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 $[0,1)$.
 A: I'm not sure though how to show this for $[0,1)$.
Hint. For $x \in [0,1]$, one has
$$
\frac{x}{n(n+x^2)}\leq \frac{1}{n(n+0^2)}=\frac1{n^2}
$$ giving
$$
\sum_{n=1}^{\infty}\frac{x}{n(n+x^2)} \leq \sum_{n=1}^{\infty}\frac{1}{n^2}.
$$
A: Now that Olivier has provided a simple solution, let us go for the extreme overkill.
By the properties of the digamma function (the logarithmic derivative of the $\Gamma$ function)
$$ \sum_{n\geq 1}\frac{x}{n(n+x^2)} = \frac{\gamma+\psi(1+x^2)}{x} \tag{1}$$
holds for any $x\geq 0$. The Taylor series of the RHS in a neighbourhood of the origin is given by:
$$ \zeta(2) x -\zeta(3) x^2 + O(x^5) \tag{2} $$
while the asymptotic behaviour as $x\to +\infty$ is given by:
$$ \frac{\gamma+2\log(x)}{x}+\frac{1}{2x^3}+O\left(\frac{1}{x^5}\right). \tag{3}$$
A: 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 $x$,
$$
\begin{align}
\sum_{n=N+1}^\infty\frac x{n(n+x^2)}
&\le\sum_{n=N+1}^\infty\frac1{2n^{3/2}}\\
&\le\frac12\int_N^\infty\frac{\mathrm{d}x}{x^{3/2}}\\[3pt]
&=\frac1{\sqrt{N}}
\end{align}
$$
