 + ⋯ using a term-by-term subtraction similar to Ramanujan's argument.[32] Numberphile also released a 21-minute version of the video featuring Nottingham physicist Ed Copeland, who describes in more detail how 1 − 2 + 3 − 4 + ⋯ = 1/4 as an Abel sum and 1 + 2 + 3 + 4 + ⋯ =  1/12 as ζ(−1).[33] After receiving complaints about the lack of rigour in the first video, Padilla also wrote an explanation on his webpage relating the manipulations in the video to identities between the analytic continuations of the relevant Dirichlet series.[34]

In The New York Times coverage of the Numberphile video, mathematician Edward Frenkel commented, "This calculation is one of the best-kept secrets in math. No one on the outside knows about it."[31]

Coverage of this topic in Smithsonian magazine describes the Numberphile video as misleading, and notes that the interpretation of the sum as  1/12 relies on a specialized meaning for the equals sign, from the techniques of analytic continuation, in which equals means is associated with.[35]

References

  1. ^ a b c d Tao, Terence (April 10, 2010), The Euler–Maclaurin formula, Bernoulli numbers, the zeta function, and real-variable analytic continuation, retrieved January 30, 2014
  2. ^ Lepowsky, J. (1999), Naihuan Jing and Kailash C. Misra (ed.), Vertex operator algebras and the zeta function, Contemporary Mathematics, 248, pp. 327–340, arXiv:math/9909178, Bibcode:1999math......9178L
  3. ^ Tong, David (February 23, 2012). "String Theory". p. 28-48. arXiv:0908.0333.
  4. ^ Gannon, Terry (April 2010), Moonshine Beyond the Monster: The Bridge Connecting Algebra, Modular Forms and Physics, Cambridge University Press, p. 140, ISBN 978-0521141888
  5. ^ Pengelley, David J. (2002), Otto Bekken; et al. (eds.), The bridge between the continuous and the discrete via original sources, National Center for Mathematics Education, University of Gothenburg, Sweden, p. 3, ISBN 978-9185143009
  6. ^ Hardy p.10 (More detail on the source needed)
  7. ^ Ramanujan's Notebooks, retrieved January 26, 2014
  8. ^ Abdi, Wazir Hasan (1992), Toils and triumphs of Srinivasa Ramanujan, the man and the mathematician, National, p. 41
  9. ^ Berndt, Bruce C. (1985), Ramanujan's Notebooks: Part 1, Springer-Verlag, pp. 135–136
  10. ^ Euler, Leonhard (2006). "Translation with notes of Euler's paper: Remarks on a beautiful relation between direct as well as reciprocal power series". Translated by Willis, Lucas; Osler, Thomas J. The Euler Archive. Retrieved 2007-03-22. Originally published as Euler, Leonhard (1768). "Remarques sur un beau rapport entre les séries des puissances tant directes que réciproques". Mémoires de l'Académie des Sciences de Berlin. 17: 83–106.
  11. ^ Promoting numbers to functions is identified as one of two broad classes of summation methods, including Abel and Borel summation, by Knopp, Konrad (1990) [1922]. Theory and Application of Infinite Series. Dover. pp. 475–476. ISBN 0-486-66165-2.
  12. ^ Stopple, Jeffrey (2003), A Primer of Analytic Number Theory: From Pythagoras to Riemann, p. 202, ISBN 0-521-81309-3
  13. ^ Knopp, Konrad (1990) [1922]. Theory and Application of Infinite Series. Dover. pp. 490–492. ISBN 0-486-66165-2.
  14. ^ Berndt et al. p.53.
  15. ^ a b Berndt, Bruce C. (1985), Ramanujan's Notebooks: Part 1, Springer-Verlag, pp. 13, 134
  16. ^ Hardy p.346
