How to prove $\lim_{n \to +\infty} \sqrt{n}\int_0^\pi{\cos(\frac{t}{2})^n}dt>0$ I want to prove 
$$\lim_{n \to +\infty}\sqrt{n}\int_0^\pi{\cos\left(\frac{t}{2}\right)^n}dt>0.$$
First, I consider $$\lim_{n \to +\infty}\sqrt{n}\int_0^\pi{\cos\left(\frac{t}{2}\right)^n}\sin\left(\frac{t}{2}\right)dt,$$ which is smaller than what I want, but the second integral leads to $$\lim_{n \to +\infty}\frac{\sqrt{n}}{2(n+1)}.$$ So it does not work.
 A: HINT
In fact, it is not hard to compute the limit exactly. Let $I_n = \displaystyle \int_0^{\pi} \cos^n(t/2) dt$. Compute $I_n$ and use Stirling's formula to obtain asymptotics of $I_n$.
Move your cursor over the gray area for the complete solution.

Setting $t = 2x$, we get $$I_n = 2 \int_{0}^{\pi/2} \cos^n(x) dx$$ For $n = 2k+1$, we get that \begin{align} I_{2k+1} & = 2 \left( \dfrac{4^k (k!)^2}{(2k+1)!} \right)\\ & \sim 2 \cdot 4^k \times (2 \pi k) \times \left(\dfrac{k}e \right)^{2k} \times \dfrac1{\sqrt{2 \pi (2k+1)}} \left(\dfrac{e}{2k+1} \right)^{2k+1}\\ & = \dfrac{2^{2k+2} \pi k^{2k+1} e^{2k+1}}{e^{2k} \sqrt{2 \pi (2k+1)} (2k+1)^{2k+1}}\\ & = \dfrac{2 \pi (2k)^{2k+1} e}{\sqrt{2 \pi (2k+1)} (2k+1)^{2k+1}}\\ & = \dfrac{\sqrt{2\pi} e}{\sqrt{2k+1}} \left(\dfrac{2k}{2k+1} \right)^{2k+1}\\ & = \dfrac{\sqrt{2\pi} e}{\sqrt{2k+1}} \left(1-\dfrac1{2k+1} \right)^{2k+1}\\ & \sim \dfrac{\sqrt{2\pi}e}{\sqrt{2k+1}} \times e^{-1}\\ & = \sqrt{\dfrac{2 \pi}{2k+1}} \end{align} Hence, $$\sqrt{2k+1} I_{2k+1} \sim \sqrt{2 \pi}$$ You will get the same result for even $n$ as well courtesy Wallis formula.

A: By a change of variables $t\mapsto 2x$, you want to show $$\mathop {\lim }\limits_{n \to  + \infty } 2\sqrt n \int_0^{\frac{\pi }{2}} {{{\cos }^n}x} dx > 0.$$
Now, you need to evaluate $$I\left( n \right) = \int_0^{\frac{\pi }{2}} {{{\cos }^n}x} dx$$
For $n\geq 2$ we can reduce this integral with integration by parts
$$\eqalign{
  & {\cos ^{n - 1}}x = u  \cr 
  & \cos xdx = dv \cr} $$
then $$\eqalign{
  &  - \left( {n - 1} \right){\cos ^{n - 2}}x\sin xdx = du  \cr 
  & \sin x = v \cr} $$
thus $$\int_0^{\frac{\pi }{2}} {{{\cos }^n}x} dx = \int_0^{\frac{\pi }{2}} {\left( {n - 1} \right){{\cos }^{n - 2}}x{{\sin }^2}xdx} $$ for the other term vanishes. But $$\displaylines{
  \int_0^{\frac{\pi }{2}} {\left( {n - 1} \right){{\cos }^{n - 2}}x{{\sin }^2}xdx}  = \left( {n - 1} \right)\int_0^{\frac{\pi }{2}} {{{\cos }^{n - 2}}x\left( {1 - {{\cos }^2}x} \right)dx}  \cr 
   = \left( {n - 1} \right)\int_0^{\frac{\pi }{2}} {{{\cos }^{n - 2}}xdx}  - \left( {n - 1} \right)\int_0^{\frac{\pi }{2}} {{{\cos }^{n - 2}}x{{\cos }^2}xdx}  \cr 
   = \left( {n - 1} \right)I\left( {n - 2} \right) - \left( {n - 1} \right)I\left( n \right) \cr} $$
which means $$I\left( n \right) = \left( {\frac{{n - 1}}{n}} \right)I\left( {n - 2} \right)$$
If $n=2m$ is even, or if $n=2m+1$ is odd we have
$$\displaylines{
  I\left( n \right) = \frac{{2m - 1}}{{2m}}\frac{{2m - 3}}{{2m - 2}} \cdots \frac{3}{4}\frac{1}{2}I\left( 0 \right) \cr 
  I\left( n \right) = \frac{{2m}}{{2m + 1}}\frac{{2m - 2}}{{2m - 1}} \cdots \frac{4}{5}\frac{2}{3}I\left( 1 \right) \cr} $$
for we reduce the number by $2$ each time, so if we start with an odd number, we'll be left with $1$.
So finally since
$$\displaylines{
  I\left( 0 \right) = \frac{\pi }{2} \cr 
  I\left( 1 \right) = 1 \cr} $$
we get for $n$ even $$I\left( n \right) = \frac{{\left( {n - 1} \right)!!}}{{n!!}}\frac{\pi }{2}$$ and for $n$ odd  $$I\left( n \right) = \frac{{\left( {n - 1} \right)!!}}{{n!!}}$$
Thus we have two sequences to consider
$$\displaylines{
  {a_{2m}} = \sqrt {2m} \frac{{\left( {2m - 1} \right)!!}}{{\left( {2m} \right)!!}}\frac{\pi }{2} \cr 
  {a_{2m + 1}} = \sqrt {2m + 1} \frac{{\left( {2m} \right)!!}}{{\left( {2m + 1} \right)!!}} \cr} $$
But Wallis' approximation says $$\mathop {\lim }\limits_{n \to \infty } {\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]^2}\frac{1}{2n} = \frac{\pi }{2}$$
so that
$$\mathop {\lim }\limits_{n \to \infty } \frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}\frac{1}{{\sqrt n }} = \sqrt \pi $$
All things considered $$\displaylines{
  \mathop {\lim }\limits_{m \to \infty } {a_{2m + 1}} = \mathop {\lim }\limits_{m \to \infty } \sqrt {2m + 1} \frac{{\left( {2m} \right)!!}}{{\left( {2m + 1} \right)!!}} \cr 
   = \mathop {\lim }\limits_{m \to \infty } \frac{{\sqrt {2m + 1} }}{{2m + 1}}\frac{{\left( {2m} \right)!!}}{{\left( {2m - 1} \right)!!}} \cr 
   = \mathop {\lim }\limits_{m \to \infty } \frac{1}{{\sqrt {2m + 1} }}\frac{{\sqrt m }}{{\sqrt m }}\frac{{\left( {2m} \right)!!}}{{\left( {2m - 1} \right)!!}} \cr 
   = \mathop {\lim }\limits_{m \to \infty } \sqrt {\frac{m}{{2m + 1}}} \mathop {\lim }\limits_{m \to \infty } \frac{1}{{\sqrt m }}\frac{{\left( {2m} \right)!!}}{{\left( {2m - 1} \right)!!}} = \frac{1}{{\sqrt 2 }}\sqrt \pi  = \sqrt\frac\pi2 \cr} $$

