The series:

$$\sum_{n \, \text{odd}}^{\infty} \frac{n^2}{(4-n^2)^2} = \pi^2/16$$

showed up in my quantum mechanics homework. The problem was solved using a method that avoids evaluating the series and then by equivalence the value of the series was calculated.

How do I prove this directly?

  • 1
    $\begingroup$ integrate $\frac{z^2}{(4-z^2)^2}\tan(z)$ over a big circle in the complex plane $\endgroup$ – tired Feb 3 '17 at 8:28

Wolfram Alpha gives me the partial fraction expansion:

$$\frac{n^2}{(4 - n^2)^2} = \frac{1}{8}\left(\frac 1{n-2}-\frac{1}{n + 2}\right) + \frac{1}{4}\left(\frac{1}{(n-2)^2}+\frac{1}{(n+2)^2}\right)$$

So the first part telescopes, and the second part will be some modified version of $\zeta(2)$.

In more detail: $$\begin{align}\sum_{n\text{ odd}} \left(\frac 1{n-2}-\frac{1}{n + 2}\right)&=\frac{1}{-1}-\frac{1}{3}+\frac{1}{1}-\frac{1}{5}+\frac{1}{3}-\frac{1}{7}+\cdots\\ &=-1+1=0\end{align}$$ since all the other terms cancel out.

And $$\begin{align}\sum_{n\text{ odd}}\left(\frac{1}{(n-2)^2}+\frac{1}{(n+2)^2}\right)&=\frac{1}{(-1)^2}+\frac{1}{3^2}+\frac{1}{1^2}+\frac{1}{5^2}+\frac{1}{3^2}+\frac{1}{7^2}+\cdots\\ &=2\left(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\cdots\right) \end{align}$$

It's a famous result that $\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\cdots=\frac{\pi^2}{8}$. You can can prove it if you know:

$$\frac{\pi^2}{6}=\zeta(2) = \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots$$

and thus $$\zeta(2)-\frac1{2^2}\zeta(2) = \frac{1}{1^2}+\frac{1}{3^3}+\frac{1}{5^2}+\cdots$$

So $$\sum \frac{n^2}{(4-n^2)^2}=0 + \frac{1}{4}\cdot 2\cdot \frac{\pi^2}{8}=\frac{\pi^2}{16}$$

  • $\begingroup$ Hi Thomas. I had assumed that the sum over odd integers included the negative ones. Naturally, the answer in that case is twice the answer if the summation extends to the positive odds only. Anyway, (+1) for your well-written post. -Mark $\endgroup$ – Mark Viola Feb 3 '17 at 4:30
  • $\begingroup$ @Dr.MV Yeah, I wondered about that. I might have read it your way if it had been $\sum_{n\text{ odd}}$ rather than $\sum_{n\text{ odd}}^{\infty}$. $\endgroup$ – Thomas Andrews Feb 3 '17 at 4:37
  • $\begingroup$ Ah, yes. I see your point. Well, I did address the issue at the end of the post. ;-)) $\endgroup$ – Mark Viola Feb 3 '17 at 4:40

First, the partial fraction of the summand can be written

$$\begin{align} \frac{n^2}{(4-n^2)^2}&=\frac14\left(\frac{1}{n-2}+\frac{1}{n+2}\right)^2\\\\ &=\frac14 \left(\frac{1}{(n-2)^2}+\frac{1}{(n+2)^2}+\frac{1/2}{n-2}-\frac{1/2}{n+2}\right) \end{align}$$

Second, we note that

$$\begin{align} \sum_{n\,\,\text{odd}}\frac{1}{(n\pm 2)^2}&=\sum_{n=-\infty}^\infty \frac{1}{(2n-1)^2}\\\\ &=2\sum_{n=1}^\infty \frac{1}{(2n-1)^2}\\\\ &=2\left(\sum_{n=1}^\infty \frac{1}{n^2}-\sum_{n=1}^\infty \frac{1}{(2n)^2}\right)\\\\ &=\frac32 \sum_{n=1}^\infty \frac{1}{n^2}\\\\ &=\frac{\pi^2}{4} \end{align}$$

Third, it is easy to show that

$$\sum_{n=-\infty}^\infty \left(\frac{1}{2n-3}-\frac{1}{2n+1}\right)=0$$

Putting it all together we have


If we sum over the positive odd only, then the answer is $(1/2)\pi^2/8=\pi^2/16$



$$\sum_{n \, \text{odd}}^{\infty} \frac{n^2}{(n^2-4)^2}=\sum_{n=1}^{\infty} \frac{(2n-1)^2}{((2n-1)^2-4)^2}$$ Using partial fraction expansion, note $$\frac{(2n-1)^2}{((2n-1)^2-4)^2}=\left(\frac{1}{4(2n+1)^2}+\frac{1}{4(2n-3)^2}\right)-\left(\frac{1}{8(2n+1)}-\frac{1}{8(2n-3)}\right)$$ Note that the second part has cancelling terms.

  • $\begingroup$ Why the downvote? $\endgroup$ – S.C.B. Feb 3 '17 at 4:40
  • $\begingroup$ @ThomasAndrews I fixed the signs. $\endgroup$ – S.C.B. Feb 3 '17 at 4:40

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