Integral of $\sin^n(x)$, recurrence relation, some properties

Question

Practicing the manipulation of recurrence relations, I'm stuck on this :

Defining $I(n)=\int_{0}^{\pi/2}sin^n(x)dx$, I got the recurrence relation $nI(n)=(n-1)I(n-2)$ for $n\ge2$.

Now I'm also trying to prove that $I(n) \le I(n-1)$ for all $n\ge 1$.

[I've tried by induction on the odd and even possibilities but it doesn't give anything concluent.]

and that $2n/2n+1 \le I(2n+1)/I(2n) \le 1$

The second part of the equality can be obtained easily from above but I have no idea for the first one.

Can you hint me? I've never been good with these...

Answer 1

With your induction relation, you can write $I(2n)=\frac{(2n-1)(2n-3)...}{2n(2n-2)...}I(0)$. You can symplify this in $I(2n)=\frac{(2n)!}{4^n(n!)^2}I(0)$. The same kind of relation exists for I(2n+1).

Answer 2

I finally found something =)

I(n) is a decreasing sequence so $\frac{I(n+1)}{I(n)} \ge \frac{I(n+2)}{I(n)} = \frac{n+1}{n+2}$.

Then $\frac{I(2n+1)}{I(2n)} \ge \frac{2n+1}{2n+2}$!!!

and $\frac{2n+1}{2n+2} \ge \frac{2n}{2n+1}$.