Uniform convergence of this series I'm studying the convergence of this series on $ \mathbb{R}_{\geqslant 0}$:
$$ \sum_{n=1}^\infty \frac{\sqrt{nx}}{1+n^2x} $$
It's clear that series converges pointwise, since it's $0$ at $x=0$, otherwise it's equal to:
$$ \frac{1}{\sqrt{x}} \sum_{n=1}^\infty \frac{\sqrt{n}}{\frac{1}{x}+n^2}\leqslant\frac{1}{\sqrt{x}} \sum_{n=1}^\infty \frac{\sqrt{n}}{n^2}=\frac{1}{\sqrt{x}} \sum_{n=1}^\infty \frac{1}{n^{\frac{3}{2}}}$$.
The function $\frac{\sqrt{nx}}{1+n^2x}$ has a maximum at $\frac{1}{n^2},$ and if we sum those values we get the sum $\sum \frac{1}{2\sqrt{n}}$, which diverges, so total convergence fails.
Uniform convergence holds on $[1, \infty)$:
$$ \sum_{n=k}^\infty \frac{\sqrt{nx}}{1+n^2x} = \frac{1}{\sqrt{x}} \sum_{n=k}^\infty \frac{1}{n^{\frac{3}{2}}} \leqslant \sum_{n=k}^\infty \frac{1}{n^{\frac{3}{2}}} \to 0 \ \text{ as } \ k \to \infty$$
But I believe that uniform convergence fails on $[0,1]$, but I can't prove why. Does anyone know how to show this?
 A: To show directly that the series fails to converge uniformly on $[0,1]$ note that
$$\sup_{x \in [0,1]}\left|\sum_{k=n+1}^\infty \frac{\sqrt{kx}}{1 + k^2x}\right| =\sup_{x \in [0,1]}\sum_{k=n+1}^\infty \frac{\sqrt{kx}}{1 + k^2x}\geqslant \sup_{x \in [0,1]}\sum_{k=n+1}^{2n} \frac{\sqrt{kx}}{1 + k^2x} \\\geqslant \sup_{x \in [0,1]}n \cdot \frac{\sqrt{nx}}{1 + (2n)^2 x}\geqslant n \cdot \frac{\sqrt{n\frac1{n^2}}}{1 + (2n)^2 \frac1{n^2}} = \frac{\sqrt{n}}{5} \underset{n \to \infty}\longrightarrow \infty \neq 0,$$
and the limit of the LHS does not converge to $0$.
Recall the general result that a sequence of partial sums $S_n(x)=\sum_{k=1}^nf_k(x)$ converges uniformly to $S(x) = \sum_{k=1}^\infty f_k(x)$  for $x \in D$ if and only if
$$\lim_{n \to \infty}\sup_{x \in D}|S_n(x) - S(x)|= \lim_{n \to \infty}\sup_{x \in D}\left|\sum_{k=n+1}^\infty f_k(x)\right|=0$$
A: Let's substitute $x = \delta^2$. Then the series becomes
\begin{align*}
\sum_{n=1}^{\infty} \frac{\sqrt{n\delta^2}}{1 + n^2\delta^2}
&= \sum_{n=1}^{\infty} \frac{\sqrt{n\delta}}{1 + (n\delta)^2} \sqrt{\delta}
= \frac{1}{\sqrt{\delta}} \sum_{n=1}^{\infty} \frac{\sqrt{n\delta}}{1 + (n\delta)^2} \delta.
\end{align*}
However, thinking of the last series as a Riemann sum, it is reasonable to expect that
$$ \sum_{n=1}^{\infty} \frac{\sqrt{n\delta}}{1 + (n\delta)^2} \delta \quad \xrightarrow{\delta \to 0^+} \quad \int_{0}^{\infty} \frac{\sqrt{x}}{1 + x^2} \, \mathrm{d}x > 0 $$
and hence the overall quantity will diverge as $\delta \to 0^+$. This claim can be verified relatively easily by a simple comparison between the "Riemann sum" and the actual integral.
This means that the sum is unbounded on $(0, \infty)$, and this cannot happen for a uniform limit of bounded functions. Therefore the sum does not converge uniformly.
