Why is $\sum_{k=0}^{\infty} \frac{k!}{\prod_{j=0}^{k} \left(2j+3\right)} = 2-\frac{\pi}{2}$ and how is this solution derived?

Question

I recently came across the problem:

$$\sum_{k=0}^{\infty} \frac{k!}{\prod_{j=0}^{k} \left(2j+3\right)}$$ and decided to try finding its solution. I started off by writing a program that gave me the answer $0.4292036732051...$ which I found probably meant the solution was $2-\frac{\pi}{2}$, but I wanted to see if I could prove it. I started by manipulating the denominator into a double factorial resulting in: $$\sum_{k=0}^{\infty} \frac{k!}{\left(2k+3\right)!!}$$ Then, I thought things would be easier to work with if I only had regular factorials. This gave me: $$\sum_{k=0}^{\infty} \frac{4\cdot2^k k! \left(k+2\right)!}{\left(2k+4\right)!}$$ I unfortunately had little clue to go from here considering I haven't dealt with factorials often, much less in infinite series such as this one. I am curious how it is possible to go forth from here and also how to solve infinite series that are similarly structured. Any help would be appreciated.

Answer 1

We obtain \begin{align*} \sum_{k=0}^\infty\frac{k!}{(2k+3)!!}&=\sum_{k=0}^\infty\frac{k!(2k+2)!!}{(2k+3)!}\\ &=\sum_{k=0}^\infty\frac{k!2^{k+1}(k+1)!}{(2k+3)!}\\ &=\sum_{k=0}^\infty\binom{2k}{k}^{-1}\frac{2^k}{(2k+1)(2k+3)}\\ &=\sum_{k=0}^\infty\binom{2k}{k}^{-1}\frac{2^{k-1}}{2k+1}-\sum_{k=0}^\infty\binom{2k}{k}^{-1}\frac{2^{k-1}}{2k+3}\tag{1} \end{align*}

We use a representation of reciprocal binomial coefficients via the Beta function:

\begin{align*} \binom{n}{k}^{-1}=(n+1)\int_0^1z^k(1-z)^{n-k}\,dz\tag{2} \end{align*}

and the left-hand series of (1) can be calculated as

\begin{align*} \color{blue}{\sum_{k=0}^\infty\binom{2k}{k}^{-1}\frac{2^{k-1}}{2k+1}} &=\sum_{k=0}^\infty 2^{k-1}\int_0^1z^k(1-z)^k\,dz\tag{3}\\ &=\frac{1}{2}\int_{0}^{1}\sum_{k=0}^\infty \left(2z(1-z)\right)^k\,dz\\ &=\frac{1}{2}\int_{0}^1\frac{dz}{1-2z(1-z)}\tag{4}\\ &=\frac{1}{2}\int_{0}^{1}\frac{dz}{z^2+(1-z)^2}\\ &=\frac{1}{2}\int_{0}^{\infty}\frac{du}{1+u^2}\tag{5}\\ &\,\,\color{blue}{=\frac{\pi}{4}}\tag{6} \end{align*}

Comment:

• In (3) we use the identity (2).

• In (4) we apply the geometric series expansion.

• In (5) we use the substitution $u=\frac{1-z}{z}, du=-\frac{1}{z^2}dz$.

We also want to apply (2) to the right-hand series of (1). To do this conveniently we need some preparatory work: \begin{align*} \sum_{k=0}^\infty&\binom{2k}{k}^{-1}\frac{2^{k-1}}{2k+3}\\ &=\sum_{k=0}^\infty\frac{k!k!}{(2k)!}\cdot\frac{2^{k-1}}{2k+3}\\ &=\sum_{k=0}^\infty\frac{k!(k+1)!(2k+1)}{(2k+1)!(k+1)}\cdot\frac{2^{k-1}}{2k+3}\\ &=\sum_{k=0}^\infty\frac{(k+1)!(k+1)!}{(2k+2)!}\cdot\frac{2^{k+1}}{2k+3}-\sum_{k=0}^\infty\frac{k!(k+1)!}{(2k+1)!(k+1)}\cdot\frac{2^{k-1}}{2k+3}\\ &=\sum_{k=0}^\infty\binom{2k+2}{k+1}^{-1}\frac{2^{k+1}}{2k+3}-\sum_{k=0}^\infty\binom{2k}{k}^{-1}\frac{2^{k-1}}{(2k+1)(2k+3)}\\ &=\sum_{k=0}^\infty\binom{2k+2}{k+1}^{-1}\frac{2^{k+1}}{2k+3} -\sum_{k=0}^\infty\binom{2k}{k}^{-1}\frac{2^{k-2}}{2k+1} +\sum_{k=0}^\infty\binom{2k}{k}^{-1}\frac{2^{k-2}}{2k+3}\tag{7}\\ \end{align*} In the last line (7) we use a partial fraction decomposition as we did in (1).

