Question

If $\sum _{r=1}^{\infty }\frac{1}{(2\pi -1{)}^2}=\frac{{\pi }^2}{8}$, 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. $\frac{{\pi }^2}{8}$.

Answer 1

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

$\sum _{r=1}^{\infty }\frac{1}{(2r-1{)}^2}=\frac{{\pi }^2}{\theta }$

$\Rightarrow \frac{1}{1^2}+\frac{1}{(2\times 2-1{)}^2}+\frac{1}{(2\times 3-1{)}^2}+.....+\infty =\frac{{\pi }^2}{8}$ ...........$\left(1\right)$

$I=\sum _{r=1}^{\infty }\frac{1}{r^2}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+......\infty$

$=(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+....)+(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}-\infty )$

$\Rightarrow$ from equation $\left(1\right)$

$\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+.....\infty =\frac{{\pi }^2}{8}$

$I\Rightarrow \left(\frac{{\pi }^2}{8}\right)+\frac{1}{2^2}(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+.....\infty )$

$I\Rightarrow \frac{{\pi }^2}{8}+\frac{1}{4}I$

$\frac{3I}{4}=\frac{{\pi }^2}{8}$

$I=\frac{{\pi }^2}{6}$.