<s>[INST] 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.
 [/INST] 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.

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