Evaluation of limit $ \lim_{n\rightarrow \infty}\frac{(2n+1)(2n+3)\cdots (4n+1)}{(2n)(2n+2)\cdots (4n)}$

Question

Evaluation of limit $\displaystyle \lim_{n\rightarrow \infty}\frac{(2n+1)(2n+3)\cdots (4n+1)}{(2n)(2n+2)\cdots (4n)}$

What I try :

$\displaystyle \lim_{n\rightarrow \infty}\frac{(2n-1)!}{(2n-1)!}\frac{(2n)(2n+1)(2n+2)\cdots (4n+1)}{\bigg((2n)(2n+2)(2n+4)\cdots (4n)\bigg)^2}$

$\displaystyle \lim_{n\rightarrow \infty}\frac{(4n+1)!\cdot (2n-1)!}{((4n)!)^2}$

How do solve it, please have a look on that problem, Thanks

Answer 1

Here is a method without Stirling's formula. Just use an inequality: $$0<x-\log(1+x)<\frac{x^2}{2},\quad \forall x>0.$$

Let $$x_n=\frac{(2n+1)(2n+3)\cdots (4n+1)}{(2n)(2n+2)\cdots(4n)},$$ then $$\log x_n=\sum_{k=0}^{n}\log\left(1+\frac{1}{2n+2k}\right).$$ Since $$0<\frac{1}{2n+2k}-\log\left(1+\frac{1}{2n+2k}\right)<\frac1{8(n+k)^2},k=0,1,\cdots,n,$$ we have $$0<\sum_{k=0}^{n}\frac{1}{2n+2k}-\sum_{k=0}^{n}\log\left(1+\frac{1}{2n+2k}\right)<\sum_{k=0}^{n}\frac1{8(n+k)^2} =\frac{1}{8n}\cdot\frac{1}{n}\sum_{k=0}^{n}\frac{1}{\left(1+\frac{k}{n}\right)^2}\to0.$$ So $$\lim_{n\to\infty}\log x_n=\lim_{n\to\infty}\sum_{k=0}^{n}\frac{1}{2n+2k} =\frac12\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n}\frac{1}{1+\frac{k}{n}} =\frac12\log2.$$ Hence $$\lim_{n\to\infty}\frac{(2n+1)(2n+3)\cdots (4n+1)}{(2n)(2n+2)\cdots(4n)} =\lim_{n\to\infty}e^{\log x_n}=\sqrt2.$$

Answer 2

First recall that as $x\to\infty$, $$\frac{\Gamma\left(x+\frac12\right)}{\Gamma(x)}\sim\sqrt x.$$ Therefore, as $n\to\infty$, $$\begin{align}\frac{(2n+1)(2n+3)\cdots (4n+1)}{(2n)(2n+2)\cdots (4n)}&=\frac{\left(n+\frac12\right)\left(n+\frac32\right)\cdots\left(2n+\frac12\right)}{n(n+1)\cdots (2n)}\\&=\frac{\Gamma\left(2n-\frac12\right)/\Gamma\left(n+\frac12\right)}{\Gamma(2n-1)/\Gamma(n)}\\ &\sim\sqrt{2n-1}/\sqrt n \\&\to\sqrt2. \end{align}$$

Answer 3

$$\begin{gathered} \mathop {\lim }\limits_{n \to \infty } \frac{{\prod\limits_{k = 0}^n {\left( {2n + 2k + 1} \right)} }}{{\prod\limits_{k = 0}^n {\left( {2n + 2k} \right)} }} = \mathop {\lim }\limits_{n \to \infty } \frac{{{2^{n + 1}}\Gamma \left( {2n + \frac{3}{2}} \right)}}{{\Gamma \left( {n + \frac{1}{2}} \right)}}\frac{{\sqrt \pi }}{{{2^{3n + 1}}n\Gamma \left( {n + \frac{1}{2}} \right)}} \hfill \\ {\text{As }}n \to \infty :\Gamma \left( {n + a} \right) \approx \Gamma \left( n \right){n^a} \hfill \\ \Rightarrow L = \mathop {\lim }\limits_{n \to \infty } \frac{{{2^{n + \frac{5}{2}}}\sqrt \pi }}{{{2^{3n + 1}}}}\frac{{\Gamma \left( {2n} \right)}}{{\sqrt n \Gamma \left( n \right)\Gamma \left( n \right)}}{\text{. Gamma duplication formula}}:\Gamma \left( {2n} \right) = \frac{{{2^{2n - 1}}\Gamma \left( n \right)\Gamma \left( {n + \frac{1}{2}} \right)}}{{\sqrt \pi }} \hfill \\ \Rightarrow L = \mathop {\lim }\limits_{n \to \infty } \frac{{{2^{3n + \frac{3}{2}}}}}{{{2^{3n + 1}}}}\frac{{\Gamma \left( n \right)\Gamma \left( {n + \frac{1}{2}} \right)}}{{\sqrt n \Gamma \left( n \right)\Gamma \left( n \right)}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{2^{3n + \frac{3}{2}}}}}{{{2^{3n + 1}}}}\frac{{\Gamma \left( n \right)\Gamma \left( n \right)\sqrt n }}{{\sqrt n \Gamma \left( n \right)\Gamma \left( n \right)}} = \frac{{{2^{\frac{3}{2}}}}}{2} = {2^{\frac{1}{2}}} = \sqrt 2 \hfill \\ \end{gathered}$$