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

Question

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.

Need your help ! :)

Answer 1

HINT:

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

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

for $-1\le x<1$.

Answer 2

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 $x$.