Prove that the sum of series $\sum_{k=0}^{\infty}\frac{k!}{(2k+2)!!}$ is equal to $\log 2$

Question

Prove that the sum of series $\sum_{k=0}^{\infty}\frac{k!}{(2k+2)!!}$ is equal to $\log 2$-

I tried to change the double factorial to single but I failed to do this. Thanks for any help.

Answer 1

If $n$ is even then $n!!=2^{n/2}\cdot (n/2)!$. Hence $$\sum_{k=0}^{\infty}\frac{k!}{(2k+2)!!}=\sum_{k=0}^{\infty}\frac{k!}{2^{k+1}(k+1)!}=\sum_{k=0}^{\infty}\frac{1}{2^{k+1}(k+1)}=\sum_{k=1}^{\infty}\frac{(1/2)^k}{k}=-\log(1-1/2)=\log(2).$$

Answer 2

Another way is observing that $$S=\sum_{k\geq1}\frac{\left(k-1\right)!}{\left(2k\right)!!}=\sum_{k\geq1}\frac{1}{\left(2k\right)!!}\int_{0}^{\infty}x^{k-1}e^{-x}dx=\int_{0}^{\infty}\sum_{k\geq1}\frac{x^{k-1}}{\left(2k\right)!!}e^{-x}dx $$ but the power series inside the integral has a well known closed form $$\sum_{k\geq0}\frac{x^{k}}{\left(2k\right)!!}=e^{x/2} $$ so $$S=\int_{0}^{\infty}\frac{1}{x}\left(e^{x/2}-1\right)e^{-x}dx=\int_{0}^{\infty}\frac{e^{-x/2}-e^{-x}}{x}dx $$ and now we can use the Frullani's theorem and get $$S=\color{red}{\log\left(2\right)}$$ as wanted.

Answer 3

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$\ds{\pars{2k + 2}!! = \pars{2k + 2}\pars{2k}\pars{2k - 2}\ldots 2 = 2^{k + 1}\pars{k + 1}k \pars{k - 1}\ldots 1 = 2^{k + 1}\pars{k + 1}!}$

\begin{align} \color{#f00}{\sum_{k = 0}^{\infty}{k! \over \pars{2k + 2}!!}} & = \sum_{k = 0}^{\infty}{1 \over 2^{k + 1}\pars{k + 1}} = \sum_{k = 0}^{\infty}{1 \over 2^{k + 1}}\int_{0}^{1}x^{k}\,\dd x = \half\int_{0}^{1}\sum_{k = 0}^{\infty}\pars{x \over 2}^{k}\,\dd x \\[5mm] & = \half\int_{0}^{1}{\dd x \over 1 - x/2} = \color{#f00}{\ln\pars{2}} \end{align}