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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.

user42819
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  • 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? – davidlowryduda Sep 21 '17 at 14:33
  • I checked that it is convergent and want to evaluate the infinite sum. – user42819 Sep 21 '17 at 14:35
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    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). – achille hui Sep 21 '17 at 14:58

3 Answers3

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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 $$\begin{align}\sum_{r=1}^\infty \frac{(2r)!!}{(2r+3)!!} &= \frac12\sum_{r=1}^\infty \int_0^1 t^r\sqrt{1-t} dt = \frac12 \int_0^1 \left(\sum_{r=1}^\infty t^r\right)\sqrt{1-t} dt\\ &= \frac12 \int_0^1 \frac{t}{\sqrt{1-t}}dt = \frac23 \end{align} $$

achille hui
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  • Yes the sum exactly converges into 2/3, but I wonder if there's an easier way to solve it. – user42819 Sep 21 '17 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). – achille hui Sep 21 '17 at 15:34
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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.

Sungjin Kim
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0

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}

Leucippus
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