# [SOLVED] Possibly simple series question

• Jan 15th 2010, 12:20 AM
student1
[SOLVED] Possibly simple series question
http://latex.codecogs.com/gif.latex?...e^{n}+e^{-n})}

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

(Evaluate)
• Jan 15th 2010, 02:01 AM
Shanks
It is a alternating series and $\displaystyle \frac{1}{log(e^n+e^{-n})}\to 0\text{ as }n\to \infty$, thus converges.
• Jan 15th 2010, 08:32 AM
student1
Thanks, good point. What about absolutely?
• Jan 15th 2010, 04:08 PM
mr fantastic
Quote:

Originally Posted by student1
http://latex.codecogs.com/gif.latex?...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.
• Jan 15th 2010, 11:25 PM
tonio
Quote:

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

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

Tonio