Question

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

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

Answer 1

Answer: (B)

$\frac{{\pi }^2}{6}$

Given that, $\sum _{r=1}^{\infty }\frac{1}{{(2r-1)}^2}=\frac{{\pi }^2}{8}$
let $S_{\infty }=\sum _{r=1}^{\infty }\frac{1}{r^2}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....\infty$
$\Rightarrow S_{\infty }=\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+...\infty +\frac{1}{2^2}(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...\infty )$
$\Rightarrow S_{\infty }=\sum _{r=1}^{\infty }\frac{1}{{(2r-1)}^2}+\frac{S_{\infty }}{4}$
$\Rightarrow \frac{3S_{\infty }}{4}=\frac{{\pi }^2}{8}$
$\Rightarrow S_{\infty }=\frac{{\pi }^2}{6}$
Ans: B