Sum of convergent series

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 $

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

so

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

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