# Thread: [SOLVED] Possibly simple series question

1. ## [SOLVED] Possibly simple series question

$\sum_{n=1}^{\infty}\frac{(-1)^{n}}{\log(e^{n}+e^{-n})}$

code: \sum_{n=1}^{\infty}\frac{(-1)^{n}}{\log (e^{n}+e^{-n})}

(Evaluate)

2. It is a alternating series and $\frac{1}{log(e^n+e^{-n})}\to 0\text{ as }n\to \infty$, thus converges.

3. Thanks, good point. What about absolutely?

4. Originally Posted by student1
$\sum_{n=1}^{\infty}\frac{(-1)^{n}}{\log(e^{n}+e^{-n})}$

code: \sum_{n=1}^{\infty}\frac{(-1)^{n}}{\log (e^{n}+e^{-n})}

(Evaluate)
It would appear that the stuff in red is NOT what you wanted to do.

It would be a big help if you actually said clearly and correctly what you did want to do with this series. eg. Find whether it converges or diverges. etc. We are not mind readers.

5. Originally Posted by student1
Thanks, good point. What about absolutely?

$\frac{1}{\ln(e^n+e^{-n})}\geq \frac{1}{\ln(2e^n)}=\frac{1}{n+\ln 2}$ and thus the series doesn't converge absolutely.

Tonio