$$\mathop {\lim }\limits_{m \to \infty } {a_{2m}} = \frac{\pi }{{\sqrt 2 }}\mathop {\lim }\limits_{m \to \infty } \sqrt m \frac{{\left( {2m - 1} \right)!!}}{{\left( {2m} \right)!!}} = \frac{\pi }{{\sqrt 2 }}\frac1{\sqrt\pi} = \sqrt\frac\pi2  $$
Hence, the limit exists and is positive, as desired.
A: This integral can be analyzed using Laplace's Method.  Write 
$$\cos{\left(\frac{t}{2}\right)}^n = e^{n \log{\cos{(t/2)}}}$$
Note that $\log{\cos{(t/2)}}$ has a maximum at $t=0$; as $n \to \infty$ there is an increasingly narrow and high peak about $t=0$.  Thus we may Taylor expand about $t=0$ and ignore contributions away from a small neighborhood of $t=0$:
$$\int_0^{\pi} dt \, \cos{\left(\frac{t}{2}\right)}^n \sim \int_0^{\epsilon} dt \, e^{n \log{(1-t^2/8)}} $$
where $\epsilon = O\left(n^{-1/2}\right)$.  Because contributions away from $[0,\epsilon]$ are exponentially small, we may extend the integral out to infinity within the error of the leading-order approximation.  Taylor expanding the log term, we find
$$\int_0^{\pi} dt \, \cos{\left(\frac{t}{2}\right)}^n \sim \int_0^{\infty} dt \, e^{-n t^2/8}  = \frac12 \sqrt{\frac{8 \pi}{n}}$$
Therefore 
$$\lim_{n \to \infty} \sqrt{n} \int_0^{\pi} dt \, \cos{\left(\frac{t}{2}\right)}^n =\sqrt{2 \pi}$$
A: Your first attempt is a good idea, in that it reduces the problem to something you can compute explicitly. Too bad it didn't work out. Perhaps you should instead try to analyze the problem from a different angle. After all, $\cos(t/2)<1$ most places, so the $n$th power goes to zero really fast (geometrically, that is) as $n\to\infty$. That is why you might expect the limit to be zero. You only have the $\sqrt{n}$ factor to stand against this. But it can't win against a geometric descent to zero! So you need to concentrate on the places where $\cos(t/2)$ is close to $1$, that is, the very small $t$. Now you may see why your attempt failed, for $\sin(t/2$ is also very small where $t$ is small, and that blows it for you.
So maybe you should look for another function that is smaller than $\cos(t/2)^n$, yet is close to it when $t$ is small, and which you can integrate explicitly.
