# Does $\sum_{n=1}^\infty (-1)^n\frac{n\log n}{e^n}$ converge absolutely?

How to check whether the series converges absolutely: $$\sum_{n=1}^\infty (-1)^n\frac{n\log n}{e^n}$$

I tried tests like Ratio, Raabe but not working

• Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta. – AlexR Dec 8 '14 at 15:07
• This is the derivative of the polylogarithm. – Lucian Dec 8 '14 at 18:51

The ratio test should work just fine. If $a_n = (-1)^n \frac{n \log n}{e^n}$ then $$\left| \frac{a_{n+1}}{a_n} \right| = \frac{n+1}{n} \frac{\log(n+1)}{\log n} \frac 1e$$ which converges to $\dfrac 1e$ as $n \to \infty$.
• why is $\frac{log(n+1)}{log n}$ =1 as $n \rightarrow \infty$ – Learnmore Dec 8 '14 at 15:17
• One explanation is that $\log(n+1) - \log n = \log(1+1/n) \to 1$, so that $\log(n+1)/\log n - 1 \to 0$. – Umberto P. Dec 8 '14 at 15:25
$a_n=(-1)^n\dfrac{n\log n}{e^n}\implies\displaystyle\lim_{n\to\infty}\sqrt[n]{\left \lvert a_n\right \rvert}=\displaystyle\lim_{n\to\infty}\sqrt[\large n]{\dfrac{n\log n}{e^n}}=\dfrac{1}{e}\left(\displaystyle\lim_{n\to\infty}{\sqrt[n]{n\log n}}\right)=\dfrac{1}{e}<1$