Having misread the recent question here as $$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+3)}$$ and having solved it, I thought that I would post it as a question instead. It has a rather interesting answer!
Edit
Now that we have nice solutions from Jack D'Aurizio and Michael Biro, I'd like to point out that what struck me was the fact that
$$\begin{align} \sum_{n=0}^\infty=\frac \pi 4-\frac 12&=\sum_{n=0}^\infty (-1)^n\frac 1{2n+1}-\frac 1{2^{n+2}}\\ \color{red}{\frac 1{1\color{black}{\cdot 3}}-\frac 1{3\color{black}{\cdot 5}}+\frac 1{5\color{black}{\cdot 7}}-\frac 1{7\color{black}{\cdot 9}}+\cdots }&= \left(\color{red}{\frac 11-\frac 13+\frac15-\frac17+\cdots}\right)-\frac 12\\ &=\left(\color{red}{\frac 11-\frac 13+\frac15-\frac17+\cdots}\right)-\left(\frac 14+\frac 18+\frac 1{16}+\frac 1{32}+\cdots \right) \end{align}$$ Is it possible to reduce the LHS expansion to the RHS expansion directly without first knowing the answer? If so then this would be another solution method.