<s>[INST] A Series for $\pi$ 
Question: Can we show that $$\sum_{n=0}^\infty(-1)^{n+1}\frac{(2n-3)!!}{(2n+3)!!}=\frac{\pi}{8} $$ ?


According to wolfram alpha this result is true.
Just amateur curiosity, not sure of a starting point to show if this true.
 [/INST] Note that after cancellation we have
$$\frac{(2n-3)!!}{(2n+3)!!}=\frac{1}{(2n-1)(2n+1)(2n+3)}$$
Taking partial fractions gives
$$
\frac{1}{(2n-1)(2n+1)(2n+3)}=\frac{1}{8}\left(-\frac{1}{2n-1}+\frac{2}{2n+1}-\frac{1}{2n+3}\right)
$$
Now, the standard arctangent series for $\pi$ gives $\sum_{n=0}^\infty (-1)^n \frac{1}{2n+1}=\frac{\pi}{4}$. So 
$$\sum_{n=0}^\infty (-1)^n \frac{1}{2n-1}=-\sum_{n=-1}^\infty (-1)^n \frac{1}{2n+1}=-1-\frac{\pi}{4}$$
$$
\sum_{n=0}^\infty (-1)^n \frac{1}{2n-3}=-\sum_{n=1}^\infty (-1)^n \frac{1}{2n+1}=1-\frac{\pi}{4}
$$
which means that when we combine all three infinite sums we get
$$
\frac{1}{8}\left[-\left(-1-\frac{\pi}{4}\right)+2\left(\frac{\pi}{4}\right)-\left(1-\frac{\pi}{4}\right)\right]=\frac{\pi}{8}
$$
as desired.
(This solution makes it obvious that the sum should be of the form $a+b\pi$ with $a$ and $b$ rational, but makes the cancellation of the rational term look somewhat coincidental. I would be curious to know if there's a different way of doing it which more clearly demonstrates why the sum ends up being a rational multiple of $\pi$.)
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