Compute the following product limit

Question

From another problem, I got stuck trying to solve this limit: $$\lim_{k\to +\infty}\prod_{v=k+1}^{2k}\left(1-\frac1{va}\right)$$ where $a>1$ is a positive integer. I tried to take $\log$ on both sides and expanded each term in series, but it doesn't look easier. The result turns out to be $2^{-1/a}$, according to my book. What's the trick behind this? One can rewrite this following the hint in the comments as $$\lim_{k\to +\infty}\frac{(\frac{a-1}a+k)(\frac{a-1}a+k+1)\cdots(\frac{a-1}a+2k-1)}{(k+1)(k+2)\cdots(2k)}.$$ What is this value?

Answer 1

It is not bad with Taylor series $$P_k=\prod_{v=k+1}^{2k}\left(1-\frac1{va}\right)\implies \log(P_k)=\sum_{v=k+1}^{2k}\log\left(1-\frac1{va}\right)$$ $$\log\left(1-\frac1{va}\right)=-\frac{1}{a v}-\frac{1}{2 a^2 v^2}+O\left(\frac{1}{v^3}\right)$$ $$\log(P_k)=\sum_{v=k+1}^{2k}\log\left(1-\frac1{va}\right)=\frac{2 a( H_k- H_{2 k})-\psi ^{(1)}(k+1)+\psi ^{(1)}(2 k+1)}{2 a^2}$$ Now, using asymptotics $$\log(P_k)=-\frac{\log (2)}{a}+\frac{a-1}{4 a^2 k}+\frac{3-a}{16 a^2 k^2}+O\left(\frac{1}{k^3}\right)$$ $$P_k=e^{\log(P_k)}=2^{-1/a}\left(1+\frac{a-1}{4 a^2 k}+O\left(\frac{1}{k^2}\right) \right)$$

Answer 2

Let's start with $$ \prod _{v=1}^{m} 1-\frac{1}{va} = \frac{\left(\frac{a-1}{a}\right)_m}{m!} $$ Then the limit you seek is $$ \lim_{k\to\infty} \frac{\left(\frac{a-1}{a}\right)_{2k}}{(2k)!}\cdot \frac{k!}{\left(\frac{a-1}{a}\right)_k} $$Let's turn Pochhammer symbols and factorial into gamma functions using $(b)_m = \Gamma(b+m)/\Gamma(m)$, $m!=\Gamma(m+1)$: $$= \lim_{k\to\infty} \frac{\Gamma\left(2k+\frac{a-1}{a}\right)}{\Gamma(2k+1) \Gamma\left(\frac{a-1}{a}\right)}\cdot \frac{\Gamma(k+1)\Gamma\left(\frac{a-1}{a}\right)}{\Gamma\left(k+\frac{a-1}{a}\right)} $$ $$= \lim_{k\to\infty} \frac{\Gamma\left(2k+1-\frac{1}{a}\right)}{\Gamma(2k+1) }\cdot \frac{\Gamma(k+1)}{\Gamma\left(k+1-\frac{1}{a}\right)} $$Now we use the handy asymptotic $$ \lim_{z\to\infty}\frac{\Gamma(z+\alpha)}{\Gamma(z+\beta)}z^{\beta-\alpha}=1; $$matching arguments, we have $$= \color{green}{2^{-1/a}} \lim_{k\to\infty} \frac{\Gamma\left(2k+1-\frac{1}{a}\right)\color{red}{(2k)^{1/a}}}{\Gamma(2k+1) }\cdot \frac{\Gamma(k+1)}{\Gamma\left(k+1-\frac{1}{a}\right)\color{blue}{k^{1/a}}} $$ $$ =\color{green}{2^{-1/a}} \cdot \color{red}{1}\cdot \color{blue}{1} = 2^{-1/a} $$

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The starting point $$ \prod _{v=1}^{m} 1-\frac{1}{va} = \frac{\left(\frac{a-1}{a}\right)_m}{m!} $$ can be shown by induction on $m$.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\lim_{k \to \infty}\prod_{v = k + 1}^{2k} \pars{1 - {1 \over va}}} = \lim_{k \to \infty}\prod_{v = k + 1}^{2k}\pars{v - 1/a \over v} = \lim_{k \to \infty}{\pars{k + 1 - 1/a}^{\,\overline{k}} \over \pars{k + 1}^{\,\overline{k}}} \\[5mm] = &\ \lim_{k \to \infty}{\Gamma\pars{2k + 1 - 1/a}/\Gamma\pars{k + 1 - 1/a} \over \Gamma\pars{2k +1}/\Gamma\pars{k + 1}} = \lim_{k \to \infty}{\pars{2k - 1/a}! \over \pars{k - 1/a}!} {k! \over \pars{2k}!} \\[5mm] = &\ \lim_{k \to \infty}{\root{2\pi}\pars{2k - 1/a}^{2k + 1/2 - 1/a} \,\expo{-2k + 1/a} \over \root{2\pi}\pars{k - 1/a}^{k + 1/2 - 1/a}\,\expo{-k + 1/a}}\, {\root{2\pi}k^{k + 1/2}\,\expo{-k} \over \root{2\pi}\pars{2k}^{2k + 1/2}\,\expo{-2k}} \\[5mm] = &\ \lim_{k \to \infty}{\pars{2k}^{2k + 1/2 - 1/a}\, \bracks{1 - \pars{1/a}/\pars{2k}}^{\, 2k} \over k^{k + 1/2 - 1/a}\,\bracks{1 - \pars{1/a}/k}^{\, k}}\, {k^{k + 1/2} \over \pars{2k}^{2k + 1/2}} \\[5mm] = &\ \lim_{k \to \infty} {\pars{2k}^{-1/a}\expo{-1/a}k^{1/a} \over \expo{-1/a}} = \bbx{\large{1 \over 2^{1/a}}} \end{align}