$x\mapsto \sum_{n=1}^{\infty}ne^{-nx}$ is continuous but not uniform continuous on $(0,+\infty)$. 
Prove that $f(x)=\sum_{n=1}^{\infty}ne^{-nx}$ is continuous but not uniform continuous on $(0,+\infty)$.

Is there any good ways to deal with this kind of functions?
 A: Yes. To prove the continuity we work locally and we use the uniform convergence:
$$ne^{-nx}\le ne^{-n a},\quad\forall x\ge a$$
and the series $\sum_n ne^{-na}$ is convergent because $ne^{-na}=_\infty o\left(\frac1{n^2}\right)$  hence we have the uniform convergence of the given series on every interval $[a,+\infty),\;a>0$ and since the functions $x\mapsto ne^{-nx}$ are continuous we conclude the desired result.
This series isn't uniformly convergent on $(0,\infty)$ because $\lim\limits_{x\to0}f(x)$ doesn't exist.
Edit We can prove that the series isn't uniformly convergent using:
$$\sup_{x>0}\sum_{k=n+1}^\infty ne^{-nx}\ge \sup_{x>0}\sum_{k=n+1}^{2n} ne^{-nx}=2n^2\xrightarrow{n\to\infty}\infty$$
A: *

*$f$ is continuous over $(0,\infty)$


It suffices to prove that this series of functions converges uniformly over $(a,\infty)$ where $a>0$ is arbitrary.
For $x\in (a,\infty)$, $|n\exp(-nx)|\leq n\exp(-na)$
Conclude with Weierstrass M test.
Then let $a \to 0$


*

*f is not uniformly continous


Without accounting for term-by-term differentiation, $f'(x)=\sum -n^2\exp(-nx)$
The derivative of $f$ is not bounded, hence $f$ is not uniformly continous.
A: Your problem is equivalent to showing the non-uniform convergence of $\sum_{n=1}^\infty n y^n$ on $(0,1)$ if we let $y=e^{-x}$. This sum has a closed form expression given by
$$\sum_{n=1}^\infty ny^n = \frac{1}{(1-y)^2}$$
since it is the derivative of $\sum_{n=0}^\infty y^n = \frac{1}{1-y}$. Okay. Consider a partial sum $S_N(y) = \sum_{n=0}^N ny^n$. We can bound $|S_N(y)| \leq n(n+1)/2$. If we had uniform convergence, then, we should have for a particular $N$ large
$$\left| \frac{1}{(1-y)^2} - S_N(y) \right| < \epsilon$$
independent of $y$. But this is impossible because $|S_N(y)|$ is bounded and hence the expression in parentheses can become arbitrarily large as $y \to 1$.
To show continuity, use the closed form expression already mentioned. Remember to use the substitution $y=e^{-x}$ as needed.
A: An alternative method of showing this is by seeing that
$$
\sum_{n=0}^\infty e^{-nx} = \frac{1}{1-e^{-x}} \forall x > 0
$$
Now we can differentiate term by term (only for $x>0$) to see that:
$$
\sum_{n=0}^\infty - n e^{-nx} = -1(1-e^{-x})^{-2} (e^{-x}) = \frac{-e^{-x}}{(1-e^{-x})^2} = \frac{-e^{x}}{(e^x-1)^2} \\
\implies \sum_{n=0}^\infty n e^{-nx} = \frac{e^{x}}{(e^x-1)^2}
$$
Now you can prove the righthand is continuous for $x > 0$ in an assortment of ways, the easiest of which uses the continuity of $e^x$ and $(\cdot)^2$.
To show it's not uniformly continuous since:
$$
\lim_{n \to \infty} \frac{e^{1/n}}{(e^{1/n}-1)^2}
$$
is non convergent and so the function composed with a cauchy sequence isn't a cauchy sequence, showing that the function isn't uniformly continuous.
