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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.

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If a power series is evaluated within its radius of convergence, it is not only continuous but differentiable there as well. So do you see a way to consider $f(x)$ as a power series? – hardmath Jan 18 '13 at 23:16
If $x=1/n$, then $ne^{-nx}=ne^{-1}$. The terms of the sum do not uniformly converge to $0$; so, the sum can't converge uniformly on $(0,\infty)$. – David Mitra Jan 18 '13 at 23:18
@hardmath I'm not evaluating the sum at $x=1/n$, I'm evaluating the $n$'th term of the sum ($ne^{-nx}$) at $x=1/n$. One has the result: If the series $\sum g_n(x)$ converges uniformly on $I$, then the sequence $(g_n)$ converges uniformly to the zero function on $I$. – David Mitra Jan 18 '13 at 23:26
Try substituting $y = e^{-x}$ and follow hardmath's suggestion. – A Blumenthal Jan 18 '13 at 23:26
$f(x) = \frac{e^{-x}}{(1-e^{-x})^2}$ certainly is continuous. – Siméon Jan 18 '13 at 23:28
up vote 0 down vote accepted

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.

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Very clever! Thank you for your help. – James Evans Jan 19 '13 at 0:12
@Ju'x has been cleverer in giving a closed form in the comment above. That is, $g(y) = y(1-y)^{-2}$ and radius of convergence reaches to the nearest singularity. – hardmath Jan 19 '13 at 0:17

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 $\mathbb{R}_+$

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Nice, in that you rehabilitate the OP's original approach. – hardmath Jan 19 '13 at 2:22

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