  17. ^ Natiello, Mario A.; Solari, Hernan Gustavo (July 2015), "On the removal of infinities from divergent series", Philosophy of Mathematics Education Journal, 29: 1–11
  18. ^ Barbiellini, Bernardo (1987), "The Casimir effect in conformal field theories", Physics Letters B, 190 (1–2): 137–139, Bibcode:1987PhLB..190..137B, doi:10.1016/0370-2693(87)90854-9
  19. ^ See v:Quantum mechanics/Casimir effect in one dimension
  20. ^ Zee pp.65–67
  21. ^ Zeidler, Eberhard (2007), "Quantum Field Theory I: Basics in Mathematics and Physics: A Bridge between Mathematicians and Physicists", Quantum Field Theory I: Basics in Mathematics and Physics. A bridge between mathematicians and physicists, Springer: 305–306, Bibcode:2006qftb.book.....Z, ISBN 9783540347644
  22. ^ Kline, Morris (November 1983), "Euler and Infinite Series", Mathematics Magazine, 56 (5): 307–314, doi:10.2307/2690371, JSTOR 2690371
  23. ^ Ayoub, Raymond (December 1974), "Euler and the Zeta Function" (PDF), The American Mathematical Monthly, 81 (10): 1067–1086, doi:10.2307/2319041, JSTOR 2319041, retrieved February 14, 2014
  24. ^ Lefort, Jean, "Les séries divergentes chez Euler" (PDF), L'Ouvert, IREM de Strasbourg (31): 15–25, archived from the original (PDF) on February 22, 2014, retrieved February 14, 2014
  25. ^ Kaneko, Masanobu; Kurokawa, Nobushige; Wakayama, Masato (2003), "A variation of Euler's approach to values of the Riemann zeta function" (PDF), Kyushu Journal of Mathematics, 57 (1): 175–192, arXiv:math/0206171, doi:10.2206/kyushujm.57.175, archived from the original (PDF) on 2014-02-02, retrieved January 31, 2014
  26. ^ Sondow, Jonathan (February 1994), "Analytic continuation of Riemann's zeta function and values at negative integers via Euler's transformation of series", Proceedings of the American Mathematical Society, 120 (4): 421–424, doi:10.1090/S0002-9939-1994-1172954-7, retrieved February 14, 2014
  27. ^ Barbeau, E.J.; Leah, P.J. (May 1976), "Euler's 1760 paper on divergent series", Historia Mathematica, 3 (2): 141–160, doi:10.1016/0315-0860(76)90030-6
  28. ^ Leavitt, David (2007), The Indian Clerk, Bloomsbury, pp. 61–62
  29. ^ Complicite (April 2012), A Disappearing Number, Oberon, ISBN 9781849432993
  30. ^ Thomas, Rachel (December 1, 2008), "A disappearing number", Plus, retrieved February 5, 2014
  31. ^ a b Overbye, Dennis (February 3, 2014), "In the End, It All Adds Up to –1/12", The New York Times, retrieved February 3, 2014
  32. ^ ASTOUNDING: 1 + 2 + 3 + 4 + 5 + ... = –1/12 on YouTube
  33. ^ Sum of Natural Numbers (second proof and extra footage) on YouTube
  34. ^ Padilla, Tony, What do we get if we sum all the natural numbers?, retrieved February 3, 2014
  35. ^ Schultz, Colin (2014-01-31). "The Great Debate Over Whether 1+2+3+4..+ ∞ = −1/12". Smithsonian. Retrieved 2016-05-16.

Bibliography

Further reading

  • Zwiebach, Barton (2004). A First Course in String Theory. Cambridge UP. ISBN 0-521-83143-1. See p. 293.
  • Elizalde, Emilio (2004). "Cosmology: Techniques and Applications". Proceedings of the II International Conference on Fundamental Interactions. arXiv:gr-qc/0409076. Bibcode:2004gr.qc.....9076E.
  • Watson, G. N. (April 1929), "Theorems stated by Ramanujan (VIII): Theorems on Divergent Series", Journal of the London Mathematical Society, 1, 4 (2): 82–86, doi:10.1112/jlms/s1-4.14.82

External links

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