Absolutely Converges, but why?

Mar 2013
United States
((-1)ne1/n)/n3 converges absolutely apparently, but I don't see how. When I applied the ratio test, it came out to be one, but obviously that is incorrect. Here is what I did: | [((-1)n+1 e1/n+1)/(n+1)3] * [n3/(-1)n e1/n] | = | [n3/(n+1)3] * [e1/n+1/e1/n] | = 1, as n -> infinity
Sep 2012
Planet Earth
Its easier to apply a limit comparison first.

\(\displaystyle \frac{e^{\frac{1}{n}}}{n^3} < \frac{3}{n^3}\)

Do you see why the latter converges absolutely? If so, do you understand why the former has to as well?

The ratio IS indeed 1. But this just means the test is inconclusive. The ratio test for the simplified series on the right side of my inequality would also give 1. They both converge absolutely.
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Prove It

MHF Helper
Aug 2008
We can see that \(\displaystyle \displaystyle e^{\frac{1}{n}} \leq e \) for all \(\displaystyle \displaystyle n \geq 1 \), so that means

\(\displaystyle \displaystyle \begin{align*} \sum_{n = 1}^{\infty} \left| \frac{ (-1)^n \, e^{ \frac{1}{n} } }{n^3} \right| &= \sum_{n = 1}^{\infty} \frac{ e^{ \frac{1}{n} } }{ n^3 } \\ &\leq \sum_{n = 1}^{\infty} \frac{e}{n^3} \\ &= e \sum_{n = 1}^{\infty} \frac{1}{n^3} \end{align*}\)

which is a convergent p-series. Since your absolute value series is less than a convergent series, your series is absolutely convergent.
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