1. ## Stirling's formula manipulation

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 $
{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

2. 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 $
{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.

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

4. 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...

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

6. 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...