Calculating $\lim\limits_{j \to \infty} \frac{j^{j/2}}{j!}$

Question

\begin{align*} & \lim\limits_{j \to \infty}{j^{\,j/2} \over j!} \\ \end{align*}

This problem is from a real analysis textbook in the chapter on the natural log and properties of exponents. I'm struggling with how to approach this. I don't think you would use L'Hopital's rule.

Answer 1

Let $L$ be the limit. Then, it is :

$$L = \lim_{n \to \infty} \frac{n^{n/2}}{n!} \Leftrightarrow \ln L = \lim_{n \to \infty} \left(\ln n^{n/2}-\ln n! \right)$$ $$\Leftrightarrow$$ $$\ln L =\lim_{n \to \infty} \left(\frac{n}{2}\ln n - \ln n!\right)$$

Now we will use the fact

$$\ln n! = n \ln n - n + \mathcal{O}\left(\ln n\right)$$

which is called Stirling's Approximation (actually it is a consquence of the original formula, more information can be found in the link and credits to Jack D'Aurizio for mentioning it as well).

$$\begin{align*}\ln L &=\lim_{n \to \infty} \left(\frac{n}{2}\ln n - n \ln n + n - \mathcal{O}\left(\ln n\right)\right)\\&=\lim_{n \to \infty} \left(n-\frac{n}{2}\ln n\; -\mathcal{O}\left(\ln n\right)\right) \end{align*}$$

But $\mathcal{O}\left(\ln n\right) \to 0$ and you should be able to finish now.

Answer 2

Without invoking Stirling's approximation, the Hermite-Hadamard inequality, the $\log$-convexity of $\Gamma$ or the trapezoid rule (all viable approaches), one may simply notice that by defining $$ a_n = \frac{n^n}{n!^2} $$ we have $$ \frac{a_{n+1}}{a_n} = \frac{1}{n+1}\underbrace{\left(1+\frac{1}{n}\right)^n}_{\text{bounded}} \to 0 $$ as $n\to +\infty$, hence your limit is zero as well.

Answer 3

The ratio test is a more elementary tool: $$\frac{(j+1)^\tfrac{j+1}2}{(j+1)!}\,\frac{j!}{j^{\tfrac j2}}=\frac{(j+1)^{\tfrac{j}2}\sqrt{j+1}}{(j+1)j^{\tfrac j2}}=\underbrace{\sqrt{\Bigl(1+\frac1j\Bigr)^j}}_{\begin{array}{c}\downarrow\\\sqrt{\mathrm e}\end{array}}\,\frac1{\sqrt{j+1}}\to 0$$