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I am trying to prove the following series converges and find its limit if indeed it does:

$$ \sum_{i=1}^{\infty}\frac{2^{i-1}(i-1)!}{\prod_{j=1}^{i}(2j+1)}$$

Intuitively, I can see this is converging since the $i$-term product on the denominator, when expanded, should yield something at least as great as $2^i$ as a result of the $2$ coefficient, as well as something at least as great as the $(i-1)!$ due to $j$ itself. However, I am not sure how to formally prove this. Some preliminary calculations show this may converge to something close to but less than $1$ (maybe $1$ itself?) but how can I obtain the exact limit?

Winther
  • 24,478
Julia
  • 288

3 Answers3

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Simplifying the Sum $$ \begin{align} \sum_{n=1}^\infty\frac{(n-1)!\,2^{n-1}}{\prod_{k=1}^n(2k+1)} &=\sum_{n=1}^\infty\frac{(n-1)!\,2^{n-1}}{\frac{(2n+1)!}{n!\,2^n}}\\ &=\sum_{n=1}^\infty\frac1{2n(2n+1)}\frac{n!\,2^n}{\frac{(2n)!}{n!\,2^n}}\\ &=\sum_{n=1}^\infty\frac1{2n(2n+1)}\frac{4^n}{\binom{2n}{n}}\tag{1} \end{align} $$ Since $\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}$, we have $\frac1{2n(2n+1)}\frac{4^n}{\binom{2n}{n}}\sim\frac{\sqrt\pi}{4n^{3/2}}$ and the sum converges since $\frac32\gt1$.

Evaluation $\boldsymbol{1}$

Equation $(2)$ from this answer says $$ \sum_{n=1}^\infty\frac{4^nx^{2n}}{\binom{2n}{n}} =\frac1{1-x^2}\left[x^2+\frac{x}{\sqrt{1-x^2}}\sin^{-1}(x)\right]\tag{2} $$ Dividing $(2)$ by $x$ and integrating twice gives $$ \sum_{n=1}^\infty\frac1{2n(2n+1)}\frac{4^nx^{2n+1}}{\binom{2n}{n}} =x-\sqrt{1-x^2}\sin^{-1}(x)\tag{3} $$ Evaluating $(3)$ at $x=1$ gives $$ \sum_{n=1}^\infty\frac{(n-1)!\,2^{n-1}}{\prod_{k=1}^n(2k+1)}=1\tag{4} $$

Evaluation $\boldsymbol{2}$

Using the Beta Function integral, we get $$ \frac1{\binom{2n}{n}}=(2n+1)\int_0^1x^n(1-x)^n\,\mathrm{d}x\tag{5} $$ Therefore, $$ \begin{align} \sum_{n=1}^\infty\frac1{2n(2n+1)}\frac{4^n}{\binom{2n}{n}} &=\int_0^1\sum_{n=1}^\infty\frac{4^n}{2n}x^n(1-x)^n\,\mathrm{d}x\tag{6a}\\ &=-\frac12\int_0^1\log(1-4x(1-x))\,\mathrm{d}x\tag{6b}\\ &=-\int_0^1\log|1-2x|\,\mathrm{d}x\tag{6c}\\ &=-\frac12\int_{-1}^1\log|x|\,\mathrm{d}x\tag{6d}\\ &=-\int_0^1\log(x)\,\mathrm{d}x\tag{6e}\\[6pt] &=1\tag{6f} \end{align} $$ Explanation:
$\text{(6a)}$: apply $(5)$
$\text{(6b)}$: use the Taylor series for $\log(1-x)$
$\text{(6c)}$: $\sqrt{1-4x+4x^2}=|1-2x|$
$\text{(6d)}$: substitute $x\mapsto\frac{1-x}2$
$\text{(6e)}$: symmetry
$\text{(6f)}$: $\int\log(x)\,\mathrm{d}x=x\log(x)-x$

