Question

It is known that ${\sum }_{r=1}^{\infty }\frac{1}{{(2r-1)}^2}=\frac{{\pi }^2}{8}$, then ${\sum }_{r=1}^{\infty }\frac{1}{r^2}$ is equal to

• A. $\frac{{\pi }^2}{24}$
• B. $\frac{{\pi }^2}{3}$
• C. $\frac{{\pi }^2}{6}$
• D. None of these

Answer 1

The correct option is C $\frac{{\pi }^2}{6}$

We know $\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+.....\infty =\frac{x^2}{8}$
Let $x=\frac{1}{1^2}+\frac{1}{2^2}+.....\infty$
$=(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+.....\infty )+(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2})$
$=\frac{{\pi }^2}{8}+\frac{1}{4}(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+.....\infty )$
$=\frac{{\pi }^2}{8}+\frac{1}{4}x\Rightarrow \frac{3x}{4}=\frac{{\pi }^2}{8}\Rightarrow x=\frac{{\pi }^2}{6}$