Is series $\sum_{n=1}^{\infty} xe^{-nx}$ uniformly convergent?

Question

As in title, is series $\sum_{n=1}^{\infty} xe^{-nx}$ uniformly convergent on $x \in (0, \infty)$? I'm stuggling with this example, because normal methods don't work for me. The sequence $f_n = xe^{-nx}$ uniformly converges to $0$, but $\sum_{n=1}^{\infty} ||f_n||$ does not.

Answer 1

If this series would be uniformly convergent, then $|\sum_{k=m}^n x \exp(-x k) | \leq \varepsilon$ for all $n > m > N(\varepsilon)$ and all $x\in(0,\infty)$. On the other hand, taking $x_n= 1/n$ we find for all $\lceil n/2 \rceil > N(\varepsilon)$ $$\exp(-1/2) \frac{1}{2} \leftarrow \exp(-1/2) \frac{n - \lceil n/2 \rceil}{n} \leq \sum_{k=\lceil n/2 \rceil}^n x_n \exp(-x_n k) < \varepsilon.$$ That is a contradiction!

Update: In fact, it is easy to see that this function is uniformly convergent on any interval $[\delta,\infty)$. Moreover, using the geometric series, we get the identity $$x \sum_{n=1}^\infty e^{-nx} = \frac{x}{e^x-1}.$$

Answer 2

Evaluating the geometric series, we find that the pointwise limit is $$ x\sum_{n=1}^\infty e^{-nx}=\lim_{N\to \infty}xe^{-x}\frac{1-e^{-Nx}}{1-e^{-x}}=\frac{xe^{-x}}{1-e^{-x}} $$ our question is if that limit is a uniform one. By writing the limit in the above way, we can see precisely what our error term is as $N\to \infty$: $$ \left|\frac{xe^{-(N+1)x}}{1-e^{-x}}\right| $$ Now note that examining the sequence $x=\frac{1}{N+1}$ we see that $$ \sup_{x\in (0,\infty)}\left|\frac{xe^{-(N+1)x}}{1-e^{-x}}\right|\geq \left|\frac{e^{-1}}{(N+1)(1-e^{-\frac{1}{N+1}})}\right|\to e^{-1} $$ with $N\to \infty$, as can be seen by taking $$ 1-e^{-1/(N+1)}=-\frac{1}{N+1}+O\left(\frac{1}{(N+1)^2}\right) $$

Answer 3

For $x\gt0$, $$ \sum_{k=1}^\infty xe^{-kx}=\frac{x}{e^x-1} $$ which is decreasing since its derivative is $\frac{\left(1-x-e^{-x}\right)\,e^x}{\left(e^x-1\right)^2}$. Thus, since $e\lt3$, for $0\lt x\le 1$, $$ \sum_{k=1}^\infty xe^{-kx}\ge\frac1{e-1}\gt\frac12 $$ However, for $x\gt0$, $$ \sum_{k=1}^nxe^{-kx}\le nx $$ Thus, for any $n$, for $0\lt x\le\frac1{4n}$, $$ \left|\,\sum_{k=1}^n xe^{-kx}-\sum_{k=1}^\infty xe^{-kx}\,\right|\gt\frac14 $$ Therefore, the sequence of partial sums does not converge uniformly; i.e. for any $n$, there is some point so that $|f_n(x)-f(x)|\ge\frac14$.