Prove that function $f(x)=\sum_{n=1}^{\infty}ne^{-nx}$ is continuous.

Question

I have to prove that the following function is continuous. $$ f(x)=\sum_{n=1}^{\infty}u_n(x),\ \ u_n(x)=ne^{-nx},\ \ x\in E=(0,+\infty) $$

In order to do that, I tried to show that $u_n(x)$ is continuous on $E$ for any $n\in\mathbb{N}$ and also that the series $\sum_{n=1}^{\infty}u_n(x)$ is uniformly convergent on $E$. The former is, of course, true. However, the latter condition is not met: $$ 0<x<1: |u(x)-u_n(x)|=\sum_{k=n+1}^{\infty}ke^{-kx}=\left[x=\frac{1}{k}\in E\right]=\sum_{k=n+1}^{\infty}\frac{k}{e}\\ \lim_{n\rightarrow\infty}\sup_{x\in E}\sum_{k=n+1}^{\infty}\frac{k}{e}=+\infty\ne 0\Rightarrow\sum_{n=1}^{\infty}u_n(x)\ \ \text{is not uniformly convergent on}\ \ E \Rightarrow\\ \Rightarrow f(x)\ \ \text{is not continuous on}\ \ (0,1). $$ Thus, I thought that the problem statement is not correct. But I am not sure.

I would be very grateful if someone clarified my confusion!

P.S. I need to prove continuity of $f(x)$ in order to find $\int_{\ln2}^{\ln5}f(x)dx$.

Answer 1

For each $a>0$, the convergence is uniform on $(a,\infty)$. So, $f|_{(a,\infty)}$ is continuous and, since this takes place for each $a>0$, $f$ is continuous.

Answer 2

Note that for $x>0$, $$\sum_{n=0}^{\infty} e^{-nx} = \frac{1}{1-e^{-x}}.$$ Differentiating w. r. t. $x$ pn both sides we get $$ f(x)=\sum_{n=1}^{\infty} n e^{-nx} = \frac{e^{-x}}{(1-e^{-x})^2}.$$ Finally, check that this function $f(x)$ is real and finite for all values of $x> 0.$