Pointwise convergence of series $\sum_{n=1}^{\infty}\frac{1}{n}e^{-nx}$ For my math course I have to show that the series $$\sum_{n=1}^{\infty}\frac{1}{n}e^{-nx}$$ converges pointwise on $I=(0,\infty)$ and find the limit.
I think I should prove that $f_n$ is a Cauchy sequence which would show that the series converges. So what I have so far is this:
We want to show that $\forall \varepsilon>0$ there exists an $N\in\mathbb{N}$ such that for $m,n\geqslant N$ we have $|{f_m-f_n}|<\varepsilon$. Also take $m>n$. 
Filling this in gives
$$|f_m-f_n|=|\frac{e^{-mx}}{m}-\frac{e^{-nx}}{n}|=\frac{e^{-nx}}{n}-\frac{e^{-mx}}{m}\leqslant\frac{e^{-nx}}{N}-\frac{e^{-mx}}{N}=\frac{e^{-nx}-e^{-mx}}{N}\leqslant\frac{1}{N}<\varepsilon$$
So we pick our $N$ to be $N>\frac{1}{\varepsilon}$. And then we have proven the convergence.
I am however not really sure if the step where I go from $n$ and $m$ in the denominator to $N$ is actually true.
Also, I've tried writing out this sequence to see what this limit might be, but I'm getting nowhere, really. So any tips on either proving the pointwise convergence and finding the limit is very much appreciated!

In order to show that the given series converges pointwise on $]0,+\infty[$, just fix $x$ in such interval and consider the series. The general term is
$$
\frac1{ne^{nx}}
$$
thus clearly converges; root test sounds good.
Let's now find the sum. Observe that
$$
\left(-\frac{e^{-nx}}{n}\right)'=e^{-nx}
$$
then
$$
\sum_{n=1}^{+\infty}e^{-nx}=\frac{e^{-x}}{1-e^{-x}}
$$
but since
$$
\sum_{n=1}^{+\infty}e^{-nx}
=\sum_{n=1}^{+\infty}\left(-\frac{e^{-nx}}{n}\right)'
=-\left(\sum_{n=1}^{+\infty}\frac{e^{-nx}}{n}\right)'
$$
we get that
$$
\sum_{n=1}^{+\infty}\frac{e^{-nx}}{n}
=-\int\frac{e^{-t}}{1-e^{-t}}\,dt
=-\log(1-e^{-x})+C
$$
passing to the limit $x\to+\infty$ both series and $\log$ vanish, thus $C=0$.
The only fact to check is this last passage of the limit for $x\to+\infty$ under the sum; but this is allowed because, for any fixed $M>0$, the series above is uniformly convergent in $[M,+\infty[$.
