Question

If $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+....upto\infty =\frac{{\pi }^2}{6}$, then find $1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+....upto\infty$

• A. $\frac{{\pi }^2}{12}$
• B. $\frac{{\pi }^2}{8}$
• C. $\frac{{\pi }^2}{6}$
• D. none of these

Answer 1

The correct option is A $\frac{{\pi }^2}{12}$

$S$ $=$ $1+\frac{1}{2^2}+\frac{1}{3^2}+....upto\infty =\frac{{\pi }^2}{6}$ -------(1)
Let
$x$ $=$ $1-\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....upto\infty$ ----- (2)
$\left(1\right)-\left(2\right)$, gives
$S-x$ $=$ $\frac{2}{2^2}+\frac{2}{4^2}+....upto\infty$
$=\frac{2}{2^2}(1+\frac{1}{2^2}+\frac{1}{3^2}+....upto\infty )$
$=\frac{1}{2}\left(\frac{{\pi }^2}{6}\right)$
$\Rightarrow x=\frac{{\pi }^2}{6}-\frac{{\pi }^2}{12}$
$\therefore x=\frac{{\pi }^2}{12}$
Hence, option A.