How can I show that $\left|\sum_{n=1}^\infty\frac{x}{n^2+x^2}\right|\leq\frac{\pi}{2}$ for any $x\in{\bf R}$? 
Show that
  $$\left|\sum_{n=1}^\infty\frac{x}{n^2+x^2}\right|\leq\frac{\pi}{2}$$ 


It is true for $x=0$. But I don't see how it is true for any $x\in{\bf R}$. The identity $\frac{\pi}{2}=\int_0^\infty\frac{1}{1+x^2}dx$ may help, I think. But I don't know how to go on. 

We know that $$\left|\sum_{n=1}^\infty\frac{x}{n^2+x^2}\right| < \int_0^\infty{\frac{x}{n^2+x^2}dn}$$ for any $x\in{\Bbb R}$
Evaluating the integral we get $$\begin{align}
\int_0^\infty{\frac{x}{n^2+x^2}dn} &= tan^{-1}\bigg(\frac{n}{x}\bigg)\bigg|_0^\infty \\
&= \lim_{t\to\infty} tan^{-1}\bigg(\frac{t}{x}\bigg) \\
&= \frac{\pi}{2}
\end{align}$$
Therefore $$\left|\sum_{n=1}^\infty\frac{x}{n^2+x^2}\right| < \frac{\pi}{2}$$
