TITLE: Estimating the integral $\sqrt{n}\cdot \int\limits_0^\pi \left( \frac{1 + \cos t}{2} \right)^n dt$
QUESTION [11 upvotes]: Consider the sequence $\{a_n\}$ defined by 
$$ a_n = \sqrt{n}\cdot \int_{0}^{\pi} \left( \frac{1 + \cos t}{2} \right)^n dt.$$
An exercise in Rudin, Real and Complex Analysis, requires showing that this sequence is
convergent to a real number $a$, with $ a > 0$.
I don't have any idea of how to prove this. 
I only obtained the following estimation
$$ \begin{align*} 
\int_{0}^{\pi} \left( \frac{1 + \cos t}{2} \right)^n  dt &= 2 \int_{0}^{\frac{\pi}{2}} \left( 1 - \sin^2 t \right)^n dt \\
 &> 2 \int_{0}^{\frac{1}{\sqrt{n}}} (1 - t^2)^n dt \\ 
&> 2 \int_{0}^{\frac{1}{\sqrt{n}}} (1 - n t^2) dt \\ &
= \frac{4}{3 \sqrt{n}}, 
\end{align*}$$
which shows that $ a_n > \frac{4}{3}$.
Thank you very much in advance for any help.

REPLY [10 votes]: Edited. The original answer is appended.
Put
$$c_m:=\int_0^{\pi/2} \cos^m s\ ds\qquad(m\geq0)\ .$$
Then $c_1=1$, $\> c_2={\pi\over4}$, and an easy partial integration shows that
$$c_m={m-1\over m} c_{m-2}\qquad(m\geq2)\ .\tag{1}$$ Now
$$a_n:=\sqrt{\mathstrut n}\int_0^\pi\left({1+\cos t\over2}\right)^n\ dt=2\sqrt{\mathstrut n}\int_0^{\pi/2}\cos^{2n}s\ ds$$
and therefore
$$a^2_n=4 n\> c_{2n}^2\ .$$
I claim that
$$a_n^2={c_{2n}\over c_{2n-1}}\pi\qquad(n\geq1)\ .\tag{2}$$
Proof by induction: $(2)$ is true for $n=1$. Assume that it is true for $n-1$. Using $(1)$ we then have
$$\eqalign{a_n^2 &= {n\over n-1}\left({c_{2n}\over c_{2n-2}}\right)^2 a_{n-1}^2 =
{n\over n-1}{2n-1\over 2n}{c_{2n}\over c_{2n-2}}{c_{2n-2}\over c_{2n-3}}\pi\cr &={n\over n-1}{2n-1\over 2n}{2n-2\over 2n-1}{c_{2n}\over c_{2n-1}}\pi={c_{2n}\over c_{2n-1}}\pi\ .\qquad\square\cr}$$
Since $c_m\leq c_{m-1}\leq c_{m-2}$ it follows easily from $(1)$ that $$\lim_{m\to\infty}{c_m\over c_{m-1}}=1\ .$$
From $(2)$ we therefore conclude that $\lim_{n\to\infty} a_n=\sqrt{\mathstrut \pi}$.
Original answer:
Let
$$b_n:=\int_0^\pi\left({1+\cos t\over2}\right)^n\ dt=2\int_0^{\pi/2}\cos^{2n}s\ ds\ .$$
Then
$$b_n =2\int_0^{\pi/2}\cos^{2n-2}s(1-\sin^2 s)\ ds=b_{n-1}-2\int_0^{\pi/2}\bigl(\cos^{2n-1}s\>\sin s\bigr)\cdot\sin s\ ds\ .$$
We integrate the last integral by parts and obtain
$$b_n=b_{n-1}-{1\over 2n-1} b_n\ .$$
Therefore the $b_n$ satisfy the recursion
$$b_n={2n-1\over 2n}\>b_{n-1}\qquad(n\geq1)\ ,$$
and the $a_n=\sqrt{\mathstrut n}\>b_n$ satisfy the recursion
$$a_n^2={(2n-1)^2\over 2n\>(2n-2)}\>a^2_{n-1}\qquad(n\geq2)\ .$$
As $a_1={\pi\over2}$ we therefore have
$$a_n^2=2\cdot\left({1\over2}\cdot{3\over2}\cdot{3\over4}\cdot{5\over4}\cdot{5\over6}\cdots{2n-1\over2n-2}{2n-1\over2n}\right)\cdot{\pi^2\over4}\qquad(n\geq2)\ .$$
In the large parenthesis the reciprocal of Wallis' product appears. It follows that this parenthesis  converges to ${2\over\pi}$, and we obtain
$$\lim_{n\to\infty}a_n=\sqrt{\mathstrut \pi}\ .$$