Use Stirling's Formula to find constants

Question

I'm trying to use Stirling's formula to find constants $a, b, c$ such that

$\frac{1}{n+1}$${2n}\choose{n}$ ~ $ca^{n}n^{b}$

So far, I've worked the left hand side like this:

$\frac{1}{n+1}$${2n}\choose{n}$$\;= \frac{1}{n+1} \frac{(2n)!}{n!(2n-n)!} \;= \frac{(2n)!}{n!n!(n+1)} \;= \frac{(2n)!}{n!(n+1)!}$

Using Stirling's formula $n!=n^ne^{-n}\sqrt{2\pi n}$
$\frac{2n^{2n}e^{-2n}\sqrt{2\pi 2n}}{n^ne^{-n}\sqrt{2\pi n} \;*\; (n+1)^(n+1)e^{-(n+1)}\sqrt{2\pi (n+1)}}$

I try to simplify the equation above, but is there a better way to set this up to get it closer to the $ca^{n}n^{b}$ form?

Answer 1

In Is it true that for $n \ge5$, ${{3n} \choose {2n}} > \frac{6^n}{n}$ I prove this result:

$\binom{an}{bn} \sim \sqrt{\dfrac{ a}{2\pi bn(a-b)}}\left(\dfrac{a^a}{b^b(a-b)^{a-b}}\right)^n $. Explicit error bounds are also given there.

If $ b=1$, this becomes $\binom{an}{n} \sim \sqrt{\dfrac{ a}{2\pi n(a-1)}}\left(\dfrac{a^a}{(a-1)^{a-1}}\right)^n $.

Finally, if $a=2$ this becomes the well-known $\binom{2n}{n} \sim 4^n\sqrt{\dfrac{ 1}{2\pi n}} $.

Therefore $\dfrac1{n+1}\binom{2n}{n} \sim \dfrac{4^n}{(n+1)\sqrt{n}}\sqrt{\dfrac{ 1}{2\pi }} = \dfrac{4^n}{n^{3/2}(1+1/n)}\sqrt{\dfrac{ 1}{2\pi }} = \dfrac{4^n}{n^{3/2}}(1-\dfrac1{n}+O(\dfrac1{n^2}))\sqrt{\dfrac{ 1}{2\pi }} $.

Answer 2

The only real improvement I can suggest is to write $(n+1)!\approx n\cdot n!$. That's definitely worth doing so you don't mistakenly approximate $(n+1)^{n+1}$ as $n^{n+1}$ instead of $e$ times that since $(1+1/n)^{n+1}\approx e$. You might also find taking logarithms throughout helps.