Series Converging Uniformly Given $$\sum\limits_{n=1}^\mathbb{∞}\frac1{ne^{nx}}$$ why does this series not converge uniformly for $x$ in $(0,∞)$ but converges uniformly for $x$ in $[,∞)$ for $ > 0$?
 A: 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
$$\sup_{x>0}\left\{\sum_{n=N+1}^\infty\frac1{ne^{nx}}\right\}\ge\sup_{x>0}\sum_{n=N+1}^{2N}\frac{1}{ne^{nx}}=\sum_{n=N+1}^{2N}\frac1n\ge N\frac1{2N}=\frac12\not\xrightarrow{N\to\infty}0$$
A: 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 $N \to \infty$.
A: Denote $$\xi_{n,m} (\theta ) =\sum_{k=m+1}^{n+m} \frac{1}{ke^{k\theta}}$$ then $$\sup_{\theta >0} |\xi_{2n, n} (\theta ) |\geqslant \xi_{2n, n} (n^{-1} )\geqslant e^{-2} n\frac{1}{2n} =2^{-1}e^{-2}$$ thus the series does not converges uniformly.
A: Suppose $x\le\frac1n$, then
$$
\begin{align}
\sum_{k=n}^\infty\frac1{ke^{kx}}
&\ge\sum_{k=n}^{2n}\frac1{ke^{kx}}\\
&\ge\sum_{k=n}^{2n}\frac1{ke^2}\\
&\ge\frac{\log(2)}{e^2}
\end{align}
$$
Thus, there is an $\epsilon\gt0$, i.e. $\frac{\log(2)}{e^2}$, so that for any $n$, we can find an $x\gt0$, i.e. $\frac1n$, so that
$$
\sum_{k=n}^\infty\frac1{ke^{kx}}\ge\epsilon
$$

Therefore,
  $$
\sum_{k=1}^\infty\frac1{ke^{kx}}
$$
  does not converge uniformly on $(0,\infty)$.


For all $x\ge\delta$,
$$
\begin{align}
\sum_{k=n}^\infty\frac1{ke^{kx}}
&\le\sum_{k=n}^\infty\frac1{ke^{k\delta}}\\
&=\frac1n\frac{e^{-n\delta}}{1-e^{-\delta}}
\end{align}
$$
and
$$
\lim_{n\to\infty}\frac1n\frac{e^{-n\delta}}{1-e^{-\delta}}=0
$$

Therefore,
  $$
\sum_{k=1}^\infty\frac1{ke^{kx}}
$$
  converges uniformly for $x\ge\delta$.

