A series

[Tolaso J Kos]

Evaluate the series:

$$\sum_{n=1}^{\infty} \frac{n^2}{(4n^2-1)^2}$$

Possible hint

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Hint: Expand \( \cos x , \;\; x \in (0, \frac{\pi}{2} )\) in a Fourier sin series.

[galactus]

This one can be done pretty slick by using the residues.

Consider:

$$f(z)=\frac{\pi \cot(\pi z) z^{2}}{(4z^{2}-1)^{2}}$$

Note that $$\sum_{-\infty}^{\infty}f(n)=2\sum_{n=1}^{\infty}f(n)$$

$$\lim_{z\to \pm 1/2}\frac{d}{dz}\left[\frac{\pi \cot(\pi z)z^{2}}{4(2z\pm 1)^{2}}\right]=\frac{-\pi^{2}}{64}$$

$$Res(f(z), \;\ z=1/2)=\frac{-\pi^{2}}{64}$$

$$Res(f(z), \;\ z=-1/2)=\frac{-\pi^{2}}{64}$$

$$Res(f(z), \;\ z=n)=\sum_{n=1}^{\infty}\frac{n^{2}}{(4n^{2}-1)^{2}}$$

$$2\sum_{n=1}^{\infty}\frac{n^{2}}{(4n^{2}-1)^{2}}=\frac{\pi^{2}}{32}$$

[Tolaso J Kos]

Hi C:

here are two other solutions.

1st:

$$\frac{n^2}{(4n^2-1)^2}=\frac{1}{16}\left(\frac{1}{2n-1}+\frac{1}{2n+1}\right)^2=\frac{1}{16}\left(\frac{1}{(2n-1)^2}+\frac{1}{(2n+1)^2}+ \\
\frac{1}{2n-1}-\frac{1}{2n+1}\right)$$

Then the sum in question is:

$$\frac{1}{8}\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2}=\dfrac{\pi^2}{64}$$

2nd: (using Fourier series)

We easily note that Fourier series of \( \cos x, \; x \in (0, \pi/2) \) is:

$$\cos x = \frac{8}{\pi}\sum_{n=1}^{\infty} \frac{n \sin 2nx}{4n^2-1}$$

Then we appy Parseval's identity leading us to the wanted result.