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I'm trying to show the following identity,

$$ \sum_{n=1}^{m} \prod_{k=1}^n \frac{2k-1}{2k} = \frac{2(m+1)\Gamma(m+\frac{3}{2})}{\sqrt{\pi}\Gamma(m+2)}.$$

First, simplify the product within the sum,

$$ \sum_{n=1}^{m} \frac{\prod_{k=1}^n (2k-1)}{\prod_{k=1}^n (2k)} \\ = \sum_{n=1}^{m} \frac{\frac{(2n)!}{2^n n!}}{2^n n!} \\ = \sum_{n=1}^{m} \frac{(2n)!}{2^{2n}(n!)^2} $$

I'm not sure where to continue from here. I know the factorials will be rewritten as the Gamma function and I'm guessing the $\sqrt{\pi}$ will come from Stirling's approximation. Where should I go from here?

Dando18
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3 Answers3

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{n = 1}^{m}\prod_{k = 1}^{n}{2k - 1 \over 2k} & = \sum_{n = 1}^{m}{\prod_{k = 1}^{n}\pars{k - 1/2} \over \prod_{k = 1}^{n}k} = \sum_{n = 1}^{m}{\pars{1/2}^{\overline{n}} \over n!} = \sum_{n = 1}^{m}{\Gamma\pars{1/2 + n}/\Gamma\pars{1/2} \over n!} \\[5mm] & = \sum_{n = 1}^{m}{\pars{n - 1/2}! \over n!\pars{-1/2}!} = \sum_{n = 1}^{m}{n - 1/2 \choose n} = \bracks{z^{m}}\sum_{\ell = 0}^{\infty}z^{\ell} \bracks{\sum_{n = 1}^{\infty}{n - 1/2 \choose n}\bracks{n \leq \ell}} \\[5mm] & = \bracks{z^{m}}\sum_{n = 1}^{\infty}{n - 1/2 \choose n} \sum_{\ell = n}^{\infty}z^{\ell} = \bracks{z^{m}}\sum_{n = 1}^{\infty}{n - 1/2 \choose n} \sum_{\ell = 0}^{\infty}z^{\ell + n} \\[5mm] & = \bracks{z^{m}}{1 \over 1 - z}\sum_{n = 1}^{\infty}{n - 1/2 \choose n}z^{n} = \bracks{z^{m}}{1 \over 1 - z}\sum_{n = 1}^{\infty}{-1/2 \choose n}\pars{-z}^{n} \\[5mm] & = \bracks{z^{m}}{1 \over 1 - z}\bracks{\pars{1 - z}^{-1/2} - 1} = \bracks{z^{m}}\pars{1 - z}^{-3/2} - \bracks{z^{m}}\pars{1 - z}^{-1} \\[5mm] & = {-3/2 \choose m}\pars{-1}^{m} - 1 = {m + 1/2 \choose m} - 1 = {\pars{m + 1/2}! \over m!\pars{1/2}!} - 1 \\[5mm] & = {\Gamma\pars{m + 3/2} \over \Gamma\pars{m + 1}\Gamma\pars{3/2}} - 1 = {\Gamma\pars{m + 3/2} \over \bracks{\Gamma\pars{m + 2}/\pars{m + 1}}\bracks{\Gamma\pars{1/2}/2}} - 1 \\[5mm] & = \bbx{{2\pars{m + 1}\Gamma\pars{m + 3/2} \over \root{\pi}\Gamma\pars{m + 2}} - 1} \end{align}

