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Prove that $\sum_{n=1}^{\infty}$ $\frac{(-n)^{n}}{(n+1)^{n+1}} $ converges conditionally.

I have been able to prove that the series converges normally. But when it comes to the absolute series, I've been struggling to show divergence. One thing that I've tried is to compare it with the Harmonic Series ($\sum_{n=1}^{\infty}$ $\frac{1}{n} $) which I know diverges. But the series turns out to be smaller; so, the comparison test fails. The root and ratio tests are also inconclusive because the value I get with the former is $\alpha = 1$. Do you have any advice?

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  • $\begingroup$ In absolute value, the terms are like $(1/e)\cdot(1/n)$ $\endgroup$
    – zhw.
    Mar 24 '20 at 17:20
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$\frac{n^n}{(n+1)^n}=\frac1{(1+\frac1n)^n}$ converges to $\mathrm e^{-1}$. Thus you can compare the terms to $\frac1{3n}$.

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You can compare use the limit comparison test with $\sum_{n=1}^{\infty}\frac{1}{n+1}$, which is well-known that diverges by the integral test. Now comparing to the limit we have:

$\rm{Lim}_{n\rightarrow \infty}\frac{\frac{n^n}{(n+1)^{n+1}}}{\frac{1}{n+1}}=\rm{Lim}_{n\rightarrow \infty}\frac{n^n(n+1)}{(n+1)^{n+1}}=\rm{Lim}_{n\rightarrow \infty}\frac{n^n}{(n+1)^n}=\rm{Lim}_{n\rightarrow \infty}\frac{1}{(1+1/n)^{n}}=\frac{1}{\rm{e}}$.

Given that the two series are comparable because the quotient of their n-terms tends to a value when n tends to infinite then we deduce by the limit comparison test that the original series does not converge absolutely.

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You have right idea to compare it to harmonic series.

$\lim_{n \to \infty} \frac{n^n}{(n+1)^{n+1}} = \frac{n^{n+1}}{n(n+1)^{n+1}} = \frac{1}{n}(\frac{n}{n+1})^{n+1} = \frac{1}{n}(1 - \frac{1}{n+1})^{n+1} = \frac{1}{n}e^{-1}$

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