# 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 $\epsilon = n_{o}-x$.

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Do you expect the same behavior as $x\to \infty$ and $x\to -\infty$? How might they differ? –  Jonas Meyer Sep 28 '12 at 0:43
@Jonas Meyer: I just want to check if both limits are nonzero. –  Cherly Sep 28 '12 at 0:45
Cherly: I think I understand that. Interesting. Would you mind sharing where the problem came from? –  Jonas Meyer Sep 28 '12 at 0:49
It came up during the class, discussion about interchanging the order of limit and summation. –  Cherly Sep 28 '12 at 0:59
Note that the assertion in your last two lines of the question is false: the term on the left goes to infinity when $n$ does, while the right hand side is fixed. –  Martin Argerami Sep 28 '12 at 1:33

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 $0$.

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Nice! What about the other limit at $+\infty$? –  Cherly Sep 28 '12 at 1:36
@Cherly: I am inviting you to investigate how the $+\infty$ case is handled by the observation in the first paragraph of my answer. –  Jonas Meyer Sep 28 '12 at 1:37

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

$$\sum_{n=-\infty}^{\infty}\frac{1}{(x-n)^{2}+1}={\frac {\pi \,\cosh \left( \pi \right) \sinh \left( \pi \right) }{ \left( \cosh \left( \pi \right) \right) ^{2}- \left( \cos \left( \pi \,x \right) \right) ^{2}}}$$

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So for the second form the limit at $\pm\infty$ is zero, this is also interesting, thanks! How did you get the $\cosh$ and $\sinh$ functions to be involved here? –  Cherly Sep 28 '12 at 2:53
@Cherly: The second function is periodic, and has no limits at $\pm \infty$. –  Jonas Meyer Sep 28 '12 at 5:09
Actually, the second function can be used to help answer the question of what happens to the original function at $+\infty$. Note that $f(x)+f(-x) +\frac{1}{x^2+1}$ is periodic and nonconstant, hence has no limits at $\pm\infty$, hence $f$ cannot have limits at both $+\infty$ and $-\infty$. (What I have said here doesn't quite show that $f$ has a positive lower bound in one direction, however, for which the method hinted in my answer, or perhaps use of the Psi functions, may be of more use.) –  Jonas Meyer Sep 28 '12 at 5:50