Behaviour of the sequence $u_n = \frac{\sqrt{n}}{4^n}\binom{2n}{n}$

Question

Let $n\in \mathbb{N}^{*}$, let ${\displaystyle u_{n}={2n \choose n}\sqrt{n}\ 4^{-n}}$

Show that $(u_{n})_{n}$ is convergent and ${l.e^{-\frac{1}{8n}}<u_{n}<l}$

The original text

i'm trying to compute $\dfrac{u_{n+1}}{u_n}$ but i failed even with wolframe

{u(n) = (binomial(2 n, n) sqrt(n))×4^(-n), (u(n+1))/(u(n))=? }

any help would be appreciated

Answer 1

We have:

$$ u_n = \sqrt{n}\frac{(2n)!}{4^n n!^2} = \sqrt{n}\frac{(2n-1)!!}{(2n)!!} = \sqrt{n}\prod_{k=1}^{n}\left(1-\frac{1}{2k}\right),$$ so: $$ u_n^2 = \frac{n}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}+\frac{1}{4k^2}\right)=\frac{n}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right) $$ or just : $$ u_n^2 = \frac{1}{4}\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right). $$ Since the product, as $n\to +\infty$, is convergent, we have: $$ u_n^2 \leq \frac{1}{4}\prod_{k=1}^{+\infty}\left(1+\frac{1}{4k(k+1)}\right)=C.$$ On the other hand: $$ u_n^2 = \frac{C}{\prod_{k=n}^{+\infty}\left(1+\frac{1}{4k(k+1)}\right)}\geq\frac{C}{\exp\left(\frac{1}{4}\sum_{k=n}^{+\infty}\left(\frac{1}{k}-\frac{1}{k+1}\right)\right)}=C e^{-\frac{1}{4n}}. $$