How to show that $\sum _{i=1}^{\infty } \frac{i (i+1)!}{(2 i+1)\text{!!}}=2+\frac{\pi }{2}$?

Question

When I was working on my high school homework, I found a series, which is $$ \sum _{i=1}^{\infty } \frac{i (i+1)!}{(2 i+1)\text{!!}} $$ It is obviously out of my knowledge. Though I solved my homework in an other way, I want to figure out the answer of this series. So, I used mathematica and I get the following answer $$ \sum _{i=1}^{\infty } \frac{i (i+1)!}{(2 i+1)\text{!!}}=2+\frac{\pi }{2}. $$ Knowing the answer of this series, I still couldn't know why and did not have any thought. But I really want to see how to proof this series. Can anyone help me. Thanks!

Answer 1

I do not know how to give a simple solution without using hypergeometric functions. First, note that $$f(x):={_2F_1}\left(1,1;\frac12;x\right)=\sum_{k=0}^\infty x^k\prod_{j=1}^k\left(\frac{j}{j-\frac12}\right)$$ has a closed form given by $$f(x)=\frac{1}{1-x}+\frac{\sqrt{x}\arcsin(\sqrt{x})}{(1-x)^{3/2}}$$ for $0\leq x<1$.

Let $g(x)$ denote $$x^2\frac{d}{dx}\left(\frac{f(x)-1}{x}\right)=\frac{x(1+2x)}{2(1-x)^2}-\frac{x^{1/2}(1-4x)\arcsin(\sqrt{x})}{2(1-x)^{5/2}}\tag{1}$$ for $0<x<1$. Observe that $$g(x)=\sum_{k=2}^\infty (k-1)x^k\prod_{j=1}^k\left(\frac{j}{j-\frac{1}{2}}\right)=\sum_{k=2}^\infty\frac{(k-1)\cdot k!}{(2k-1)!!}(2x)^k$$ for $0<x<1$. That is, $$g\left(\frac{1}{2}\right)=\sum_{k=2}^\infty\,\frac{(k-1)\cdot k!}{(2k-1)!!}=\sum_{i=1}^\infty\,\frac{i\cdot(i+1)!}{(2i+1)!}.$$ But we have from (1) that $$g\left(\frac12\right)=2+\frac{\pi}{2}.$$

Answer 2

We obtain \begin{align*} \color{blue}{\sum_{i=1}^\infty}&\color{blue}{\frac{i(i+1)!}{(2i+1)!!}}\\ &=\sum_{i=1}^\infty\frac{i(i+1)!(2i)!!}{(2i+1)!}\tag{1}\\ &=\sum_{i=1}^\infty i2^i\binom{2i+1}{i}^{-1}\tag{2}\\ &=\sum_{i=1}^\infty i2^i(2i+2)\int_0^1x^i(1-x)^{i+1}\,dx\tag{3}\\ &=2\int_{0}^{1}(1-x)\sum_{i=1}^\infty i(i+1)(2x(1-x))^i\,dx\tag{4}\\ &=2\int_{0}^{1}(1-x)\frac{4x(1-x)}{(2x^2-2x+1)^3}\,dx\qquad\quad |x(1-x)|<\frac{1}{2}\tag{5}\\ &=8\int_{0}^{1}\frac{x(1-x)^2}{(2x^2-2x+1)^3}\,dx\tag{6}\\ &=\left.\frac{4x^3-10x^2+10x-3}{2(2x^2-2x+1)}\right|_0^1-\left.\arctan(1-2x)\right|_0^1\tag{7}\\ &=\left(\frac{1}{2}+\frac{3}{2}\right)-(\arctan(-1)-\arctan(1))\\ &=2-\left(-\frac{\pi}{4}-\frac{\pi}{4}\right)\\ &\,\,\color{blue}{=2+\frac{\pi}{2}} \end{align*}

Comment:

• In (1) we use $(2i+1)!=(2i+1)!!(2i)!!$.

• In (2) we apply $(2i)!!=2^i i!$ and write the expression using binomial coefficients.

• In (3) we apply the beta function identity $$ \binom{n}{r}^{-1}=(n+1)\int_0^1x^r(1-x)^{n-r}\,dx $$

• In (4) we do a rearrangement.

• In (5) we calculate \begin{align*} f(z)&=\sum_{i=1}^\infty i(i+1)z^i\\ &=z^2\sum_{i=1}^\infty i(i-1)z^{i-2}+2z\sum_{i=1}^\infty iz^{i-1}\\ &=z^2\frac{d^2}{dz^2}\left(\frac{1}{1-z}\right)+2z\frac{d}{dz}\left(\frac{1}{1-z}\right)\qquad\qquad |z|<1\\ &=\frac{2z^2}{(1-z)^3}+\frac{2z}{(1-z)^2}\\ &=\frac{2z}{(1-z)^3} \end{align*} and take $f(2x(1-x))$ which makes the given region of convergence $|x(1-x)|<\frac{1}{2}$ reasonable when compared with $|z|<1$.

• In (6) we collect terms.

• In (7) we use the help of Wolfram Alpha, since partial fraction decomposition and using some standard techniques to solve the integral looks rather cumbersome.