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Test the series for convergence or divergence.

$ \displaystyle \sum_{n = 1}^{\infty} \frac {e^n}{n^2} $

Diverges

Calculus 2 / BC

Chapter 11

Infinite Sequences and Series

Section 7

Strategy for Testing Series

Sequences

Series

Campbell University

Harvey Mudd College

Baylor University

Idaho State University

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something to observe here is that at the top we have even the end. That's exponential growth at the bottom. We just have n squared. So that's polynomial growth. The exponential growth is faster than polynomial growth, which means that the terms that we're looking at here aren't even going to approach zero. That's Ah, necessary condition for convergence is not a sufficient condition, but it is a necessary condition, so that fails, and we certainly have divergence. If you want to be rigorous about why this does not approach zero as in goes to infinity, you could take the limit is n goes to infinity of this and you'd get infinity over infinity if you just plugged in and equals infinity. But since that's not allowed ah, you know usual open towels rule. That's one of the situations where local towers rule is applicable. Solo petals roll. We see that this is the same as limited in approach infinity of the derivative of the top, divided by the derivative of the bottom. And then we're in the same type of situation we were in before. Now we just repeat this. Oops! We get even the end on top again and then at the bottom we get to, and now we're in a situation where if we just plug in and equals infinity, so to speak, we're going to get infinity divided by two, and that's that's just infinity. So not only do the terms not approach zero, they in fact blow up to infinity, so certainly going to be divergence here.

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