How to find $ \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2}$? Let $f$ be a $2\pi$-periodic function whose restriction on $[-\pi, \pi]$ is $f(x)_{[-\pi, \pi]} = |x|$
It is easy to see that its fourier series converges uniformily to $f$ and is $$f(x) = \frac \pi2 - \frac 4\pi \sum_{m=0}^\infty \frac{\cos(2m+1)x}{(2m+1)^2}$$
Next it asks me three questions:

Find $\sum_{m=0}^\infty \frac 1{(2m+1)^2}$

This is easy, just set $x=0$ to get at once $\sum_{m=0}^\infty \frac 1{(2m+1)^2} = \frac {\pi^2} 8$

Find $\sum_{m=0}^\infty \frac{1}{(2m+1)^4}$

Again, very easy. Using the parseval identity one gets $\sum_{m=0}^\infty \frac{1}{(2m+1)^4} = \frac{\pi^4}{96}$

Finally, find $$\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2}$$

This is where I am stuck. I thought about calculating the series in a suitable point $x$, but I can't get the cosine to behave like $(-1)^n$.
I tried writing it like $$\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2} = \sum_{k=0}^\infty \frac 1{(2k+1)^2} - 2\sum_{k=0}^\infty \frac 1{(4k+3)^2}$$ but I don't know how to calculate the latter sum.
Hints are appreciated! :)

The value of the series is known as Catalan's constant. It's value is approximately 0.915 965 594. At the moment there is no known method to evaluate the series in closed form.
