Divergent geometric series

Transcription

We have three different geometric series here. And what I want you to think about is, which of these converge and which of these diverge? Or another way of thinking about it, it converges if the sum comes up to a finite value. It diverges if the sum goes to infinity, or even possibly negative infinity. So pause the video and think about that. Well, the general way to think about it-- and we've seen it in a previous video-- is that you're going to have a converging geometric series. Infinite geometric series is if the absolute value of your common ratio is greater than zero and less than one. So we just have to think about, what is the absolute value of the common ratios over here? And it's not as obvious, because they didn't just write one term or one number to an exponent. They wrote several numbers to an exponents. So we'll have to do a little bit of algebraic manipulation. So 5 n to the negative 1, that's the same thing. Let me do that over here. 5 n to the negative 1 is the same thing as 5 to the n over 5. Or you could write it as 5 to the n times 5 to the negative 1, which is the same thing as 5 to the n over 5. And so 5 to the n over 5, 5 to the n over 5 times 9 10 to the n. 9 over 10 to the n. This is the same thing. This is equal to 1 over 5 times 5 times 9 over 10 to the nth power. And now the common ratio becomes a little bit more clear. This right over here would be our-- well, it won't be our first term anymore, because we're starting at n equals 2. So we won't view it that way. But if we look at our common ratio, this right over here is going to be our common ratio. And this is absolute value greater than or less than 1. Well, 5 times 9 is 45, divided by 10. That's going to give you 4 and 1/2. So this is greater than 1. The absolute value is greater than 1. So this one is going to diverge. This sum is going to go to infinity. Now let's think about this one over here. I encourage you to pause it if that one helped you a little bit. Let's try to rewrite this in a way that the common ratio is a little bit more obvious. So 3/2 to the n times 1 over 9 to the n plus 2. Well, we could think about this as 1-- let me write it this way. 3/2 to the n times 1 over 9 to the n plus 2 is the same thing as-- let me write it this way-- 9 to the n times 9 squared. And if we write it that way, then this is going to be the same thing as-- let's write it this way-- 1 over 9 squared, which is 81, times 3/2 to the n times 1 over 9 to the n is the same thing as 1 over 9 to the n. Notice, 1 over 9 to the n, if I raise 1 to the n power, it's not going to change the value of that 1. And so that's going to be the same thing as 1/81 times-- let's see. 3 divided by 9, we're going to get 1 over 6 to the n. And so here it's a little bit clearer that our common ratio is 1/6. Its absolute value is clearly less than 1. So this is going to converge. This is actually going to give you a finite value for the sum. Finally, let's go on to this one right over here. So let's see. 1 over 3 to the n minus 1. So let me rewrite the whole thing. This is 2 to the n. And I'm going to write it over 3 to the n times 3 to the negative 1. Same exact thing, which is the same thing as 3 to the negative 1 in the denominator. I could write it in the numerator. So it's going to be 3 times 2 to the n over 3 to the n, which is equal to 3 times 2/3 to the n. So all this business is the same thing as 3 times 2/3 to the n. Common ratio-- its absolute value of 2/3-- is clearly less than 1. So once again, this infinite geometric series will converge. It will give you a finite value.