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If we know that $I_{n}=\int_0^\frac{\pi}{2}\sin^n(x)\,{\rm d}x$ we need to evaluate:

$\lim_{n\rightarrow\infty}\left[\frac{(2n)!!}{(2n-1)!!}\right]^2\cdot\frac{1}{2n+1}$ ( !! means double factorial ).

Here is all my steps to arrive at a squeeze theorem:

$I_{_{n}}=\frac{n-1}{n}\cdot I_{n-2}$, $\forall x\geq 2$. Therefore:

$I_{_{2k}}=\frac{\pi}{2}\cdot (\prod_{k=0}^{n}\frac{2k+1}{2k+2})=\frac{\pi}{2}\cdot(\prod_{k=1}^{n}\frac{2k-1}{2k})$

$I_{_{2k+1}}=\prod_{k=0}^{n}\frac{2k+2}{2k+3}=\prod_{k=1}^{n}\frac{2k}{2k+1}$

$\Rightarrow I_{2k}\geq I_{2k+1}$ , $\forall x\in[0,\frac{\pi}{2}]$

$\Rightarrow \prod_{k=1}^{n}\frac{2k}{2k+1}\leq\frac{\pi}{2}(\prod_{k=1}^{n}\frac{2k-1}{2k})\mid\cdot\prod_{k=1}^{n}\frac{2k}{2k-1}$

$\Rightarrow \prod_{k=1}^{n}\frac{2k^2}{4k^2-1}\leq\frac{\pi}{2}$

I don't know how can I arrive at $\left[\frac{(2n)!!}{(2n-1)!!}\right]^2\cdot\frac{1}{2n+1}$ and after use squeeze theorem.

Is something $\frac{(2n)!!}{(2n-1)!!}=\prod_{k=1}^{n}\frac{2k}{2k-1}$ ?

I want to continue with this method, if is something who can help me to finish I'll apreciate.

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Lucas
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2 Answers2

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Write things in terms of the Gamma function $$ \frac{(2n)!!}{(2n-1)!!}=\frac{\Gamma(n+1)\,\Gamma(\frac12)}{\Gamma(n+\frac12)} $$ Gautschi's Inequality, proven in this answer, says $$ n^{1/2}\lt\frac{\Gamma(n+1)}{\Gamma(n+\frac12)}\lt(n+1)^{1/2} $$ Since $\Gamma(\frac12)=\sqrt\pi$, $$ \frac{\pi n}{2n+1}\lt\left[\frac{(2n)!!}{(2n-1)!!}\right]^2\frac1{2n+1}\lt\frac{\pi(n+1)}{2n+1} $$ Therefore, using the Squeeze Theorem, $$ \lim_{n\to\infty}\left[\frac{(2n)!!}{(2n-1)!!}\right]^2\frac1{2n+1}=\frac\pi2 $$

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robjohn
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This is the Wallis formula and its derivation can be found here. Also, note that double factorial is a terrible notation to use and kindly refrain from using it in future.

Adhvaitha
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  • @Lucas Did you read the second part on the link? Do not comment without reading it completely. – Adhvaitha May 08 '15 at 15:49
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    the double factorial is often more compact than trying to write the same thing with factorials. I see nothing wrong with it. – robjohn May 08 '15 at 16:03
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    @robjohn There are much more elegant and clear ways to express double factorial. $n!! = 2^{n/2} (n/2)!$ when $n$ is even and $n!! = \dfrac{n!}{2^{(n-1)/2} ((n-1)/2)!}$. Ofcourse, notions and more important than notations but that doesn't mean we can abuse notations. When you read typically we parse from left to right or right to left, a single character by character. Hence, a typical interpretation is $(n!)!$. And an even better notation would be $$\prod_{k =0}^{(n-1)/2} (n-2k) \text{ instead of }n!!$$ – Adhvaitha May 08 '15 at 16:06