Differentiability of $s=\sum\limits_{-\infty}^{\infty} {1\over (x-n)^2}$

How do I show that $s=\sum\limits_{-\infty}^{\infty} {1\over (x-n)^2}$ on $x\not\in \mathbb Z$ is differentiable without using its compact form? I realize that the sequence of sums $s_a=\sum\limits_{-a}^{a} {1\over (x-n)^2}$ is not uniformly convergent.

I also tried to prove that it is continuous by using the usual $\varepsilon \over 3$ method. And it seems to apply because each $s_a$ is continuous and they converge pointwise to $s$. But then I realized that this should not be the right proof because I didn't use any special property of the given functions and the general case only works for uniform convergence. I am very confused. Please help!

Actually I do realize that if I could prove differentiability, continuity follows.

Thanks.

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There's a theorem they teach in undergraduate analysis classes which says that if $\{f_n(x)\}$ are $C^1$ functions on an interval $[a,b]$ such that $|f_n(x)| \leq M_n$ and $|f_n'(x)| \leq N_n$ where $\sum_n M_n$ and $\sum_n N_n$ are both finite, then $\sum_n f_n(x)$ is a differentiable function whose derivative is $\sum_n f_n'(x)$.

You can apply this result to your series on any interval $[k + \epsilon, k + 1 - \epsilon]$; if $n \geq 2|k| + 2$ for example ${1 \over (x - n)^2} \leq {4 \over n^2}$ and similarly the derivative ${2 \over (x - n)^3}$ is of absolute value at most ${16 \over |n|^3}$; these inequalities follow from the fact that $|x| \leq {n \over 2}$ and therefore $|x - n| \geq {n \over 2}$ when $x$ is in the interval $[k + \epsilon, k + 1 - \epsilon]$. (The terms for either series when $n < 2|k| + 2$ have uniform bounds on the interval simply because they're continuous. Hence these earlier terms don't affect the applicability of the result.)

Note the same argument applied repeatedly shows the limit function is $C^{\infty}$, and there is an analytic version of this that shows the resulting function is analytic except at the integers.

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Indeed, there is such a theorem in complex analysis as well. The domain of this function is $\mathbb C \setminus \mathbb Z$. It converges uniformly on compact subsets of that domain. So it is analytic on that domain. – GEdgar Sep 20 '11 at 0:15

OK, just to be exotic (?), let's see if we if we can get this from Morera's theorem. Let $C$ be a simple closed curve neither winds around any integer nor passes through any integer. Then $$\int\limits_C \sum_{n=-\infty}^\infty \frac{1}{(x-n)^2}\;dx = \sum_{n=-\infty}^\infty\ \int\limits_C \frac{1}{(x-n)^2}\;dx = \sum_{n=-\infty}^\infty 0 = 0.$$ The first equality follows from Fubini's theorem.*

The second equality follows from the fact that $C$ does not wind around any point where the holomorphic function $x\mapsto 1/(x-n)^2$ behaves badly (fails to be holomorphic).

Morera's theorem says that if the integral of a function along every simple closed curve that does not wind around any point not in the domain is $0$, then the function is holomorphic.

* Later note: Are the hypotheses of Fubini's theorem satisfied? The terms of the sum are nonnegative and the sum converges to a finite number. If one can show that it depends continuously on $x$, then the integral is that of a continuous function on a compact set, so that is also finite.

However, it now occurs to me that we don't need to go into that, because nonnegativity means we can cite Tonelli's theorem instead. That says that one can interchange the order of two Lebesgue integrations for functions that are everywhere nonnegative, regardless of whether the value of the integral is finite or infinite.

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Any proof that invokes Morera's Theorem gets a +1 from me. But could you elaborate on how the first equality "follows from Fubini's theorem"? That is, can you explain how we know that the hypotheses of Fubini's Theorem are satisfied? – Jesse Madnick Sep 20 '11 at 6:25
+1. Just a little remark on notation... I think it is better to denote by $z$ rather than $x$ the integration variable. – Giuseppe Negro Sep 20 '11 at 17:38
@Jesse: I've added a later note on the Fubini step. – Michael Hardy Sep 20 '11 at 17:42
@Giuseppe: I appreciate your point, but it was called $x$ in the posted question, and changing the notation is sometimes confusing. – Michael Hardy Sep 20 '11 at 17:43