Uniform convergence of $\sum_{n=1}^{\infty}\frac{1-e^{-\frac{x}{n}}}{nx}$ I would like to check if the series
$$
\sum_{n=1}^{\infty}\frac{1-e^{-\frac{x}{n}}}{nx}
$$
converges uniformly on $(0, +\infty)$ and if the sum of this series is differentiable function.
My approach:
I see that for specified $x$ and all $n\in\mathbb{N}$
$$
1-\frac{x}{n}\leq e^{-\frac{x}{n}}
$$
so I would like to compare the given series with 
$$
\sum_{n=1}^{\infty}\frac{1}{n^2}
$$
but I don't know if this gives me the uniform convergence or only in some point.
I know the Weierstrass test but don't know how to use it there and I have no idea how to check differentiability of the function defined as series.
Can you give any hints or the ways how can I answer to this question? 
 A: Since (using convexity)
$$0 < 1 - e^{-x/n}  \le x/n\ \text{for $x > 0$}$$ 
we have 
$$ \frac{1-e^{-x/n}}{nx} \le \frac{1}{n^2}$$
By the Weierstrass M-test your series converges absolutely and uniformly on $(0,\infty)$,
and the limit is a continuous function.
Moreover, 
$$0 > \dfrac{d}{dx} \frac{1-e^{-x/n}}{nx} = \frac{(n+x) e^{-x/n} - n}{n^2 x^2}> -\frac{1}{n^3}  $$
so the series of derivatives converges absolutely and uniformly as well, which 
means the sum of your series is differentiable.
A: Hint: write $e^{-x/n}$ as a series to estimate $\dfrac{1-e^{-x/n}}{nx}$ and its derivative.
A: I think your approach is right. Let's call $f(x)=\sum_{n=1}^\infty \frac{1-e^{-x/n}}{nx}$. You already have proved that this series converges by comparison test, so $f(x)$ is well defined. 
You want to prove that $f_m(x)=\sum_{n=1}^m\frac{1-e^{-x/n}}{nx}$. Converges uniformly to $f$ as $m\to\infty$. Just note that by the same inequality you wrote, we have $$0\leq f-f_m(x)=\sum_{n=m+1}^\infty\frac{1-e^{-x/n}}{nx}\leq\sum_{n=m+1}^\infty\frac{1}{n^2}$$
Note that this bound is independent of $x$ and converges to $0$ as $m\to \infty$ since it is the tail of a convergent series. So, $f_m\to f$ uniformly.
