Limit of $ \int_{0}^{\frac{\pi}{2}}\sin^n xdx$ and probability I begin with this problem:

Calculate the limit of $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^n xdx$ when $n\rightarrow \infty$.

It's natural to think of recurrence relation. Let $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^n xdx=I_n$. By integration by part: $$\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^n xdx=-\sin^{n-1}x\cos x|_{0}^{\pi/2}+(n-1)\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{n-2}\cos^2 x xdx$$
or $I_n=(n-1)(I_{n-2}-I_n)$. Thus $$I_n=\frac{n}{n-1}\cdot I_{n-2}$$
So we can find a formula for $I_n$ with odd $n$ : $$I_n=\frac{2\cdot 4\cdot 6\cdots (n-1)}{3\cdot 5\cdot 7\cdots n} $$
From the recurrence relation, we have $(I_n)$ decreases and is bounded below by $0$, then it has a limit $L\ge 0$. We will prove that $L=0$.
Suppose $L>0$. First, consider the series $$1+\frac{1}{2}+\frac{1}{3}+\cdots$$ 
This series diverges. Suppose the series $\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots$ converges. Since $\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots>\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots$, the series $\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots$ also converges. This yields that $1+\frac{1}{2}+\frac{1}{3}+\cdots$ converges, which is a contradiction. So $\frac{1}{3}+\frac{1}{5}+\frac{1}{7}\cdots$ diverges.
By Lagrange theorem, there exists $c_n\in [n-1,n]$ ($n\ge 3$) such that
$$\ln (n-1)-\ln n=-\frac{1}{c_n}$$
For all $n\ge 3$: $$\frac{1}{c_3}+\frac{1}{c_5}+\frac{1}{c_7}+\cdots>\frac{1}{3}+\frac{1}
{5}+\frac{1}{7}+\cdots$$
so $\frac{1}{c_3}+\frac{1}{c_5}+\frac{1}{c_7}+\cdots$ diverges by comparison test. Thus $\left[-\left(\frac{1}{c_3}+\frac{1}{c_5}+\frac{1}{c_7}+\cdots\right)\right]$ also diverges.
Back to our problem. Since $L>0$, then $\lim \ln I_n=\ln L$. This means $\ln(2/3)+\ln(5/4)+\cdots=\ln L$, or $-\frac{1}{c_3}-\frac{1}{c_5}-\cdots=\ln L$, contradiction. Thus $L=0$.
I have some questions for the above problem:


*

*Is there another way to compute the limit of $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^n xdx$? My solution is quite complicated

*Rewrite $I_n$ as $\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\cdots \left(1-\frac{1}{n}\right)$. Consider a probability problem: in a test of multiple choice questions, each question has exactly one correct answer. The first question has $3$ choices, the second has $5$ choices, etc. By the above result, if the tests has more and more questions, then the probability for the student to be false at all is smaller and smaller. This sounds interesting, because the questions have more and more choices. Is there any generalized result for this (fun) fact?  
 A: Here is a possibly simpler way for proving your theorem. Since $\sin x$ is strictly increasing in the range $[0,\pi/2]$, for each $\epsilon > 0$ we can bound
$$
\begin{align*}
\int_0^{\pi/2} \sin^n x \, dx &= \int_0^{\pi/2-\epsilon} \sin^n x \, dx + \int_{\pi/2-\epsilon}^{\pi/2} \sin^n x \, dx \\ &\leq \frac{\pi}{2} \sin^n(\tfrac{\pi}{2}-\epsilon) + \epsilon.
\end{align*}
$$
Since $\sin(\tfrac{\pi}{2}-\epsilon) < 1$, we conclude that $\int_0^{\pi/2} \sin^n x \, dx \leq 2\epsilon$ for all large enough $n$. In particular, $\limsup_{n\to\infty} \int_0^{\pi/2} \sin^n x \, dx \leq 0$. Since $\int_0^{\pi/2} \sin^n x \, dx \geq 0$, we conclude that the limit exists and equals $0$.
A: You asked if there were other ways. 
You may use the dominated convergence theorem since $\sin^n(x)$ converges pointwise to $1_{\{0\}}$.
You may also write an $\epsilon-\delta$ proof, that I can detail if you want me to.
A: There is a simple way of proving that the limit is zero. Since:
$$\forall x\in[-\pi/2,\pi/2],\qquad \cos x \leq 1-\frac{4x^2}{\pi^2}, $$
we have:
$$ I_n = \int_{0}^{\pi/2}\sin^n(x)\,dx = \int_{0}^{\pi/2}\cos^n x\,dx \leq \frac{\pi}{2}\int_{0}^{1}(1-x^2)^n\,dx\leq\frac{\pi}{2}\int_{0}^{1}e^{-nx^2}\,dx $$
so:
$$ I_n \leq \frac{\pi}{2}\cdot\sqrt{\frac{\pi}{4n}}=O\left(\frac{1}{\sqrt{n}}\right). $$

As an alternative technique, notice that:
$$ I_n^2 = \frac{\pi}{2n}\cdot\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)\Gamma\left(1+\frac{n}{2}\right)}\leq\frac{\pi}{2n}$$
since $\Gamma$ is a log-convex function due to the Bohr-Mollerup theorem.
