# Summation involving double factorials

How can we evaluate exact infinite sum of the following? $${ \sum_{r=1}^{\infty}\frac{(2r)!!}{(2r+3)!!}}$$ where $k!! = \begin{cases} 2\cdot4\cdot6 . . . k, & \text{if$k$is even} \\ 1\cdot3\cdot5 . . . k, & \text{if$k$is odd} \end{cases}$

What approach do we need to solve such type of summations?

My attempt: \begin{align} \sum_{r=1}^{\infty}\frac{(2r)!!}{(2r+3)!!} &= \sum_{r=1}^{\infty}\frac{(2r)!!\cdot (2r+2)!!} {(2r+3)!!(2r+2)!!} \\ &= \sum_{r=1}^{\infty}\frac{2^r\cdot (r)!\cdot 2^{r+1}\cdot (r+1)!}{(2r+3)!} \\ &= \sum_{r=1}^{\infty}\frac{2^{2r+1}\cdot (r)!\cdot (r+1)!}{(2r+3)!} \end{align}

I also tried using the generalized binomial theorem but it doesn't seem to follow the pattern.

EDIT: I suspect it might be solved using telescopic method, but can't figure out how.

• What are you trying to do? Are you determining if the series converges? Do you want to approximate the value of the series? Do you want to try to evaluate it exactly? Sep 21, 2017 at 14:33
• I checked that it is convergent and want to evaluate the infinite sum. Sep 21, 2017 at 14:35
• If one cast this to an hypergeometric series, it become $\frac{1}{3}\left[{}_2F_1(1,1; \frac52; 1) - 1\right]$ and WA evaluate this to $\frac23$ exactly. Solving it using telescoping methods seems a good approach (not that I know how to do that). Sep 21, 2017 at 14:58

Notice

\begin{align} \frac{(2r)!!}{(2r+3)!!} &= \frac{2\cdot 4 \cdots \cdot 2r}{3\cdot 5 \cdots (2r+3)}= \frac13 \frac{\prod\limits_{s=0}^{r-1}(1+s)}{\prod\limits_{s=0}^{r-1}(\frac52+s)}\\ &= \frac13 \frac{\Gamma(r+1)/\Gamma(1)}{\Gamma(r+\frac52)/\Gamma(\frac52)} = \frac12\frac{\Gamma(r+1)\Gamma(\frac32)}{\Gamma(r+\frac52)}\\ &= \frac12 \int_0^1 t^{(r+1)-1} (1-t)^{3/2-1} dt = \frac12\int_0^1 t^r \sqrt{1-t}dt \end{align} We have \begin{align}\sum_{r=1}^\infty \frac{(2r)!!}{(2r+3)!!} &= \frac12\sum_{r=1}^\infty \int_0^1 t^r\sqrt{1-t} dt = \frac12 \int_0^1 \left(\sum_{r=1}^\infty t^r\right)\sqrt{1-t} dt\\ &= \frac12 \int_0^1 \frac{t}{\sqrt{1-t}}dt = \frac23 \end{align}

• Yes the sum exactly converges into 2/3, but I wonder if there's an easier way to solve it. Sep 21, 2017 at 15:31
• @user42819 I have looked at the partial sums, I cannot figure out a simple expression for that (if one can do that, one can prove the sum using telescoping methods). Sep 21, 2017 at 15:34

Let $A_r = \frac23 \frac45 \cdots \frac{2r}{2r+1}$. Then $$\frac{(2r)!!}{(2r+3)!!} = A_r \cdot \frac1{2r+3}.$$ Use $\frac1{2r+3} = 1 - \frac{2r+2}{2r+3}$. Then the sum telescopes. So, we are to evaluate $$\sum_{r=1}^{\infty} (A_r - A_{r+1}).$$ To complete the solution, we need the proof of $\lim_{r\rightarrow\infty} A_r = 0$. Your expressions and Stirling's formula will do the job.

Starting with

My attempt: \begin{align} S = \sum_{r=1}^{\infty}\frac{(2r)!!}{(2r+3)!!} &= \sum_{r=1}^{\infty}\frac{(2r)!!\cdot (2r+2)!!} {(2r+3)!!(2r+2)!!} \\ &= \sum_{r=1}^{\infty}\frac{2^r\cdot (r)!\cdot 2^{r+1}\cdot (r+1)!}{(2r+3)!} \\ &= \sum_{r=1}^{\infty}\frac{2^{2r+1}\cdot (r)!\cdot (r+1)!}{(2r+3)!} \end{align}

One can continue as follows: \begin{align} S &= \sum_{r=1}^{\infty}\frac{2^{2r+1}\cdot (r)!\cdot (r+1)!}{(2r+3)!} \\ &= 2 \, \sum_{r=1}^{\infty} \frac{\Gamma(r+1) \, \Gamma(r+2)}{ \Gamma(2r + 4) } \, 4^{r} \\ &= \frac{2}{3!} \, \sum_{r=1}^{\infty} \frac{(1)_{r} \, (2)_{r}}{4^{r} \, (2)_{r} \, \left(\frac{5}{2}\right)_{r} } \, 4^{r} \\ &= \frac{1}{3} \, \sum_{r=1}^{\infty} \frac{(1)_{r} \, (1)_{r}}{r! \, \left(\frac{5}{2}\right)_{r} } = 2 \left[ {}_{2}F_{1}\left(1, 1; \frac{5}{2}; 1\right) -1 \right] \\ &= \frac{1}{3} \left[\frac{\Gamma(5/2) \, \Gamma(1/2)}{\Gamma^{2}(3/2)} - 1 \right] = \frac{2}{3}. \end{align}