<s>[INST] Infinite series of alternating reciprocals $\frac 1{1\cdot3}-\frac 1{3\cdot 5}+\frac 1{5\cdot 7}-\cdots $ 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.
 [/INST] Combine terms to get 
$\sum \frac{(-1)^n}{(2n+1)(2n+3)} = \frac{1}{1 \cdot 3} - \frac{1}{3 \cdot 5} + \dots = \frac{4}{1 \cdot 3 \cdot 5} + \frac{4}{5 \cdot 7 \cdot 9} + \dots = \sum \frac{4}{(4n+1)(4n+3)(4n+5)}$ 
Partial fractions (and some questionable rearrangement) gives:
$\sum \frac{4}{(4n+1)(4n+3)(4n+5)} = \sum \frac{1}{2(4n+1)} + \frac{1}{2(4n+5)} - \frac{1}{4n+3} $
$= \frac{1}{2}(1 + \frac{1}{5} + \frac{1}{9} + \dots) + \frac{1}{2}(\frac{1}{5} + \frac{1}{9} + \dots) - (\frac{1}{3} + \frac{1}{7} + \dots)$
$=\frac{1}{2} + (-\frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots) = \frac{1}{2} + (\arctan(1) - 1 )=\frac{1}{2} + (\frac{\pi}{4} - 1)  = \frac{\pi}{4} - \frac{1}{2}$
Rigor is a little suspect, but hey, it works! :)
</s>