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

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!

• What is the original homework question? I find it very odd that a high school problem can be this sophisticated. It would be really nice if you could show how you solved the homework problem "in an other way". – user593746 Oct 7 '18 at 14:47
• @zvi It seems, based on the way things are phrased, that the OP's homework did not ask to solve the series. I'm guessing that the school question was whether the series converged, which piqued the user's curiosity about what it actually converges to. – Cheerful Parsnip Oct 7 '18 at 14:53
• It is just a sequences problem and I got an=n(n+1)!/(2n+1)!!. My goal is to prove that an is convergence. This is actually not too difficult without knowing the limit. – M_Maxwell Oct 7 '18 at 14:57

## 2 Answers

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 $$0. Observe that $$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. 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 $$g\left(\frac12\right)=2+\frac{\pi}{2}.$$

• Isn't this too hard to understand for high School level. – Archis Welankar Oct 7 '18 at 14:48
• As I noted, I do not know any elementary method that will disentangle this sum nicely. And I would like to have more information about how a high school student gets to solve this kind of problems. – user593746 Oct 7 '18 at 14:49

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.