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

Answer: (C)

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

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

Now, $S_n=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...$

$\Rightarrow S_n=\{\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+...\}+\{\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...\}$

$\Rightarrow S_n=\frac{{\pi }^2}{8}+\frac{1}{2^2}\{\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...\}$

$S_n=\frac{{\pi }^2}{8}+\frac{1}{4}{S}_n$

$\Rightarrow \frac{3}{4}{S}_n=\frac{{\pi }^2}{8}$

$\therefore S_n=\frac{{\pi }^2}{6}$