Question

It is given 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

Answer: (C)

We have, $\sum _{r=1}^{\infty }\frac{1}{{(2r-1)}^2}=\frac{{\pi }^2}{8}$
$\Rightarrow$ $\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+...\infty =\frac{{\pi }^2}{8}$ ..... (i)

Let $x=\sum _{r=1}^{\infty }\frac{1}{r^2}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...\infty$
$\Rightarrow$ $x=\left(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+...\infty \right)$
$+\left(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...\infty \right)$
$\Rightarrow$ $x=\frac{{\pi }^2}{8}+\frac{1}{4}\left[\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...\infty \right]$
$\Rightarrow$ $x=\frac{{\pi }^2}{8}+\frac{1}{4}.x$
$\Rightarrow$ $x-\frac{x}{4}=\frac{{\pi }^2}{8}$
$\Rightarrow$ $\frac{3x}{4}=\frac{{\pi }^2}{8}$
$\Rightarrow$ $x=\frac{{\pi }^2}{6}$