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.

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