Chapter 10.6, Problem 67E

Alternating p-series Given that ${\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\frac{1}{k^2}}=\frac{{\pi }^2}{6}$, show that ${\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\frac{{(-1)}^{k+1}}{k^2}}=\frac{{\pi }^2}{12}.$ (Assume the result of Exercise 63.)