Relationship between $\zeta(2)$ and $\sum \frac{(-1)^{n-1}}{n^2}$

Question

I saw that $$\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}=\frac{1}{2}\zeta(2),$$ but I cannot deduce this.

Answer 1

$$S=\sum_{n=1}^\infty\dfrac{(-1)^{n-1}}{n^2}=\sum_{n=1}^\infty\dfrac{1}{(2n-1)^2}-\sum_{n=1}^\infty\dfrac{1}{4n^2}\\ \zeta(2)=\sum_{n=1}^\infty\dfrac{1}{(2n-1)^2}+\sum_{n=1}^\infty\dfrac{1}{4n^2}\\ \implies\zeta(2)-S=2\sum_{n=1}^\infty\dfrac{1}{4n^2}=\dfrac{\zeta(2)}{2}\\ \implies \boxed{S=\dfrac{\zeta(2)}{2}}$$

Answer 2

\begin{align} &\frac1{1^2}-\frac1{2^2}+\frac1{3^2}-\frac1{4^2}+\frac1{5^2}-\frac1{6^2}+\cdots\\ &=\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\frac1{4^2}+\frac1{5^2}+\frac1{6^2}+\cdots\\ &-\left(\frac2{2^2}+\frac2{4^2}+\frac2{6^2}+\frac2{8^2}+\frac2{10^2}+\frac2{12^2}+\cdots\right)\\ &=\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\frac1{4^2}+\frac1{5^2}+\frac1{6^2}+\cdots\\ &-\frac12\left(\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\frac1{4^2}+\frac1{5^2}+ \frac1{6^2}+\cdots\right). \end{align}