Stirling's formula manipulation

**mongrel73** · April 16th 2009, 02:07 AM

Hi, I'm stuck on part of this question involving Stirling's formula. I have to show that for $p+q=1$ (they're probabilities),
$\sum_{i=1}^\infty {2n \choose n}p^n q^n = \infty$ if and only if $p=q= \frac{1}{2}$
by using Stirling's approximation, which is $n! \approx \sqrt{2\pi n}\, \left(\frac{n}{e}\right)^{n}$ , for large $n$ .

I know that I should be aiming to show that $<br /> {2n \choose n}p^n q^n \approx \frac{(4pq)^n}{\sqrt{\pi n}}$

I've tried rearranging it but I can't seem to get rid of the terms involving $e$ (among others).

Also, anyone who can solve this is a clever clogs, so maybe they can shed some light on this one, which seems to have defeated the SOS Math community. S.O.S. Mathematics CyberBoard :: View topic - Predictors after a simple linear regression is done

Thanks VERY much

**Moo** · April 16th 2009, 08:51 AM

Hello,

Originally Posted by mongrel73

Hi, I'm stuck on part of this question involving Stirling's formula. I have to show that for $p+q=1$ (they're probabilities),
$\sum_{i=1}^\infty {2n \choose n}p^n q^n = \infty$ if and only if $p=q= \frac{1}{2}$
by using Stirling's approximation, which is $n! \approx \sqrt{2\pi n}\, \left(\frac{n}{e}\right)^{n}$ , for large $n$ .

I know that I should be aiming to show that $<br /> {2n \choose n}p^n q^n \approx \frac{(4pq)^n}{\sqrt{\pi n}}$

I've tried rearranging it but I can't seem to get rid of the terms involving $e$ (among others).

Also, anyone who can solve this is a clever clogs, so maybe they can shed some light on this one, which seems to have defeated the SOS Math community. S.O.S. Mathematics CyberBoard :: View topic - Predictors after a simple linear regression is done

Thanks VERY much

I don't understand where your problem is

${2n \choose n}=\frac{(2n)!}{n!n!}=\frac{(2n)!}{(n!)^2}$

Using Stirling's formula :

$(2n)! \sim \sqrt{4\pi n} \cdot \left(\frac{2n}{e}\right)^{2n}=2\sqrt{\pi n} \cdot 2^{2n} \cdot n^{2n} \cdot e^{-2n}$

$(n!)^2 \sim \left[\sqrt{2\pi n} \cdot \left(\frac ne\right)^n\right]^2=2 \pi n \cdot \left(\frac ne\right)^{2n}=2\pi n \cdot n^{2n} \cdot e^{-2n}$

Hence :
${2n \choose n}=\frac{{\color{red}2}\sqrt{\pi n} \cdot 2^{2n} \cdot {\color{red}n^{2n}} \cdot {\color{red}e^{-2n}}}{{\color{red}2}\pi n \cdot {\color{red}n^{2n}} \cdot {\color{red}e^{-2n}}}=\frac{2^{2n}}{\sqrt{\pi n}}$

And finally :

${2n \choose n} p^n (1-p)^n \sim \frac{(4p(1-p))^n}{\sqrt{\pi n}}$

Now, $p(1-p)=\frac 14 \Leftrightarrow p=\frac 12$ . Otherwise, $p(1-p)<\frac 14$ , for any $p \neq \frac 12$ (and in [0,1] of course). This is VERY easy to prove, by taking the derivative of $f ~:~ x \mapsto x(1-x)$ .

If $p=\frac 12$ , then the series would be equivalent to $\sum_{n\geq 1} \frac{1}{\sqrt{n}}$ , which is a divergent Riemann series.

If $p \neq \frac 12$ , then note that $\frac{1}{\sqrt{n}} \leq 1$ , for any $n \geq 1$

Thus $\frac{(4p(1-p))^n}{\sqrt{\pi n}} \leq \frac{(4p(1-p))^n}{\sqrt{\pi}}$
Since $p(1-p)<\frac 14$ , $4p(1-p)<1$ and therefore, $\sum_{n \geq 1} \frac{(4p(1-p))^n}{\sqrt{\pi n}} \leq \frac{1}{\sqrt{\pi}} \sum_{n \geq 1} (4p(1-p))^n$ is a convergent series.

**matheagle** · April 16th 2009, 10:18 PM

The only tricky part of this was trying to figure out what that subscript i meant.

**Moo** · April 16th 2009, 10:32 PM

Originally Posted by matheagle

The only tricky part of this was trying to figure out what that subscript i meant.

I guess you replied to the wrong thread...

**matheagle** · April 16th 2009, 10:40 PM

NOPE, look at the original sum, instead of using n, the OP used i.

**mongrel73** · April 17th 2009, 06:32 AM

Originally Posted by matheagle

NOPE, look at the original sum, instead of using n, the imposter used i.

Yes, oops. I also can't work out what my problem was. Thanks anyway though...

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