Uniform convergence of a function series over an open interval.

Question

Let $$f(x)=\sum_{k=1}^{\infty}(-1)^ke^{-kx}$$ for $x\in\mathbb{R}$. I am trying to prove that $f(x)$ converges uniformly over the left-open, right-closed interval $I=]0,R]$ with $R>0$. (open interval, because f(x) diverges when $x = 0$)

I already showed that (for $n\in\mathbb{N}$) $$\|\sum_{k=n+1}^{\infty}(-1)^ke^{-kx}\|_0 \leq \sup_{x\in I}\left|e^{-nx}\right|$$

So, how can I prove/disprove the uniform convergence over I ?

PS: Note that $$\|f_k(x)\|_0 := \sup_{x\in I}\left|f_k(x)\right|$$

Answer 1

It doesn't converge uniformly. Let the error in approximating $f(x)$ by its first $n$ terms be $$ \eta_n(x) := \left| \sum_{k=n+1}^\infty (-e^{-x})^k\right| = \frac{e^{-(n+1)x}}{1 + e^{-x}}.$$

Note that $\eta_n$ is continuous and $\lim_{x \to 0} \eta_n(x) = \frac{1}{2}$ for any $n$, which disproves uniform continuity: for any $n > 0$ and $\epsilon < \frac{1}{2}$ you can find some $x \approx 0$ such that $\eta_n(x) > \epsilon$.

Answer 2

Suppose the series converges uniformly on $(0,1).$ Then the summands converge uniformly to $0$ on on $(0,1).$ However,

$$\sup _{x\in (0,1)} |(-1)^ke^{-kx}| = 1\, \text { for all } k,$$

contradiction.