Felix Marin
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  • Insolito y brillante. (+1). Espero que le vaya bien en estos tiempos dificiles para su pais. – Marko Riedel May 07 '17 at 20:55
  • @MarkoRiedel Thanks. Here, the situation is quite tense. Nobody really knows what's going to happen. I'm living in a relatively peaceful place and I don't have to go to the university everyday because I'm already retired. – Felix Marin May 07 '17 at 22:13
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Observe that from the RHS we have

$$\require{cancel}\begin{align}\frac{2(m+1)\Gamma(m+3/2)}{\sqrt\pi\Gamma(m+2)}&=\frac{2\cancel{(m+1)}(m+1/2)\Gamma(m+1/2)}{\Gamma(1/2)\cancel{(m+1)}\Gamma(m+1)}\\&=\frac{2(m+1/2)^\underline {m+1}\cancel{\Gamma(1/2)}}{m!\,\cancel{\Gamma(1/2)}}\\&=\binom{m+1/2}{m}=\prod_{k=1}^m\frac{k+1/2}{k}\\&=\prod_{k=1}^m\frac{2k+1}{2k}=\frac{(2m+1)!!}{(2m)!!}\tag{1}\end{align}$$

And from the LHS we have

$$\sum_{n=1}^{m} \frac{(2n)!}{2^{2n}(n!)^2}=\sum_{n=1}^{m}\frac{(2n)!}{((2n)!!)^2}=\sum_{n=1}^{m}\frac{(2n-1)!!}{(2n)!!}\tag{2}$$

To show that $(1)$ is equivalent to $(2)$ we can do this

$$\Delta\left(\frac{(2n+1)!!}{(2n)!!}\right)=\frac{(2n+3)!!}{(2n+2)!!}-\frac{(2n+1)!!}{(2n)!!}=\left(\frac{2n+3}{2n+2}-1\right)\cdot\frac{(2n+1)!!}{(2n)!!}=\\=\frac1{2n+2}\cdot\frac{(2n+1)!!}{(2n)!!}=\frac{(2n+1)!!}{(2n+2)!!}$$

Then

$$\sum_{n=h}^{m-1}\frac{(2n+1)!!}{(2n+2)!!}=\sum_{n=h+1}^{m}\frac{(2n-1)!!}{(2n)!!}=\frac{(2n+1)!!}{(2n)!!}\bigg|_h^m\tag{3}$$

And from $(2)$ and $(3)$ we get

$$\sum_{n=1}^m\frac{(2n-1)!!}{(2n)!!}=\frac{(2n+1)!!}{(2n)!!}\bigg|_0^m=\frac{(2m+1)!!}{(2m)!!}-1$$

But there is a difference of $1$!! I double-checked the difference in Wolfram-Alpha, so you have a typo in your exercise.

Masacroso
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$$ \begin{align} \sum_{n=1}^m\prod_{k=1}^n\frac{2k-1}{2k} &=\sum_{n=1}^m\binom{n-1/2}{n}\tag{1}\\ &=\sum_{n=1}^m(-1)^n\binom{-1/2}{n}\tag{2}\\ &=\sum_{n=1}^m(-1)^m\binom{-1}{m-n}\binom{-1/2}{n}\tag{3}\\ &=(-1)^m\binom{-3/2}{m}-1\tag{4}\\ &=\binom{m+1/2}{m}-1\tag{5}\\ &=\frac{\Gamma\left(m+\frac32\right)}{\Gamma(m+1)\Gamma\left(\frac32\right)}-1\tag{6}\\ &=\frac{\Gamma\left(m+\frac32\right)}{\frac1{m+1}\Gamma(m+2)\frac12\Gamma\left(\frac12\right)}-1\tag{7}\\ &=\frac{2(m+1)}{\sqrt\pi}\frac{\Gamma\left(m+\frac32\right)}{\Gamma(m+2)}-1\tag{8} \end{align} $$ Explanation:
$(1)$: convert product to Binomial Coefficient
$(2)$: convert to Negative Binomial Coefficient
$(3)$: rewrite $(-1)^n$
$(4)$: Vandermonde's Identity
$(5)$: convert from negative binomial coefficient
$(6)$: express binomial coefficient with Gamma functions
$(7)$: apply $x\Gamma(x)=\Gamma(x+1)$
$(8)$: $\Gamma\left(\frac12\right)=\sqrt\pi$

The sum would be as given in the question if the $n=0$ term were included in the sum.

robjohn
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