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I want to compute the value of

$$\sum_{n=1}^{\infty} \frac{1}{((2n)^2 - 1)^2}.$$

I have tried evaluating the first couple partial sums, but can't find any recurrence. I don't have too many tools in my toolbox to proceed... Can anyone see how I could go on here?

Pascal Engeler
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3 Answers3

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Hint: $$\frac{1}{((2n)^2 - 1)^2} = \frac{1}{4}\left( \frac{1}{2 n+1}+\frac{1}{(2 n+1)^2}-\frac{1}{2 n-1}+\frac{1}{(2 n-1)^2}\right)$$ Then, you can start with partial sums, breaking them in four parts and doing a change of indices on 2 of them to get something which would (hopefully — I didn't go further in the details) simplify.

Edit: In particular, provided I didn't do any mistake in the process, you can show that $$ \sum_{n=1}^N \frac{1}{((2n)^2 - 1)^2} = \frac{1}{4(2N+1)} + \frac{1}{4(2N+1)^2} + \frac{1}{2}\sum_{n=1}^N \frac{1}{(2n+1)^2}$$ where the first two terms go to $0$, and the last one goes to $\frac{1}{2}\sum_{n=1}^\infty \frac{1}{(2n+1)^2}$, which can be computed exactly (knowing that $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$ and breaking the latter series' partial sum into even and odd $n$'s).

Clement C.
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    The linear terms telescope, leaving the sum of reciprocal of odd squares, so you'd need the formula for that. – Calvin Lin Jul 08 '13 at 15:24
  • Additionally, as a nice (and comforting) sanity check, doing so does give the same answer as Ron Gordon's, that is $\frac12 \left( \frac{\pi^2}{8}-1\right)$ (as $\sum_{n=0}^\infty \frac{1}{(2n+1)^2} = \frac{\pi^2}{8}$ and the summation here starts at $n=1$). – Clement C. Jul 08 '13 at 15:39
  • Alright! I figured it out with your help and with what André Nicolas said. Thanks a lot! I'd vote up if I could, but apparently I need more reputation points or something like that.. – Pascal Engeler Jul 09 '13 at 10:47
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You may use residue theory. For example, a convergent sum

$$\sum_{n=-\infty}^{\infty} f(n) = -\sum_k \text{Res}_{z=z_k} \pi \cot{\pi z} \, f(z)$$

where the $z_k$ are the non-integer poles of $f$. In your case, $f(z)=1/(4 z^2-1)^2$, so that the sum

$$\sum_{n=-\infty}^{\infty} \frac{1}{(4 n^2-1)^2} = -\frac{\pi}{16} \left (\left [\frac{d}{dz} \frac{\cot{\pi z}}{(z+1/2)^2} \right]_{z=1/2} + \left [\frac{d}{dz} \frac{\cot{\pi z}}{(z-1/2)^2} \right]_{z=-1/2}\right )$$

which, when evaluated, is $\pi^2/8$. However, this is not the sum desired; rather, it is over $1$ to $\infty$. In this case, the sum becomes

$$\sum_{n=1}^{\infty} \frac{1}{(4 n^2-1)^2} = \frac12 \left( \frac{\pi^2}{8}-1\right) \approx 0.116850$$

Ron Gordon
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  • This sounds interesting. I don't quite see yet how the first formula arises, though. I am familiar with computing integrals using the residue theorem, but I have never seen it being used in this context. Can you tell me more about this? – Pascal Engeler Jul 08 '13 at 15:44
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    @PascalEngeler: the derivation goes roughly like this: consider the integral $$\oint_C dz , \pi \cot{\pi z} , f(z)$$ where $C$ is a square contour about $\Im{z}$ and $\Re{z} \in [-(N+1/2),N+1/2]$. As $N \to \infty$, the integral goes to zero, but the integral is also equal to $i 2 \pi$ times the sum of the residues of the poles of the integrand inside $C$. Thus, this sum is zero, which leads to the relationship in the first equation. See http://www.maths.manchester.ac.uk/~cwalkden/complex-analysis/complex_analysis_part7.pdf – Ron Gordon Jul 08 '13 at 15:51
  • @PascalEngeler: note also that $\pi \cot{\pi z}$ has residue of $1$ at the poles $z_k = k$. – Ron Gordon Jul 08 '13 at 15:56
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Hint: Use partial fractions as Clement C. suggests. One part telescopes, and the other two parts are close relatives of the familiar $\sum \frac{1}{n^2}$, which maybe can be taken as known to be $\frac{\pi^2}{6}$.

A detail: $$1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots =\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots\right) -\frac{1}{4}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots\right).$$

André Nicolas
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