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Limit comparison test | Series | AP Calculus BC | Khan Academy
Limit Comparison Test for Series
Limit Comparison Test and Direct Comparison Test
Limit Comparison Test vs. Test For Divergence
Transcription
Statement
Suppose that we have two series and with for all .
Then if with , then either both series converge or both series diverge.[1]
Proof
Because we know that for every there is a positive integer such that for all we have that , or equivalently
As we can choose to be sufficiently small such that is positive.
So and by the direct comparison test, if converges then so does .
Similarly , so if diverges, again by the direct comparison test, so does .
That is, both series converge or both series diverge.
Example
We want to determine if the series converges. For this we compare it with the convergent series
As we have that the original series also converges.
One-sided version
One can state a one-sided comparison test by using limit superior. Let for all . Then if with and converges, necessarily converges.
Example
Let and for all natural numbers . Now
does not exist, so we cannot apply the standard comparison test. However,
and since converges, the one-sided comparison test implies that converges.
Converse of the one-sided comparison test
Let for all . If diverges and converges, then necessarily
, that is,
. The essential content here is that in some sense the numbers are larger than the numbers .
Example
Let be analytic in the unit disc and have image of finite area. By Parseval's formula the area of the image of is proportional to . Moreover,
diverges. Therefore, by the converse of the comparison test, we have
, that is,
.
Rinaldo B. Schinazi: From Calculus to Analysis. Springer, 2011, ISBN9780817682897, pp. 50
Michele Longo and Vincenzo Valori: The Comparison Test: Not Just for Nonnegative Series. Mathematics Magazine, Vol. 79, No. 3 (Jun., 2006), pp. 205–210 (JSTOR)
J. Marshall Ash: The Limit Comparison Test Needs Positivity. Mathematics Magazine, Vol. 85, No. 5 (December 2012), pp. 374–375 (JSTOR)