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# Test the series for convergence or divergence.$\displaystyle \sum_{n = 1}^{\infty} (-1)^{n-1} e^{2/n}$

## divergent

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

Let's test the Siri's for convergence or diversions. Now here. Although the Siri's is alternating, I won't use the all trading Siri's tests, and the reason is is I see this term over here, and I noticed that the limit does not go to zero. So that's hints that this Siri's will not converge. So this limit is n goes to infinity. The exponents goes to zero. So we just get either the zero, which is one so non zero. So this means that as n gets large, that are our term. Our end term here a N and the limit will be getting very close to the numbers. Negative one and one be due to the negative one on the outside. And the fact that even the two and goes to zero of excuse me is the one. So this means that the limit as N goes to infinity of A M does not exist in particular. If it doesn't exist, then it cannot be zero. So we conclude at the Siri's diverges, and the reason that it diverges is we have a non zero limited here. So it's the diversions test. Bye. The diversions test always good to explain your reasoning, the same watch, test or theorem you're using Anytime you ever claimed that a Siri's convergence or diverges. So that's is our final answer. Diverges by diversion says.

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp