A Series for $\pi$

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 $\pi$.)

Considering the simplicity of Micah's answer, I am ashamed to provide this complex one.

Let $$S_p(x)=\sum_{n=0}^p(-1)^{n+1}\frac{(2n-3)!!}{(2n+3)!!}x^p$$ It write $$S_p(x)=\frac{\left(8 p^3+36 p^2+46 p+15\right) \, _2F_1\left(-\frac{1}{2},1;\frac{5}{2};-x\right)+3 (-x)^{p+1} \, _2F_1\left(1,p+\frac{1}{2};p+\frac{7}{2};-x\right)}{3 \left(8 p^3+36 p^2+46 p+15\right)}$$ Using $$\, _2F_1\left(-\frac{1}{2},1;\frac{5}{2};-1\right)=\frac{3 \pi }{8}$$ then $$S_p(1)=\frac{\pi \left(8 p^3+36 p^2+46 p+15\right)-8 (-1)^p \, _2F_1\left(1,p+\frac{1}{2};p+\frac{7}{2};-1\right)}{8 \left(8 p^3+36 p^2+46 p+15\right)}$$ that it to say $$S_p(1)=\frac \pi 8-\frac{(-1)^p }{8 p^3+36 p^2+46 p+15}\, _2F_1\left(1,p+\frac{1}{2};p+\frac{7}{2};-1\right)$$ and $\, _2F_1\left(1,p+\frac{1}{2};p+\frac{7}{2};-1\right)$ looks like an hyperbolic function going asymptotically to $\frac 12$.

Then $S_\infty(1)=\frac{ \pi }{8}$.