Question

If $a={\sum }_{r=1}^{\infty }\frac{1}{r^2},b=\sum _{r=1}^{\infty }\frac{1}{(2r-1{)}^2}$, then $\frac{3a}{b}$ is equal to ...................

Answer 1

$a=\sum _{r=1}^{\infty }\frac{1}{r^2}=\frac{1}{1}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...\infty =\frac{1}{1}+\frac{1}{3^2}+\frac{1}{5^2}+....\infty +(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+....\infty )$
$\Rightarrow a=\sum _{r=1}^{\infty }\frac{1}{(2r-1{)}^2}+\frac{1}{2^2}(\frac{1}{1}+\frac{1}{2^2}+\frac{1}{3^2}+...\infty )=b+\frac{a}{4}$
Therefore, $\frac{3a}{b}=4$