Alternating sum of reciprocals of odd squares I have been struggling with this problem for a while.
May I ask what is the exact value of this series, if it’s even possible.
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2}$$
A proof of some sort would highly be appreciated.  Thank you.
Edit:
I found this by trying to evaluate the integral:
$$\int_{0}^{\infty} \frac{\ln^2(x)}{1+x^2}dx$$
What I did was I made the above integral into $u = 1/x$:
$$2\int_{0}^{1} \frac{\ln^2(x)}{1+x^2}dx$$
Then I expanded it into a geometric series and took antiderivatives, leading me to the series above.
 A: Your original question was regarding the value of the following sum:
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2}$$
The value of this sum is (by definition) given by Catalan's constant $G\approx 0.915965594$. As of today, it is not known whether $G$ is irrational, nor transcendental.

However, I see that you have updated your question regarding where this sum came from, and it seems like you have not obtained the correct sum corresponding to the value of the integral. You should have:
$$\begin{align}I=\int_0^{\infty} \frac{\ln^2(x)}{1+x^2}~dx&=2\int_0^{1} \frac{\ln^2(x)}{1+x^2}~dx\\&=2\sum_{n=0}^{\infty}(-1)^n\int_0^1 \ln^2(x)x^{2n}~dx \end{align}$$
The change in order of summation and integration can be justified by Fubini's theorem. Using integration by parts twice, we have that:
$$\int_0^1 \ln^2(x)x^{2n}~dx=-\int_0^1 \frac{x^{2n+1}}{2n+1}\cdot \frac{2\log(x)}{x}~dx=\int_0^1 \frac{x^{2n+1}}{(2n+1)^2}\cdot \frac{2}{x}~dx=\frac{2}{(2n+1)^3}$$
Thus, we obtain:
$$I=4\cdot \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^\color{red}{3}}=\frac{\pi^3}{8}$$
This sum can be evaluated using multiple methods, see here.
