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Without using the compact form of $\sum\limits_{n=-\infty}^{\infty} {1\over (x-n)^2}$, how might one show that the series is convergent? I have thought of using the Cauchy test, but am not sure what $N(\varepsilon)$ to use. Thanks.

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Suppose $x\notin \mathbb{Z}$ (otherwise we have a problem at $n=-x$) Then $$\sum_{n=-\infty}^\infty \frac{1}{(x-n)^2}=\frac{1}{x^2}+\sum_{n\neq 0} \frac{1}{n^2} \frac{1}{\left(\frac{x}{n}-1\right)^2}$$

Then since $\frac{1}{\left(\frac{x}{n}-1\right)^2}$ is bounded, and $\sum_{n\neq 0} \frac{1}{n^2} $ converges, we can use the comparison test.

(In particular, $\frac{1}{\left(\frac{x}{n}-1\right)^2}$ is bounded by either $\frac{1}{\{x\}^2}$ or $\frac{1}{(1-\{x\})^2}$.)

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Eric: What you explain, or drop the finite number of terms such that $|n|\le 2|x|$. –  Did Sep 19 '11 at 10:55
    
.....and of course "$\frac{1}{\left(\frac{x}{n}-1\right)^2}$ is bounded" means bounded as $n$ varies with $x$ fixed, not the other way around. –  Michael Hardy Sep 19 '11 at 13:24

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