# The value of the series $\frac{e^{-n}}{n}$

I have already found that the series $\sum_{n=1}^{+\infty} \frac{e^{-n}}{n}$ converges. However, I want to find its value, but I don't know how to do that. I've tried several things but without success.

• Could you find $\sum_{n=1}^{+\infty} \frac{x^n}{n}$ for a real variable $x?$ Nov 4 '16 at 22:47

HINT: Let $f(x)=\sum_{n\ge 1}\frac{x^n}n$; then

$$f'(x)=\sum_{n\ge 1}x^{n-1}=\sum_{n\ge 0}x^n\;.$$

You know a closed form for $f'(x)$, and you can integrate it. You’ll get a constant of integration, but clearly $f(0)=0$, so you can pin down the constant. Then substitute the right value of $x$.

• Heh, you beat me to that. +1 Nov 4 '16 at 22:53

HINT:

The Taylor Series for $\log(1-x)$ is given by

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

for $-1\le x<1$.

• Okay, until here I get it. After that, I want $\frac{e^{-n}}{n}$ instead of $\frac{x^{n}}{n}$. So, I would like say let's x=1/e. Then we have $\sum_{n=1}^{\infty} \frac{e^{-n}}{n} = -ln(1-\frac{1}{e})$ ? Nov 4 '16 at 23:16
• You have it! Well done. Nov 5 '16 at 1:23
• You're welcome. My pleasure. Nov 5 '16 at 12:44
• And feel free to up vote and best vote as you see fit. Nov 5 '16 at 12:45