Sum of convergent series

Submitted by Structure on Sat, 12/15/2007 - 17:21.

We know that

$\sum_{n=1}^{+\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$

Then

$\lim_{n \to +\infty}\sum_{k=1}^{n}\frac{1}{k^2}=\frac{\pi^2}{6}$

Let

$x_n=\sum_{k=1}^{n}\frac{1}{k^2},\:y_n=\sum_{k=1}^{n}\frac{1}{(2k-1)^2}$

We have

$x_{2n}=y_n+\frac{1}{4}x_n$

Then taking limit as n tends to $\infty$

$\frac{\pi^2}{6}=\lim_{n\to +\infty}y_n+\frac{1}{4}\frac{\pi^2}{6}$

$\lim_{n\to +\infty}y_n=\sum_{n=1}^{+\infty}\frac{1}{(2n-1)^2}=\frac{3}{4}\frac{\pi^2}{6}=\frac{\pi^2}{8}$