Compute the following product limit

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

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},} newcommand{braces}[1]{leftlbrace,{#1},rightrbrace} newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack} newcommand{dd}{mathrm{d}} newcommand{ds}[1]{displaystyle{#1}} newcommand{expo}[1]{,mathrm{e}^{#1},} newcommand{ic}{mathrm{i}} newcommand{mc}[1]{mathcal{#1}} newcommand{mrm}[1]{mathrm{#1}} newcommand{pars}[1]{left(,{#1},right)} newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}} newcommand{root}[2][]{,sqrt[#1]{,{#2},},} newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}} newcommand{verts}[1]{leftvert,{#1},rightvert}$ 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}{Gammapars{2k + 1 - 1/a}/Gammapars{k + 1 - 1/a} over Gammapars{2k +1}/Gammapars{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{2pi}pars{2k - 1/a}^{2k + 1/2 - 1/a} ,expo{-2k + 1/a} over root{2pi}pars{k - 1/a}^{k + 1/2 - 1/a},expo{-k + 1/a}}, {root{2pi}k^{k + 1/2},expo{-k} over root{2pi}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}

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 $$ lim_{ktoinfty} frac{left(frac{a-1}{a}right)_{2k}}{(2k)!}cdot frac{k!}{left(frac{a-1}{a}right)_k} $$Let's turn Pochhammer symbols and factorial into gamma functions using $(b)_m = Gamma(b+m)/Gamma(m)$, $m!=Gamma(m+1)$: $$= lim_{ktoinfty} frac{Gammaleft(2k+frac{a-1}{a}right)}{Gamma(2k+1) Gammaleft(frac{a-1}{a}right)}cdot frac{Gamma(k+1)Gammaleft(frac{a-1}{a}right)}{Gammaleft(k+frac{a-1}{a}right)} $$ $$= lim_{ktoinfty} frac{Gammaleft(2k+1-frac{1}{a}right)}{Gamma(2k+1) }cdot frac{Gamma(k+1)}{Gammaleft(k+1-frac{1}{a}right)} $$Now we use the handy asymptotic $$ lim_{ztoinfty}frac{Gamma(z+alpha)}{Gamma(z+beta)}z^{beta-alpha}=1; $$matching arguments, we have $$= color{green}{2^{-1/a}} lim_{ktoinfty} frac{Gammaleft(2k+1-frac{1}{a}right)color{red}{(2k)^{1/a}}}{Gamma(2k+1) }cdot frac{Gamma(k+1)}{Gammaleft(k+1-frac{1}{a}right)color{blue}{k^{1/a}}} $$ $$ =color{green}{2^{-1/a}} cdot color{red}{1}cdot color{blue}{1} = 2^{-1/a} $$

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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 $m$.