TITLE: How to find a limit of this sequence?
QUESTION [7 upvotes]: I would appreciate if somebody could help me with the following problem: 

Let the sequences $\{a_n\}$, $\{b_n\}$ be defined as
  $$a_n=\int\limits_0^{\pi/2}\sin^{3n} x \;dx,\qquad b_n=\int\limits_0^{\pi/2}\sin^{2n}x \cos^nx\;dx.$$
  Find
  $$\lim_{n\to \infty}\frac{1}{2n}\left(\frac{(2n)!\, a_n}{n!\,
b_n} \right)^
{1/n}.$$

REPLY [3 votes]: Note that, we can evaluate $a_n$ and $b_n$ explicitly using the beta function
$$ \beta(x,y) = 2\int_0^{\pi/2}(\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\,d\theta=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}, \qquad \mathrm{Re}(x)>0,\ \mathrm{Re}(y)>0 \!$$
as
$$ a_n= \frac{1}{2}\,{\frac {\sqrt {\pi }\Gamma  \left( \frac{1}{2}+\frac{3}{2}\,n \right) }{\Gamma 
 \left( 1+\frac{3}{2}n \right) }}$$
and
$$ b_n = \frac{1}{2}\,{\frac {\Gamma  \left( \frac{1}{2}n+\frac{1}{2} \right) \Gamma\left( n+\frac{1}{2}\right) }{\Gamma  \left( 1+\frac{3}{2}n \right) }}.$$
To evaluate the limit, first write the limit as
$$ \lim_{n\to \infty}\left(\frac{(2n)!\, a_n}{2^n n^n n!\,
b_n} \right)^{1/n}, $$
then use the result 
$$ \lim_{n \to \infty} c^{\frac{1}{n}}_n = \lim_{n \to \infty} \frac{c_{n+1}}{c_n}.$$
to evaluate the limit. Here is the final result
$$ \lim_{n\to \infty}\frac{1}{2n}\left(\frac{(2n)!\, a_n}{n!\,
b_n} \right)^{1/n} = \frac{3}{2}\,{{\rm e}^{-1}}\sqrt {6}\sqrt {2}. $$
Note: $$ n! = \Gamma(n+1). $$