Is the sum of this series a differentiable function? Let 
$$f(x) = \sum_{n=1}^{\infty} \frac{1}{nx} \left( 1 - \frac{1}{e^{ \frac{x}{n}}}  \right) \wedge x>0$$
Is the sum of this series a differentiable function?

my idea
For examining differentiation let:
    $$ g_n (x) :=  \frac{1}{nx} \left(1 - \frac{1}{e^{ \frac{x}{n}}} \right) $$
    then
    $$ g_n'(x) = -\frac{\left(n e^{x/n}-n-x\right)}{e^{\frac{x}{n}}n^2 x^2} $$
but $g_n'(x)$ seems to be asymptotic similar  to $\frac{1}{n}$ so I can't use theorem which can help me to eventually proof that $f$ is differentiable. What should I do in such situation?
 A: You can do some simplifications first. The factor $\frac 1 x$ has no effect on differentiability, so drop that. Next note that $1-\frac 1 {e^{x/n}}=1-e^{-x/n}$. Alos recall that $1-e^{-t} \leq t$ for all $t \geq 0$. It is now obvious that the series is uniformly convergent (by comparison with $\sum \frac 1 {n^{2}})$. If you drop $\frac  1 x$ and differentiate the series you will get $\sum\frac x {n^{2}} e^{-x/n}$. Use the fact that $e^{-x/n} \leq 1$ to conclude that the differentiated series also converges uniformly for bounded $x$. Hence the sum of the original series is a differentiable function (as I explained to you in your previous post).
A: $\mathbf{Hint}$: Why do you say $g_n'(x)$ is asymptotically similar to $\frac{1}{n}$? For $n$ large, $e^{x/n} \approx 1+\frac{x}{n}+\frac{x^2}{2n^2}$, so that 
$$n(e^{x/n}-1)-x \approx \frac{x^2}{2n}$$
And hence 
$g_n'(x) \approx -\frac{e^{-x/n}}{2n^3}$. Notice that the sum of this latter expression converges uniformly on $\mathbb{R}$. Use Taylor's theorem to make this approximation precise and show the sum of $g_n'$ converges uniformly on any bounded interval.
