Fourier Series and Summation $\sum_{n=1}^\infty \frac{1}{n^2}$ can be computed in straight-forward way by computing the Fourier co-efficients of $f(x)=x$ and applying Parseval's identity. Likewise, $\sum_{n=1}^\infty \frac{1}{n^4}$ can be computed in the same way using $f(x)=x^2$. Any thoughts on a suitable function for computing $\sum_{n=1}^\infty \frac{(-1)^k}{n^2}$? If not, how might one approach this problem?
Thanks.

The series that come out directly from Parseval's identity always have nonnegative terms. So, some reshuffling is necessary to deal with alternating series. Which is what Daniel Fischer's comment delivers:
$$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} = \sum_{n=1}^\infty \frac{1}{n^s} - 2 \sum_{n=1}^\infty \frac{1}{(2n)^s} \tag{1}$$
As an aside:  rewriting (1) as
$$
\sum_{n=1}^\infty \frac{1}{n^s} = (1-2^{1-s})^{-1}
\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}  \tag{2}$$
is a way to quickly extend   Riemann's $\zeta$-function into the halfplane $ \operatorname{Re}s>0$. From (2) one sees that $\zeta$ is holomorphic there except for a pole at $s=1$. Unfortunately, it's not as easy to see when the series on the right sums to zero...
