Positive lower bound for a function defined as a series? 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 $x<n_{o}$, so we split the summation
$$\sum_{n=1}^{n_{o}-1}\frac{1}{(x-n)^{2}+1}+\sum_{n=n_{o}}^{\infty}\frac{1}{(x-n)^{2}+1}$$
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 $\epsilon = n_{o}-x$.
 A: When $x\geq \frac{1}{2}$, there exists a positive integer $k$ such that $|x-k|\leq \frac{1}{2}$.  Thus $(x-k)^2\leq \frac{1}{4}$.  What does this imply about $f(x)$ for $x>0$, and how does it answer the $x\to \infty$ part?
When $x<0$, $x$ will no longer be so close to positive integers.  Note that $f$ is increasing on $(-\infty,1]$, so it suffices to find the sequence limit $$\lim\limits_{k\to\infty}f(-k)=\lim\limits_{k\to\infty}\sum\limits_{n=1}^\infty\dfrac{1}{(k+n)^2+1}.$$ The series $\sum\limits_{n=1}^\infty\dfrac{1}{n^2+1}$ converges, which means that the sequence of partial sums converges, and 
$$\sum\limits_{n=1}^\infty\dfrac{1}{(k+n)^2+1}\ \ =\sum\limits_{n=1}^\infty\dfrac{1}{n^2+1}-\sum\limits_{n=1}^{k}\dfrac{1}{n^2+1}\to 0.$$
That is, $f(-k)\to 0$ as $k\to \infty$ because the tail of a convergent series goes to $0$.  
A: Related problem: (I). Here is a closed form for the series, in terms of the psi function that may help you analyze your problem 
$$ \sum_{n=1}^{\infty}\frac{1}{(x-n)^{2}+1}=-\frac{1}{2}\,i \left( -\psi \left( 1-i-x \right) +\psi \left( 1+i-x \right) 
 \right)\,.$$
Another interesting closed form is when you sum the series from $-\infty$ to $\infty$, you get the following function
$$ \sum_{n=-\infty}^{\infty}\frac{1}{(x-n)^{2}+1}={\frac {\pi \,\cosh \left( \pi  \right) \sinh \left( \pi  \right) }{
 \left( \cosh \left( \pi  \right)  \right) ^{2}- \left( \cos \left( 
\pi \,x \right)  \right) ^{2}}}
 $$  
