Evaluate $\lim_{n\to \infty} \frac{5^n(n!)^2}{(2n)!}$

Question

Ive got stuck in how to show that $\lim_{n\to \infty} \frac{5^n(n!)^2}{(2n)!}$ tends to $\infty$.

If I try to develop it further, I get: $$\lim_{n\to \infty} \frac{5^n(n!)^2}{(2n)!}=\lim_{n\to \infty} \frac{5^n(n!)}{(n+1)(n+2)...(2n)}=\lim_{n\to \infty} \frac{5^n(n!)}{n^n(1+\frac{1}{n})(1+\frac{2}{n})...(2)}\ge{??}$$

and no matter what I choose to exchange the denominator, in every way I make it too big and then it tends to $0$ instead of to $\infty$.

Would appreciate your help.

Answer 1

Try to show that $(2n)! \le 4^n (n!)^2$. You can do this by considering $$(2n)! = 1 \cdot 3 \cdot 5 \cdots (2n-1) \cdot 2 \cdot 4 \cdot 6 \cdots (2n) \le 2^2 \cdot 4^2 \cdot 6^2 \cdots (2n)^2.$$

Answer 2

Recall the Stirling's formula: \begin{align*} n!\sim \sqrt{2\pi n}\frac{n^{n}}{e^{n}}. \end{align*} So \begin{align*} \frac{5^{n}(n!)^{2}}{(2n)!}\sim\frac{5^{n}(2\pi n) \frac{n^{2n}}{e^{2n}}}{\sqrt{4\pi n}\frac{(2n)^{2n}}{e^{2n}}}=\Big(\frac{5}{4}\Big)^{n}\sqrt{\pi n}, \end{align*} which tends to $\infty$ as $n\to\infty$.

Answer 3

$\binom{2n}{n}$ is the middle term for the expansion of $(1+1)^{2n}=4^n,$ so certainly $\binom{2n}{n}<4^n.$ Your term is $5^n$ divided by $\binom{2n}{n}$ making that term at least $\frac{5^n}{4^n}$ which tends to $+\infty.$

Answer 4

${u_{n+1}\over u_n}= {5(n+1)^2\over (2n+2).(2n+1)} \sim 5/4$, hence $u_n\geq C(5/4)^n$. Easier than using Stirling...

Answer 5

The sequence grows exponentially: $$ \frac{\frac{5^{n+1}((n+1)!)^2}{(2(n+1))!}}{\frac{5^n(n!)^2}{(2n)!}} = \frac{5(n+1)^2}{(2n+1)(2n+2)} = \frac{5(n+1)}{2(2n+1)} > \frac54-\epsilon. $$