We are now well prepared to do the calculation. We obtain together with (6):

\begin{align*} \color{blue}{\sum_{k=0}^\infty \binom{2k}{k}^{-1}\frac{2^{k-2}}{2k+3}} &=\sum_{k=0}^\infty\binom{2k+2}{k+1}^{-1}\frac{2^{k+1}}{2k+3}-\frac{\pi}{8}\\ &=\sum_{k=0}^\infty2^{k+1}\int_{0}^1z^{k+1}(1-z)^{k+1}\,dz-\frac{\pi}{8}\\ &=\sum_{k=1}^\infty2^k\int_{0}^1z^k(1-z)^k\,dz-\frac{\pi}{8}\\ &=\frac{\pi}{2}-2^0\int_{0}^1\,dz-\frac{\pi}{8}\\ &\,\,\color{blue}{=\frac{3}{8}\pi-1}\tag{8} \end{align*}

We finally conclude from (1) together with (6) and (8) \begin{align*} \color{blue}{\sum_{k=0}^\infty\frac{k!}{(2k+3)!!}} &=\sum_{k=0}^\infty\binom{2k}{k}^{-1}\frac{2^{k-1}}{2k+1}-\sum_{k=0}^\infty\binom{2k}{k}^{-1}\frac{2^{k-1}}{2k+3}\\ &=\frac{\pi}{4}-2\left(\frac{3}{8}\pi-1\right)\\ &\,\,\color{blue}{=2-\frac{\pi}{2}} \end{align*}

and the claim follows.

Answer 2

Starting from $$4\sum_{k=0}^{\infty} \frac{2^k k! \left(k+2\right)!}{\left(2k+4\right)!} $$Consider $$4\sum_{k=0}^{\infty}\frac{k! (k+2)!}{ (2 k+4)!}(2t)^{2k}$$ and, now, the trick is to recognize (not so obvious) that this is $$\frac{1}{t^2}-\frac{\sqrt{1-t^2} }{t^3}\sin ^{-1}(t)$$ Make $t=\frac 1 {\sqrt 2}$ and get the result.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{k = 0}^{\infty}{k! \over \prod_{j = 0}^{k}\pars{2j + 3}} & = \sum_{k = 0}^{\infty}{k! \over 2^{k + 1}\prod_{j = 0}^{k}\pars{j + 3/2}} = \sum_{k = 0}^{\infty}{k! \over 2^{k + 1}\pars{3/2}^{\overline{k + 1}}} \\[5mm] & = \sum_{k = 0}^{\infty}{1 \over 2^{k + 1}}\,{k! \over \Gamma\pars{3/2 + k + 1}/\Gamma\pars{3/2}} \\[5mm] & = \sum_{k = 0}^{\infty}\,{1 \over 2^{k + 1}}\, {\Gamma\pars{k + 1}\Gamma\pars{3/2} \over \Gamma\pars{k + 5/2}} \\[5mm] & = \sum_{k = 0}^{\infty}{1 \over 2^{k + 1}}\, \int_{0}^{1}t^{k}\pars{1 - t}^{1/2}\,\dd t \\[5mm] & = {1 \over 2}\int_{0}^{1}\root{1 - t} \sum_{k = 0}^{\infty}\pars{t \over 2}^{k}\,\dd t \\[5mm] & = \int_{0}^{1}{\root{1 - t} \over 2 - t}\,\dd t \,\,\,\stackrel{t\ =\ 1 - x^{2}}{=}\,\,\, 2\int_{0}^{1}\pars{1 - {1 \over 1 + x^{2}}}\,\dd x \\[5mm] & = \bbx{2 - {\pi \over 2}}\ \approx\ 0.4292 \end{align}