Uniform convergence of $\sum_{n=1}^{\infty}\frac{x}{n(n+x^2)}$ on [0,1] Prove that $\sum_{n=1}^{\infty}\frac{x}{n(n+x^2)}$ converges uniformly on $[0,\infty)$.
On $[1,\infty)$, we have $\frac{1}{n}\leq x$ which implies $\frac{x}{n(n+x^2)}\leq\frac{1}{n^2}$. Since $\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty$, the M-test gives that $\sum_{n=1}^{\infty}\frac{x}{n(n+x^2)}$ converges uniformly on $[1,\infty)$.
I'm not sure though how to show this for $[0,1)$.

For $x\ge0$, $x+\frac1x=\left(\sqrt{x}-\frac1{\sqrt{x}}\right)^2+2\ge2$. Therefore,
$$
\begin{align}
\frac x{n(n+x^2)}
&=\frac1{n^{3/2}}\frac1{\frac{\sqrt{n}}x+\frac x{\sqrt{n}}}\\
&\le\frac1{2n^{3/2}}
\end{align}
$$
Therefore, independent of $x$,
$$
\begin{align}
\sum_{n=N+1}^\infty\frac x{n(n+x^2)}
&\le\sum_{n=N+1}^\infty\frac1{2n^{3/2}}\\
&\le\frac12\int_N^\infty\frac{\mathrm{d}x}{x^{3/2}}\\[3pt]
&=\frac1{\sqrt{N}}
\end{align}
$$
