Continuity of an Analytic Function

Question

I am trying to show that the analytic function, $f: (0, \infty) \to \mathbb{R}$, defined by

$ f(x) = \sum \limits_{n = 1}^\infty ne^{-nx} $

is continuous.

I don't have much experience with proofs of this kind, so I'm not sure if my solution (below) is completely rigorous, but I'd welcome any advice. Cheers.

Current Approach

Examine the series of functions defined as:

$f_N(x) = \sum \limits_{n = 0}^N ne^{-nx}$.

Note that the summand can be bounded by $\frac{1}{n^2}$ as follows.

Consider the sequence $a_n = \frac{3 ln(n)}{n}.$ Since $a_n \to 0$, we can always find $N \in \mathbb{N}$ such that $n \ge N \implies |a_n| < x$ for any $x \in (0, \infty).$ For a fixed $x$, choose such an $N$. Thus,

$\begin{align*} \frac{3ln(n)}{n} &< x \\ ln(n^3) &< nx \\ n^3 &< e^nx \\ ne^{-nx} &< \frac{1}{n^2}\end{align*}$

Let $\frac{1}{n^2} = M_n$. Since $\Sigma M_n$ converges, the Weierstrass M Test gives that $f_N$ converges uniformly.

For any n, $h(x) = ne^{-nx}$ is continuous (as the product of two continuos functions). Similarly, each $f_N$ is continuous (as the sum of $N$ continuous functions). Since the sequence of continuous functions $f_N$ converges uniformly to $f(x)$, $f(x)$ must itself be continuous.

Answer 1

Consider the power series $g(y) = \sum_{n=1}^\infty n y^n$. By standard techniques this has radius of convergence $r=1$. Define $f(x) = g(e^{-x})$ for $x \in (0,\infty)$. This composition of continuous functions is continuous.

Answer 2

Another approach is to look at the interval $I = [a,\infty)$ for $a > 0$. On $I$, $|ne^{-nx}| \le ne^{-na}$, so the series converges uniformly on $I$ by Weierstrass. Hence $f$ is continuous on $I$, and by letting $a\to0$, it follows that $f$ is continuous on the whole of $\mathbb{R}_+$