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
Best Answer
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 $$\lim_{n\to\infty}\frac{(2n+1)(2n+3)\cdots (4n+1)}{(2n)(2n+2)\cdots(4n)} =\lim_{n\to\infty}e^{\log x_n}=\sqrt2.$$