Proving using squeeze principle

Question

This problem sounds very confusing. Please help me solve this problem.

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{{\cal I}_{n} \equiv \pars{2 \over 1}^{2}\pars{4 \over 3}^{2} \pars{6 \over 5}^{2}\ldots \pars{2n \over 2n - 1}^{2}{1 \over 2n + 1}\,,\qquad \lim_{n \to \infty}{\cal I}_{n}:\ {\large ?}}$

\begin{align} \ln\pars{{\cal I}_{n}} + \ln\pars{2n + 1}&= \sum_{k = 0}^{n - 1}\ln\pars{\bracks{2k + 2\over 2k + 1}^{2}} =2\sum_{k = 0}^{n - 1}\ln\pars{2k + 2 \over 2k + 1} =2\sum_{k = 0}^{n - 1}\int_{0}^{1}{\dd x \over x + 2k + 1} \\[3mm]&=\int_{0}^{1}\dd x\sum_{k = 0}^{n - 1}{1 \over k + \pars{1 + x}/2} =\int_{0}^{1}\bracks{\Psi\pars{{1 + x \over 2} + n} - \Psi\pars{x + 1 \over 2}}\dd x \end{align} where $\ds{\Psi\pars{z} = \dd\ln\pars{\Gamma\pars{z}}/\dd z}$. $\Gamma$ and $\Psi$ are the Gamma and Digamma functions, respectively.

Then \begin{align} \ln\pars{{\cal I}_{n}}&=-\ln\pars{2n + 1} \left.+2\ln\pars{\Gamma\pars{\bracks{x + 2n + 1}/2} \over \Gamma\pars{\bracks{x + 1}/2}}\right\vert_{0}^{1} \\[3mm]&=-\ln\pars{2n + 1} + 2\ln\pars{{\Gamma\pars{n + 1} \over \Gamma\pars{1}}\, {\Gamma\pars{1/2} \over \Gamma\pars{n + 1/2}}} \end{align}

Since $\Gamma\pars{1/2} = \root{\pi}$ and $\Gamma\pars{1} = 1$: \begin{align} \ln\pars{{\cal I}_{n}}&=-\ln\pars{n + \half} +2\ln\pars{\Gamma\pars{n + 1} \over \Gamma\pars{n + 1/2}} + \ln\pars{\pi \over 2} \end{align}

However, when $n \gg 1$: \begin{align} &{\Gamma\pars{n + 1} \over \Gamma\pars{n + 1/2}} ={\root{2\pi}\pars{n + 1}^{n + 3/2}\expo{-n - 1} \over \root{2\pi}\pars{n + 1/2}^{n + 1}\expo{-n - 1/2}} =n^{1/2}\,{\pars{1 + 1/n}^{n + 3/2} \over \pars{1 + 1/2n}^{n + 1}}\,\expo{-1/2} \sim n^{1/2}\,{\expo{1} \over \expo{1/2}}\,\expo{-1/2} \\[3mm]&= n^{1/2} \end{align}

Then $$ \ln\pars{{\cal I}_{n}}\sim-\ln\pars{n + \half} + 2\ln\pars{n^{1/2}} + \ln\pars{\pi \over 2} =-\ln\pars{1 + {1 \over 2n}} + \ln\pars{\pi \over 2}\,,\qquad n \gg 1 $$

Then, $$\color{#00f}{\large% \lim_{n \to \infty}\bracks{% \pars{2 \over 1}^{2}\pars{4 \over 3}^{2} \pars{6 \over 5}^{2}\ldots \pars{2n \over 2n - 1}^{2}{1 \over 2n + 1}} = {\pi \over 2}} $$