Consider the following series $$\sum_{n=2}^{\infty}\frac{xe^{-nx}}{\ln n}.$$
It is easy to see that it converges uniformly on $[a,1]$ for any $a>0$. I want to show that it does converge uniformly on $[0,1]$.
I have tried some values like $x=1/n$ and then letting $n\to\infty$ but this did not work:
$$\sum_{n=k}^{2k}\frac{1/k\cdot e^{-n/k}}{\ln n}\geq(k+1)\frac{e^{-2}}{k\ln 2k}\sim\frac{1}{e^2\ln 2k}$$
I think it is uniformly convergent: For $x \in (0,1]$ and $1\le k<l$: $$ 0 \le \sum_{n=k+1}^l \frac{xe^{-nx}}{\ln n}\le \frac{1}{\ln(k+1)}x \sum_{n=k+1}^l(e^{-x})^n = \frac{1}{\ln(k+1)} x\frac{e^{-(k+1)x}-e^{-(l+1)x}}{1-e^{-x}} $$ $$ = \frac{1}{\ln(k+1)} \frac{x}{1-e^{-x}}(e^{-(k+1)x}-e^{-(l+1)x}) \le \frac{1}{\ln(k+1)} \frac{x}{1-e^{-x}}. $$ Now $x\mapsto \frac{x}{1-e^{-x}}$ is increasing on $(0,1]$. Thus for $x \in [0,1]$: $$ 0\le \sum_{n=k+1}^l \frac{xe^{-nx}}{\ln n}\le \frac{1}{(1-e^{-1})\ln(k+1)}, $$ (for $x=0$ this inequality is trivially true).
Thus $\sum_{n=2}^\infty \frac{xe^{-nx}}{\ln n}$ is a uniform Cauchy series on $[0,1]$, hence uniformly convergent.