Find limit of $x_n=\frac{n^{\frac{n}{2}}}{n!}$

Question

Find limit of $$x_n=\frac{n^{\frac{n}{2}}}{n!}$$ I know, that it goes to $0$ by WolframAlpha, but it's not clear for me why. Also I know, that $$x_n=\frac{n^n}{n!}=+\infty$$

Answer 1

By the ratio test, since we know that

$$\lim_{n\to\infty}\frac{x_{n+1}}{x_n}=\lim_{n\to\infty}\frac{\left(1+\frac1n\right)^{n/2}}{\sqrt{n+1}}\le\lim_{n\to\infty}\frac{\left(1+\frac1n\right)^n}{\sqrt{n+1}}=\frac e\infty=0$$

Then it follows through the term test that

$$\lim_{n\to\infty}x_n=0$$

Answer 2

HINT:

Since $n!>(n/e)^n$ (this is shown two ways in the notes herein), we have

$$\frac{n^{n/2}}{n!}\le \frac{n^{n/2}}{(n/e)^n}$$

---

NOTES:

An elementary way to show $n!>(n/e)^n$ is to use induction. Then, we have

$$\left(\frac{n+1}{e}\right)^{n+1}=\underbrace{\left(\frac{n}{e}\right)^n}_{<n!\,\text{by assumption}} \underbrace{\left(1+\frac1n\right)^n}_{<e}\left(\frac{n+1}{e}\right)<(n+1)!$$

Another approach is to write

$$n!=n^n e^{n\,\frac1n\sum_{k=1}^n\log(k/n)}$$

Recognizing the Riemann sum $\frac1n\sum_{k=1}^n\log(k/n)>\int_0^1 \log(x)\,dx=-1$, we have

$$n!>(n/e)^n$$

as was to be shown!

Answer 3

You know that $n! \geq n^n/e^n$ $(*)$ hence : $$ \frac{n^{n/2}}{n!} < \frac{e^n}{n^n} n^{n/2} = \left(\frac{e\sqrt{n}}{n}\right)^n = \left(\frac{e}{\sqrt{n}}\right)^n \overset{(**)}{<} \left(\frac{e}{2e}\right)^n = \frac{1}{2^n} $$ $(**)$ Assuming $n \geq 30$

---

$(*)$ This can be derived by noticing that $$ \ln(n!) \geq \int_1^n \ln(x)\ dx = n \ln(n) -n +1 \implies n! \geq \frac{n^n}{e^n}\cdot e \geq \frac{n^n}{e^n} $$