Limit of $\frac{\frac{(\sqrt{n}+n-1)!}{(\sqrt{n})!(n-1)!}}{2^{\sqrt{n}}}$ as $n \rightarrow \infty$. I'm trying to show $$\lim_{n \rightarrow \infty} \frac{\frac{(\sqrt{n}+n-1)!}{(\sqrt{n})!(n-1)!}}{2^{\sqrt{n}}} = \infty$$ as wolfram says it should. But can't seem to get it. This is what I've tried so far:
$$\begin{align}\frac{1}{2^{\sqrt{n}}}\frac{(\sqrt{n}+n-1)!}{(\sqrt{n})!(n-1)!} & =  \frac{1}{2^{\sqrt{n}}}\frac{[\prod_{i=1}^{n-1} (\sqrt{n}+i)](\sqrt{n})!}{(\sqrt{n})!(n-1)!} \\ &= \frac{1}{2^{\sqrt{n}}} \frac{\prod_{i=1}^{n-1} (\sqrt{n}+i)}{(n-1)!}\end{align}.$$
But I'm not sure where to go from here.
 A: Without Stirling's approximation, we can do the following:
let $p \le \sqrt n \le p+1$ Then
$$\lim_{n \rightarrow \infty} \frac{\frac{(\sqrt{n}+n-1)!}{(\sqrt{n})!(n-1)!}}{2^{\sqrt{n}}} > \lim_{n \rightarrow \infty} \frac{\frac{(p+n-1)!}{(p+1)!(n-1)!}}{2^{\sqrt{n}}}= \lim_{n, p \rightarrow \infty} \frac{n\cdot (n+1) \cdot ... \cdot(n+p-1)}{2^{p+1}(p+1)!} > $$ $$\lim_{p \rightarrow \infty} \frac{p^2\cdot (p^2+1) \cdot ... \cdot(p^2+p-1)}{2^{p+1}(p+1)!} > \lim_{p \rightarrow \infty} \frac{(p^2-1)^{p}}{2^{p+1}(p+1)^{p+1}}=\lim_{p \rightarrow \infty} \frac{(p-1)^{p}}{2^{p+1}(p+1)}=\infty$$
A: $$a_n=\frac{1}{2^{\sqrt{n}}}\frac{(\sqrt{n}+n-1)!}{(\sqrt{n})!(n-1)!}$$ Take logarithms
$$\log(a_n)=-\sqrt n \log(2)+\log((\sqrt{n}+n-1)!)-\log((\sqrt{n})!)-\log((n-1)!)$$ Use three times Stirling approximation and continue with Taylor series
$$\log(a_n)=\sqrt{n} \left(\log \left(\frac{\sqrt{n}}{2}\right)+1\right)+\frac{1}{4}
   \left(2-\log \left(4 \pi ^2 n\right)\right)+O\left(\frac{1}{\sqrt n}\right)$$
So, $ \log(a_n)\to \infty$
