How can I calculate $\lim_{n\rightarrow\infty} \frac{7^{\sqrt{n}}\cdot(n/2)!\cdot(n/2)!}{n!} $?

Question

How can I calculate this limit? $$\lim_{n\rightarrow\infty} \frac{7^{\sqrt{n}}\cdot(n/2)!\cdot(n/2)!}{n!} $$

I thought to try the ratio test, but I don't know how to do it because that I get $(\frac{n+1}{2})!$ and $(\frac{n}{2})!$ such that it's not define.

Answer 1

HINT

Let apply Stirling’s approximation

$$n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$$

that is

$$\frac{7^{\sqrt{n}}\cdot(n/2)!\cdot(n/2)!}{n!}\sim \frac{7^{\sqrt{n}}\left(\sqrt{2 \pi \frac n 2}\left(\frac{n}{2e}\right)^{\frac n 2}\right)^2}{\sqrt{2 \pi n}\left(\frac{n}{e}\right)^n}=\frac{\pi}{\sqrt{2 \pi}}\frac{7^{\sqrt{n}} n^{n+1}}{2^nn^{n+\frac12}}=\frac{\pi}{\sqrt{2 \pi}}\frac{7^{\sqrt{n}} n^{\frac12}}{2^n}$$

Answer 2

Let $$ a_n=\frac{7^{\sqrt{n}}\cdot(n/2)!\cdot(n/2)!}{n!}. $$ Consider a subsequence of this sequence given by $$ a_{4k^2}=\frac{7^{\sqrt{(4k^2)}}\cdot((4k^2)/2)!\cdot((4k^2)/2)!}{(4k^2)!} =\frac{7^{2k}\cdot(2k^2)!\cdot(2k^2)!}{(4k^2)!} \leq \frac{7^{2k}(2k^2)!(2k^2)!}{(2k^2)!(2k^2)!(2k^2)!}\to 0 $$