Community
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robjohn
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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{equation} \sum_{i = 1}^{\infty}{2^{i - 1}\pars{i - 1}! \over \prod_{j = 1}^{i}\pars{2j + 1}}:\ ?\label{1}\tag{1} \end{equation}

\begin{align} &\bbox[8px,border:0.1em groove navy]{{2^{i - 1}\pars{i - 1}! \over \prod_{j = 1}^{i}\pars{2j + 1}}} = {2^{i - 1}\pars{i - 1}! \over 2^{i}\prod_{j = 1}^{i}\pars{j + 1/2}} \\[5mm] = &\ {1 \over 2}\,{\pars{i - 1}! \over \pars{3/2}_{i}} \qquad\pars{~\vphantom{\large A}\pars{a}_{n}:\ Pochhammer\ Symbol~} \\[5mm] = &\ {1 \over 2}\,{\pars{i - 1}! \over \Gamma\pars{3/2 + i}/\Gamma\pars{3/2}} = {1 \over 2}\,\ \underbrace{{\Gamma\pars{i}\Gamma\pars{3/2} \over \Gamma\pars{i + 3/2}}} _{\ds{\mrm{B}\pars{i,3/2}}}\qquad \pars{~\mrm{B}:\ Beta\ Function~} \\ = &\,\, \bbox[8px,border:0.1em groove navy]{{1 \over 2} \int_{0}^{1}x^{i - 1}\pars{1 - x}^{1/2}\,\dd x}\label{2}\tag{2} \end{align}

Note that $\ds{\pars{a}_{n} = {\Gamma\pars{a + n} \over \Gamma\pars{a}}}$ where $\ds{\Gamma}$ is the Gamma Function. By inserting the result \eqref{2} into the expression \eqref{1}:

\begin{align} \color{#f00}{\sum_{i = 1}^{\infty}{2^{i - 1}\pars{i - 1}! \over \prod_{j = 1}^{i}\pars{2j + 1}}} & = \sum_{i = 1}^{\infty}{1 \over 2} \int_{0}^{1}x^{i - 1}\pars{1 - x}^{1/2}\,\dd x = {1 \over 2}\int_{0}^{1}{\dd x \over \root{1 - x}} = \color{#f00}{1} \end{align}
Felix Marin
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0

$$2^{i-1}(i-1)! = \prod_{j=1}^{i-1} (2j)$$ so that $$ \frac{2^{i-1}(i-1)!}{\prod_{j=1}^{i} (2j+1)} = \frac{1}{2i+1}\prod_{j=1}^{i-1} \frac{2j}{2j+1} = \frac{1}{2i+1}\prod_{j=1}^{i-1} \left(1-\frac{1}{2j+1}\right) $$ Now, how does the general term $$ a_i \stackrel{\rm def}{=} \frac{1}{2i+1}\prod_{j=1}^{i-1} \left(1-\frac{1}{2j+1}\right) $$ of your series behave when $i\to\infty$?

Hint: $$ \ln \prod_{j=1}^{i-1} \left(1-\frac{1}{2j+1}\right) = \sum_{j=1}^{i-1} \ln \left(1-\frac{1}{2j+1}\right) $$ and $\ln \left(1-\frac{1}{2j+1}\right) =_{j\to\infty} -\frac{1}{2j+1} + O\left(\frac{1}{j^2}\right)$ so that, by theorems of comparison (for series with terms of constant sign) $$ \ln \prod_{j=1}^{i-1} \left(1-\frac{1}{2j+1}\right) = -\sum_{j=1}^{i-1} \frac{1}{2j+1} + O(1) = \frac{1}{2}\ln i + o(\ln i) $$ and then $a_i = \frac{1}{2i+1}\cdot \frac{1}{\sqrt{i}}\cdot f(i)$ for $f(i) = e^{o(\ln i)}$. This suffices to prove convergence, by comparison to a $p$-series (for say, $1<p<3/2$).

Clement C.
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