Show uniform convergence of a function series Show that the function series
$$\large\sum_{k = 1}^\infty \frac1{\sqrt k((x - k)^2 + 1)}$$
converges uniformly in $\mathbb{R}$.
I would like to use the Weierstrass M-test but I don't know where to start.
 A: We need to show that for all $\epsilon>0$ there exists a number $N$ that depends on $\epsilon$ only such that 
$$\sum_{k=N}^\infty \frac{1}{\sqrt k \left((x-k)^2+1\right)}<\epsilon$$
Letting $\ell =\lfloor x\rfloor -k$, we have
$$\begin{align}
\sum_{k=N}^\infty \frac{1}{\sqrt k \left((x-k)^2+1\right)}&=\sum_{\ell=-\infty}^{\lfloor x\rfloor -N} \frac{1}{\sqrt {\lfloor x\rfloor-\ell} \left((\ell +x-\lfloor x\rfloor)^2+1\right)}\\\\
&\le \frac{1}{\sqrt N}\sum_{\ell =-\infty}^{\lfloor x\rfloor -N}\frac{1}{(\ell +x-\lfloor x\rfloor)^2+1}\\\\
&\le  \frac{1}{\sqrt N}\sum_{\ell =-\infty}^{\infty}\frac{1}{\ell ^2+1}\\\\
&=\frac{1}{\sqrt N} \pi \coth(\pi)\\\\
&<\epsilon
\end{align}$$
for $N>\left(\frac{\pi \coth(\pi)}{\epsilon}\right)^2$
A: A more elegant way to do this would be to prove the normal convergence of the series on every compact interval $ [a, b] $ :
Let $ n_0 $ such that $ n_0 > b $. Then we have for $ k \geq n_0 $
$$ \displaystyle \frac{1}{\sqrt{k}((x-k)^2+1)} \leq \frac{1}{\sqrt{k}((b-k)^2+1)} = \underset{k \rightarrow \infty}o\left(\frac{1}{k^2}\right)$$
and thus 
$$ \displaystyle \sum_{k=1}^{\infty}\|\frac{1}{\sqrt{k}((x-k)^2+1)}\|_\infty \leq n_0 - 1 + \sum_{k=n_0}^\infty \frac{1}{\sqrt{k}((b-k)^2+1)} $$
with the latter series converging because of the first estimation. Therefore the series is normally convergent and we get its uniform convergence as well.
