How to sum this series for $\pi/2$ directly?

Question

The sum of the series $$ \frac{\pi}{2}=\sum_{k=0}^\infty\frac{k!}{(2k+1)!!}\tag{1} $$ can be derived by accelerating the Gregory Series $$ \frac{\pi}{4}=\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\tag{2} $$ using Euler's Series Transformation. Mathematica is able to sum $(1)$, so I assume there must be some method to sum the series in $(1)$ directly; what might that method be?

Answer 1

First, $$(2k+1)!! = (2k+1)(2k-1) \cdots (1) = \frac{(2k+1)!}{(2k)(2(k-1)) \cdots 2(1)} = \frac{(2k+1)!}{2^k k!}.$$

So your sum can be rewritten as

$$\sum_{k=0}^\infty\frac{k! \, k! \, 2^k }{(2k+1)!} = \sum_{k=0}^\infty\frac{2^k}{(2k+1)\binom{2k}{k}}.$$

Variations of the sum of reciprocals of the central binomial coefficients have been well-studied. For example, this paper by Sprugnoli (see Theorem 2.4) gives the ordinary generating function of $a_k = \frac{4^k}{(2k+1)}\binom{2k}{k}^{-1}$ to be $$A(t) = \frac{1}{t} \sqrt{\frac{t}{1-t}} \arctan \sqrt{\frac{t}{1-t}}.$$

Subbing in $t = 1/2$ says that $$\sum_{k=0}^\infty\frac{2^k}{(2k+1)\binom{2k}{k}} = 2 \arctan(1) = \frac{\pi}{2}.$$

Answer 2

We can prove this identity, as well as the corresponding power series identities by using a relation with the Beta function. Rearranging as done in Mike Spivey's answer we are looking at $$ \sum_{k=0}^\infty\frac{k! k! 2^k}{(2k+1)!}$$ Using induction or a Beta Function identity, we can show that $$\int_0^1 x^{k}(1-x)^k=\frac{k!k!}{(2k+1)!}.$$ Hence your sum becomes

$$ \sum_{k=0}^\infty 2^k \int_0^1 x^{k}(1-x)^k=\int_0^1 \left(\sum_{k=0}^\infty 2^k x^k (1-x)^k\right)dx.$$

Notice that since $0\leq x\leq 1$, $x(1-x)\leq \frac{1}{4}$ and the series converges absolutely. Summing gives

$$=\int_0^1 \frac{1}{1-2x(1-x)}dx=\int_0^1 \frac{1}{x^2+(1-x)^2}dx$$ Substituting $u=\frac{1}{x}$, and then $v=u-1$, we see that this integral is equal to $$\int_1^\infty \frac{1}{1+(u-1)^2}du=\int_0^\infty \frac{1}{1+v^2}dv=\frac{\pi}{2},$$ as desired.

Answer 3

I had intended for this to be a comment to Mike Spivey's answer, but it is too long.

One of the answers to the related question mentions a result equivalent to $$ \int_0^\frac{\pi}{2}\sin^{2k+1}(x)\;\mathrm{d}x=\frac{2k}{2k+1}\frac{2k-2}{2k-1}\cdots\frac{2}{3}=\frac{1}{2k+1}\frac{4^k}{\binom{2k}{k}}\tag{1} $$ Using $(1)$, my sum becomes $$ \begin{align} \sum_{k=0}^\infty\frac{k!}{(2k+1)!!} &=\sum_{k=0}^\infty\frac{2^k}{(2k+1)\binom{2k}{k}}\\ &=\sum_{k=0}^\infty\int_0^\frac{\pi}{2}\sqrt{2}\left(\frac{\sin(x)}{\sqrt{2}}\right)^{2k+1}\mathrm{d}x\\ &=\sqrt{2}\int_0^\frac{\pi}{2}\frac{\left(\frac{\sin(x)}{\sqrt{2}}\right)}{1-\left(\frac{\sin(x)}{\sqrt{2}}\right)^2}\;\mathrm{d}x\\ &=\int_0^\frac{\pi}{2}\frac{2\,\sin(x)}{2-\sin^2(x)}\;\mathrm{d}x\\ &=\int_\frac{\pi}{2}^0\frac{2\;\mathrm{d}\cos(x)}{1+\cos^2(x)}\\ &=\frac{\pi}{2} \end{align} $$

Answer 4

Notice that for $c_k = \frac{k!}{(2k+1)!!}$ the ratio of successive terms $\frac{c_{k+1}}{c_k} = \frac{k+1}{2k +3} = \frac{1}{2} \frac{k+1}{k+3/2}$.

This means that the series is hypergeometric with the value ${}_2 F_1(1, 1, \frac{3}{2}, \frac{1}{2})$.

This particular Gaussian hypergeometric is elementary: $$ {}_2 F_1(1, 1, \frac{3}{2}, x) = \frac{\arcsin\left(\sqrt{x}\right)}{\sqrt{1-x} \sqrt{x}} $$ Upon substitution of $x=\frac{1}{2}$ we recover the result $ 2 \arcsin(\frac{1}{\sqrt{2}}) = \frac{\pi}{2}$.

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\sum_{k = 0}^{\infty}{k! \over \pars{2k + 1}!!}} = \sum_{k = 0}^{\infty}{k! \over \prod_{j = 0}^{k}\pars{2j + 1}} \\[5mm] = &\ \sum_{k = 0}^{\infty}{k! \over 2^{k + 1} \prod_{j = 0}^{k}\pars{j + 1/2}} = {1 \over 2}\sum_{k = 0}^{\infty}{k! \over \pars{1/2}^{\overline{k + 1}}}\,\pars{1 \over 2}^{k} \\[5mm] = &\ {1 \over 2}\sum_{k = 0}^{\infty}{k! \over \Gamma\pars{k + 3/2}/\Gamma\pars{1/2}}\,\pars{1 \over 2}^{k} \\[5mm] = &\ {1 \over 2}\sum_{k = 0}^{\infty}{\Gamma\pars{1/2}\Gamma\pars{k + 1} \over \Gamma\pars{k + 3/2}}\,\pars{1 \over 2}^{k} \\[5mm] = &\ {1 \over 2}\sum_{k = 0}^{\infty}\pars{1 \over 2}^{k} \int_{0}^{1}t^{-1/2}\pars{1 - t}^{k}\,\dd t \\[5mm] = &\ {1 \over 2}\int_{0}^{1}t^{-1/2}\sum_{k = 0}^{\infty} \pars{1 - t \over 2}^{k}\,\dd t \\[5mm] = &\ {1 \over 2}\int_{0}^{1}t^{-1/2}\, {1 \over 1 - \pars{1 - t}/2}\,\dd t = \int_{0}^{1}{t^{-1/2} \over 1 + t}\,\dd t \\[5mm] \stackrel{t\ \mapsto\ t^{2}}{=}\,\,\,& 2\int_{0}^{1}{\dd t \over 1 + t^{2}}\,\dd t = 2\,{\pi \over 4} = \bbx{\pi \over 2} \\ & \end{align}