Alternating series of zeta(2)

Calculus Level 3

$\sum _{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{2}} = \frac{\pi^{2}}{A}$

find the value of $A$ .

1 solution

Jake Lai
Apr 4, 2015

$\eta(s) = (1-2^{1-s})\zeta(s)$

Since our sum is $\eta(2)$ ,

$\eta(2) = (1-2^{1-2})\zeta(2) = \frac{1}{2} \frac{\pi^{2}}{6} = \frac{\pi^{2}}{12}$

Hence, $A = \boxed{12}$ .