Can we expect to find some constant $C$; so that, $\sum_{n\in \mathbb Z} \frac{1}{1+(n-y)^{2}} Fix $y\in \mathbb R;$ and consider the series:
$$\sum_{n\in \mathbb Z}\frac{1}{1+(n-y)^{2}}.$$

My Question is: Can we expect to find some constant $C$; so that, 
  $$\sum_{n\in \mathbb Z} \frac{1}{1+(n-y)^{2}} <C$$ for all $y\in \mathbb R;$ If yes, How ?

Thanks,

\begin{align}
\sum_{n=-\infty}^{[y]} \frac{1}{1+(n-y)^2} = \sum_{n=-\infty}^{[y]-1} \frac{1}{1+(n-y)^2} + \frac{1}{1+([y]-y)^2} \leq \sum_{n=-\infty}^{1} \frac{1}{1+n^2} + 1 
\end{align}
\begin{align}
\sum_{[y]+1}^{n=+\infty} \frac{1}{1+(n-y)^2} = \sum_{[y]+2}^{n=+\infty} \frac{1}{1+(n-y)^2} + \frac{1}{1+([y]+1 -y)^2} \leq \sum_{1}^{+\infty} \frac{1}{1+n^2} + 1 
\end{align}
Thus $\sum_{n=-\infty}^{+\infty} \frac{1}{1+(n-y)^2} \leq 2 \sum_{n=1}^{+\infty}\frac{1}{n^2} + 2 = \frac{\pi^2}{3}+2$
