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. 
 A: You can rewrite the sum to
$$
\frac {1}{x}\sum_{n=1}^\infty \frac{1}{(\frac n x)^2+1}
$$
Because $f(n)=\frac1{n^2+1}$ is strictly decreasing for positive (reals) $n$, we know that 
\begin{align}
\frac 1x\sum_{n=1}^\infty \frac{1}{(\frac nx)^2+1}&\leq \frac 1x\int_{n=0}^\infty\frac1{(\frac nx)^2+1} dn\\
&=\frac 1x \left(x \frac \pi 2\right)=\frac \pi 2
\end{align}
For the integral, i used the substitution $m=\frac nx$, and because $dm=\frac 1x dn$, we get the factor $x$ in the result.
A: By showing that the sum is exactly
$$\frac{\pi}{2} \operatorname*{coth}{(\pi x)} - \frac1{2 x}$$
This may be done via, e.g., the residue theorem.  Note that the expression is bounded from above by $\pi/2$ as $x \to \infty$
A: Consider the function $f(y) = \dfrac1{y^2+x^2}$. This is a monotonically decreasing function for $y>0$. Hence, we have
$$\dfrac1{n^2+x^2} < \int_{n-1}^{n} \dfrac{dy}{y^2+x^2}$$
Hence,
$$\sum_{n=1}^{\infty}\dfrac1{n^2+x^2} < \sum_{n=1}^{\infty}\int_{n-1}^{n} \dfrac{dy}{y^2+x^2} = \int_{0}^{\infty} \dfrac{dy}{y^2+x^2} = \dfrac{\pi}{2x}$$
A: 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 $$\left|\sum_{n=1}^\infty\frac{x}{n^2+x^2}\right| < \frac{\pi}{2}$$
