# Evaluating the limit $\lim_{n\to \infty} \frac{n^n}{(2n)!}$

I'm trying to evaluate this limit. $$\lim_{n\to \infty} \frac{n^n}{(2n)!}$$ I know that $$n^n$$ grows faster than $$n!$$. So the result should be $$+\infty$$. But the real result is zero. Why?

• Try using Stirling's approximation and substituting in Commented Feb 10, 2021 at 15:07

Note that $$0 \le \dfrac{n^n}{(2n)!} = \dfrac{1}{n!}\cdot\dfrac{n \cdot n \cdots n}{(n + 1)\cdot(n+2)\cdots(2n)} \le \dfrac{1}{n!} \to 0.$$

$$n^n$$ grows faster than $$n!$$ but slower than $$(2n)!$$, since:

$$(2n)! = (2n)(2n-1)\dots(n+1)n! > n^nn!$$

Hence $$\lim_{n\to\infty}\frac {n^n}{(2n)!} \le \lim_{n\to \infty}\frac {n^n}{n^nn!} = \lim_{n\to \infty}\frac {1}{n!} = 0$$

The confusion may be from the fact that $$(2n)! \ne 2n!$$.

$$(2n)!\approx\sqrt{4\pi n}\left(\frac{2n}e\right)^{2n}=2\sqrt{\pi}\left(\frac2e\right)^{2n}n^{2n+1/2}$$ so: $$\frac{n^n}{(2n)!}=\frac{n^n}{2\sqrt{\pi}\left(\frac2e\right)^{2n}n^{2n+1/2}}=\frac{1}{2\sqrt{\pi}}\left(\frac e2\right)^{2n}n^{\frac12-n}$$ now as we take the limit notice that the $$e/2$$ term grows slower than the $$n$$ term when both raised to a power of $$n$$ and so it goes towards $$0$$

$$\dfrac{\frac{(n+1)^{n+1}}{(2n+2)!}}{\frac{n^n}{(2n)!}}=\dfrac{(2n)!\cdot (n+1)^{n+1}}{n^n\cdot (2n+2)!}=\dfrac{n+1}{(2n+1)(2n+2)}\bigg(1+\dfrac{1}{n}\bigg)^n\to0<1$$