# Given that x_0 is the unique positive solution of (2−x)^(n+1)=x(x+1)⋯(x+n), try to find the asymptotic value of M=prod^n_(k=0)((k+2)/(k+x_0))^(k+2)

Given that ${x}_{0}$ is the unique positive solution of $\left(2-x{\right)}^{n+1}=x\left(x+1\right)\cdots \left(x+n\right)$, try to find the asymptotic value of
$M=\prod _{k=0}^{n}{\left(\frac{k+2}{k+{x}_{0}}\right)}^{k+2}$
with absolute error $o\left(1\right)$ as $n\to \mathrm{\infty }$, where ${H}_{n}$ denotes n-th harmonic number $\sum _{k=1}^{n}1/k$
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periasemdy
All 0-notations and o-notations work for $n\to \mathrm{\infty }$
First we have ${2}^{n+1}\ge n!{x}_{0}$, hence ${x}_{0}\le {2}^{n+1}/n!=O\left({2}^{n}/n!\right)$. Take logarithm, we derive that
$\begin{array}{rl}\left(n+1\right)\mathrm{ln}\left(2-{x}_{0}\right)& =\mathrm{ln}{x}_{0}+\sum _{k=1}^{n}\mathrm{ln}\left({x}_{0}+k\right)\\ & =\mathrm{ln}n!+\mathrm{ln}{x}_{0}+\sum _{k=1}^{n}\mathrm{ln}\left(1+\frac{{x}_{0}}{k}\right)\end{array}$
therefore${x}_{0}=\frac{{2}^{n+1}\left(1+O\left(n{x}_{0}\right)\right)}{n!}$
We have enough stuff to estimate $\mathrm{ln}M$ now.
$\begin{array}{c}\mathrm{ln}M=\left(n+1\right)\mathrm{ln}\left(n+1\right)+\left(n+2\right)\mathrm{ln}\left(n+2\right)-2\left(n+1\right)\mathrm{ln}2\hfill \\ \hfill +\frac{{2}^{n+1}}{n!}-\frac{{4}^{n}}{n{!}^{2}}\left(n+2{H}_{n}+1\right)+O\left({n}^{3}{x}_{0}^{3}\right)\end{array}$
Finally, we have
$\begin{array}{c}\mathrm{exp}\left(\frac{{2}^{n+1}}{n!}\right)=1+\frac{{2}^{n+1}}{n!}+\frac{2\cdot {4}^{n}}{n{!}^{2}}+O\left({x}_{0}^{3}\right)\\ \mathrm{exp}\left(-\frac{{4}^{n}}{n{!}^{2}}\left(n+2{H}_{n}+1\right)\right)=1-\frac{{4}^{n}}{n{!}^{2}}\left(n+2{H}_{n}+1\right)+O\left({n}^{2}{x}_{0}^{4}\right)\end{array}$
and
$\begin{array}{rl}M& =\frac{\left(n+1{\right)}^{n+1}\left(n+2{\right)}^{n+2}}{{4}^{n+1}}\\ & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\left(1+\frac{{2}^{n+1}}{n!}-\frac{{4}^{n}}{n{!}^{2}}\left(n+2{H}_{n}-1\right)\right)\\ & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\left(1+O\left({n}^{3}{x}_{0}^{3}\right)\right)\end{array}$
Notice that the absolute error
$O\left(\frac{{n}^{3}\left(n+1{\right)}^{n+1}\left(n+2{\right)}^{n+2}}{{4}^{n+1}}{x}_{0}^{3}\right)$
approaches 0 when $n\to \mathrm{\infty }$