<s>[INST] How can a binomial coefficient can be approximated by using Stirling's formula? I've met some difficulties with such question:
How can we approximate a binomial coefficient by using a Stirling's factorial approximation.
I've evaluate a little bit and got this
How can I transform the right side of this equation for getting estimation like (1 + ?/n + O(1/n^2))
 [/INST] Thank you all for your answers, but I mentioned that I have to use slightly different factorial approximation:
$$
n!=\sqrt{2\pi n}\cdot(\frac{n}{e})^n\cdot\left(1+\frac{1}{12n}+\frac{1}{288n^2}+O(\frac{1}{n^3})\right)
$$
So, by using this approximation it's necessary to obtain approximation for the binomial coefficient $\binom{an}{n}$. Sorry that I defined the task conditions rudely:
$a$ - is a constant and $n\rightarrow \infty$. 
So, after doing some conversions I've got this one:
$$\sqrt{\frac{a}{2\pi(a-1)n}}\cdot\left(\frac{a^a}{(a-1)^{a-1}}\right)^n\cdot\frac{1+\frac{1}{12an}+\frac{1}{288(an)^2}}{\left({1+\frac{1}{12n}+\frac{1}{288n^2}}\right)\cdot\left(1+\frac{1}{12(a-1)n}+\frac{1}{288((a-1)n)^2}\right)}$$
Then let's transform the right side of the expression:
$$\frac{1+\frac{1}{12an}+\frac{1}{288(an)^2}}{\left({1+\frac{1}{12n}+\frac{1}{288n^2}}\right)\cdot\left(1+\frac{1}{12(a-1)n}+\frac{1}{288((a-1)n)^2}\right)}=\\=\displaystyle{\frac{1+\frac{1}{12an}+O(\frac{1}{n^2})}{1+\frac{a}{12(a-1)n}+O(\frac{1}{n^2})}}$$
Finally, we get:
$$\displaystyle{1+X=\frac{1+\frac{1}{12an}}{1+\frac{a}{12(a-1)n}}}\rightarrow X=\frac{a-1-a^2}{a(12(a-1)n+a)}$$So, the right side of our approximation can be submitted as:
$$1+\frac{a-1-a^2}{12a(a-1)}\cdot\frac{1}{n}+O\left(\frac{1}{n^2}\right)$$
And the whole answer is:
$$\sqrt{\frac{a}{2\pi(a-1)n}}\cdot\left(\frac{a^a}{(a-1)^{a-1}}\right)^n\cdot\left(1+\frac{a-1-a^2}{12a(a-1)}\cdot\frac{1}{n}+O\left(\frac{1}{n^2}\right)\right)$$
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