How we could know that for the function $f(x)$ below if there exist (not exist) a constant $C>0$ such that $f(x)\geq C$ for all $x\in \mathbb R$, $f(x)=\sum_{n=1}^{\infty}\frac{1}{(x-n)^{2}+1}$ (i.e., $\lim_{x\to\pm\infty}f(x)\neq 0$)
EDIT: I got something which I'm not sure if its correct or not:
If $x>0$ is too large then there is $n_{o}$ such that $x
and for any $n\geq n_{o}$ we have $(x-(n+k))^{2}< (\epsilon+k)^{2}$ in this case the second summation will be finite, where $\epsilon = n_{o}-x$.