Finding $\lim_{n\to\infty} a_n^n$ where $a_n=\frac{(n!)^2}{(n+k)!(n-k)!}$

Solution 1:

Hint We know that $\text{lim}_{n\to \infty} (1+\frac{x}{n})^n=e^x$ can you see how to proceed.

Solution 2:

We have $$a_n^n=\frac{\prod_{j=1}^{k-1}(1-j/n)^n}{\prod_{j=1}^k(1+j/n)^n}\stackrel{n\to\infty}{\to}\frac{\prod_{j=1}^{k-1}e^{-j}}{\prod_{j=1}^ke^j}=e^{-\sum_{j=1}^{k-1}j-\sum_{j=1}^kj}=e^{-k^2}.$$

Solution 3:

It was a good idea to use Stirling approximation; but for this, go to logarithms $$a_n=\frac{(n!)^2}{(n+k)!(n-k)!}\implies \log(a_n)=2\log(n!)-\log((n+k)!)-\log((n-k)!)$$ So, using the logarithmic version of Stirling approximation and continuing with Taylor expansions for large values of $n$, you should get $$\log(a_n)=-\frac{k^2}{n}+\frac{k^2}{2 n^2}+O\left(\frac{1}{n^3}\right)$$ Continue with Taylor using $$a_n^n=e^{n \log(a_n)}=e^{-k^2}\left(1+\frac{k^2}{2 n}+O\left(\frac{1}{n^2}\right) \right)$$ which shows the limit and also how it is approached.