Series Converging Uniformly

On the interval $[\delta,+\infty)$ we have

$$\sup_{x\ge\delta}\frac1{ne^{nx}}\le \frac{1}{ne^{n\delta}}=:a_n$$ and the series $\sum a_n$ is convergent so we have the uniform convergence of the given series on this interval.

And on the interval $(0,\infty)$ we have not the uniform convergence. In fact

$$\sup_{x>0}\left\{\sum_{n=N+1}^\infty\frac1{ne^{nx}}\right\}\ge\sup_{x>0}\sum_{n=N+1}^{2N}\frac{1}{ne^{nx}}=\sum_{n=N+1}^{2N}\frac1n\ge N\frac1{2N}=\frac12\not\xrightarrow{N\to\infty}0$$

Hint. One has $$ \sup_{x \in (0,\infty)}\:\sum\limits_{n=1}^N\frac{e^{-nx}}{n}=\sum\limits_{n=1}^N\frac{1}{n} $$ which tends to $\infty$ as $N \to \infty$, whereas for $\delta>0$ $$ \sup_{x \in [\delta,\infty)}\:\sum\limits_{n=1}^N\frac{e^{-nx}}{n}=\sum\limits_{n=1}^N\frac{e^{-n\delta}}{n} $$ which converges as $N \to \infty$.