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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, 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?2012-09-28
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    @Jonas Meyer: I just want to check if both limits are nonzero.2012-09-28
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    Cherly: I think I understand that. Interesting. Would you mind sharing where the problem came from?2012-09-28
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    It came up during the class, discussion about interchanging the order of limit and summation.2012-09-28
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    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.2012-09-28

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