Error of Stirling’s approximation for Binomial with central limit theorem

Question

So the question asks: Let $X_n$~Bin(2n,1/2),use Stirling’s approximation for $n!$ to show $P [X_n = n]$~ $1/√(πn)$ as $n→ ∞$, and show the error in the estimate for $P [X_n ≤ n]$, given by the central limit theorem, is approximately $1/(2√(πn))$

So far I got:

using Stirling’s approximation for $n!$:

$n! ≈(n/e)^n √(2πn)$

$\binom {2n} {1/2} = \frac {2n!} {({1/2})!(2n-{1/2})!}=\frac {(2n/e)^{2n} 2√(πn)} {(√π / 2)((2n-1/2)/e)^{2n-1/2}√(2π(2n-1/2))}$

But I can't see why it is equal to $1/√(πn)$ when n goes to infinity.

Also I am really sorry about the format which causes incovinence for reading, so I also attached the screenshot of my work below.

Also how can I use central limit theorem to find the error?

Answer 1

I get $\binom {2n} {1/2} \approx\dfrac { 2\sqrt{2n}} {\sqrt{\pi}}$.

Since $\binom {2n} {0} = 1 $ and $\binom {2n} {1} = 2n $, it seems reasonable that $\binom {2n} {1/2} $ is of order $\sqrt{n}$.

Here's your chance to find another mistake of mine.

$\begin{array}\\ \binom {2n} {1/2} &= \dfrac {(2n)!} {({1/2})!(2n-1/2)!}\\ &\approx\dfrac {(2n/e)^{2n} 2\sqrt{πn}} {(\sqrt{π} / 2)((2n-1/2)/e)^{2n-1/2}\sqrt{2π(2n-1/2)}}\\ &=\dfrac {(2n)^{2n} 4\sqrt{πn}} {e^{1/2}\sqrt{π}(2n-1/2)^{2n-1/2}√(2π(2n-1/2))}\\ &=\dfrac {(2n)^{2n} 2\sqrt{2πn}} {e^{1/2}\pi(2n-1/2)^{2n-1/2}√(2n-1/2)}\\ &=\dfrac {(2n)^{2n} 2√{(2πn)}} {e^{1/2}\pi(2n-1/2)^{2n}}\\ &=\dfrac { 2\sqrt{2n}} {e^{1/2}\sqrt{\pi}}\left(\dfrac{2n}{2n-1/2}\right)^{2n}\\ &=\dfrac { 2\sqrt{2n}} {e^{1/2}\sqrt{\pi}}\left(1-\dfrac{1/2}{2n}\right)^{-2n}\\ &=\dfrac { 2\sqrt{2n}} {e^{1/2}\sqrt{\pi}}e^{1/2} \qquad\text{since }(1-\frac{a}{x})^x \approx e^{-a} \text{ so }(1-\frac{a}{x})^{-x} \approx e^{a}\\ &=\dfrac { 2\sqrt{2n}} {\sqrt{\pi}} \end{array} $

Answer 2

As you wrote, let $$A_n=\binom{2 n}{\frac{1}{2}}=\frac{2 \Gamma (2 n+1)}{\sqrt{\pi } \Gamma \left(2 n+\frac{1}{2}\right)}$$ Take logarithms $$\log(A_n)=\log(\frac 2 \pi)+\log(\Gamma (2 n+1))-\log(\Gamma (2 n+\frac{1}{2}))$$ $$\log(A_n)=\log(\frac 2 \pi)+\log((2 n)!)-\log((2 n-\frac{1}{2})!)$$ Now, use Stirling approximation $$\log(p!) \sim p\log( p) - p + \frac12\log(2\pi p) +\frac{1}{12p}+\cdots$$ and apply it for $p=2n$ and $p=2n-\frac12$. This gives$$\log(A_n)\sim\log(\frac 2 \pi)+\Big(2n\log(2n) - 2n + \frac12\log(4\pi n) +\frac{1}{24n}+\cdots \Big)-\Big((2n-\frac 12)\log(2n-\frac 12) - (2n-\frac 12) + \frac12\log(2\pi (2n-\frac 12)) +\frac{1}{12(2n-\frac 12)}+\cdots \Big)$$ Continuing expansions writing $$\log(2n-\frac 12)=\log(2n)+\log(1-\frac 1 {4n}) $$ $$\frac{1}{12(2n-\frac 12)}=\frac 1 {24n} \frac 1 {1-\frac 1{4n}} $$ and Taylor series for large values of $n$, we finally get $$\log(A_n)=\log \left(2 \sqrt{\frac{2}{\pi }}\right)+\frac{\log (n)}{2}+\frac{1}{16 n}+O\left(\frac{1}{n^3}\right)$$ Now, using $A_n=e^{\log(A_n)}$ and continuing with Taylor expansions $$A_n =2 \sqrt{\frac{2n}{\pi }} +\frac{1}{4 \sqrt{2 \pi n} }+\frac{1}{128 \sqrt{2 \pi n^3} }+O\left(\frac{1}{n^{5/2}}\right)$$

For illustration purposes, I put below some values given using the successive approximations and the "exact" value.

$$\left( \begin{array}{ccccc} n & 1\, term & 2\,terms & 3\, terms & exact \\ 5 & 3.568248232 & 3.612851335 & 3.613130105 & 3.613112507 \\ 10 & 5.046265044 & 5.077804201 & 5.077902760 & 5.077899663 \\ 15 & 6.180387232 & 6.206138846 & 6.206192495 & 6.206191373 \\ 20 & 7.136496465 & 7.158798016 & 7.158832862 & 7.158832316 \end{array} \right)$$