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!

Hint: for $|x|\lt1$,
$$
\log\left(\frac1{1-x}\right)=\sum_{k=1}^\infty\frac{x^k}{k}
$$
To show convergence, one could try comparing the series to a Geometric Series.
