Double Factorial Summation

Question

On my calculus test there was a problem asking to determine the convergence of $$\sum_{n=1}^{\infty} (-1)^n \cdot \frac{n!}{(2n - 1)!!}.$$

I looked on WolframAlpha after the test to see the sum surprisingly had a closed form, specifically $$\frac{1}{9} \left(-3 - 2\sqrt{3} \ \sinh^{-1} \left(\frac{1}{\sqrt{2}}\right) \right).$$

I would be interested in seeing how this is derived.

Note: As usual, for positive integers $k$, $k!!$ denotes the product of the integers from $1$ to $k$ that have the same parity as $k$ (in other words, the double factorial increments down by $2.$)

Answer 1

Following @Thomas Andrews and using $\frac{(n+1)}{{n\choose k}}=\int_{0}^{1}x^k (1-x)^{n-k}\,\mathrm dx,$ we get

$$S=\sum_{k=0}^\infty (-1)^k \frac{2^k}{{2k \choose k}}=\sum_{k=0}^{\infty}(2k+1)\int_0^{1}(-2x)^k (1-x)^{k}\,\mathrm dx.$$$$\implies S=\int_{0}^{1}\sum_{k=0}^{\infty}(2k+1) (2x^2-2x)^k\,\mathrm dx.$$ Next using $\sum_{k=0}^{\infty} z^k=(1-z)^{-1}, \sum_{k=0}^{\infty} k z^k= z(1-z)^{-2}, |z|<1$, we get $$\sum_{k=0}^{\infty} (2k+1) z^k=\frac{1+z}{(1-z)^2}, z=2(x^2-x).$$ Finally, $S$ is a doable definite integral, you may take over from here.

Edit:$$S=\int_{0}^{1} \frac{1+2x^2-2x}{(1-2x^2+2x)^2} dx$$ Let $x=\sin^2 t \implies dx=\sin 2t dt$, then $$S=2\int_{0}^{\pi/2} \frac{1+\cos^2 2t}{(3-\cos^2 2t)^2} \sin 2t~ dt.$$ Let $\cos 2 t=y$, then $$S=\int_{-1}^{1} \frac{1+y^2}{(y^2-3)^2} dy=2\int_{0}^{1} \left[ \frac{1}{y^2-3}+\frac{4}{(y^2-3)^2}\right] dy.$$ Use $\int \frac{dx}{x^2-a^2}=-\frac{1}{a} \tanh^{-1}(x/a)$ and $$\int \frac{dx}{(x^2-a^2)^2}=\frac{1}{2a^3}\left[\frac{ax}{a^2-x^2}+\tanh^{-1}(x/a)\right].$$$$S=\frac{2}{3}-\frac{2}{3\sqrt{3}} \tanh^{-1}\frac{1}{\sqrt{3}}$$$$S=\frac{2}{3}-\frac{2}{3\sqrt{3}} \sinh^{-1}\frac{1}{\sqrt{2}},$$ which matches with what the Wolfram Mathematica would give. I suggest OP to check his expression obtained at Mathematica.

Answer 2

$$\frac{n!}{(2 n-1)\text{!!}}=\frac{\sqrt{\pi } \, \Gamma (n+1)}{2^n\,\Gamma\left(n+\frac{1}{2}\right)}$$ Consider $$\sum_{n=0}^\infty \frac{\sqrt{\pi } \, \Gamma (n+1)}{\Gamma\left(n+\frac{1}{2}\right)}\,x^{2n}$$ Cheating a little $$\sum_{n=0}^\infty \frac{\sqrt{\pi } \, \Gamma (n+1)}{\Gamma\left(n+\frac{1}{2}\right)}\,x^{2n}=\frac{\sqrt{1-x^2}+x \sin^{-1}(x)}{\left(1-x^2\right)^{3/2}}$$

$$x=\frac{i}{\sqrt{2}}\quad \implies \quad \sum_{n=0}^\infty (-1)^n\frac{n!}{(2 n-1)\text{!!}}=\frac{2}{3}-\frac{2 \sinh ^{-1}\left(\frac{1}{\sqrt{2}}\right)}{3 \sqrt{3}}$$ Subtract the first term which is $1$ to get the result.