# $\lim_{x \to \infty} {{(x!)^p}\over {x^x}}$

After some experimentation, I am pretty sure that $$\lim_{x \to \infty} {{(x!)^p}\over {x^x}} \to \infty$$ for $$p \gt 1$$, and that $$\lim_{x \to \infty} {{(x!)^p}\over {x^x}} \to 0$$ for $$p \le 1$$.

I do not, however, have any proof of this, and I was wondering if someone could help me. I have tried applying the ratio test, but I am not quite sure how. Thanks in advance for the help!

• I assume you mean $p\le 1$ vs $p> 1$. – J.G. Mar 26 at 14:25
• Please edit your question. First of all, you probably want to separate the cases based on $p$ and not on $x$. Second, the expression is larger if $p>1$, so the limit should also be larger... – 5xum Mar 26 at 14:29

The Stirling approximation states $$x!\sim\sqrt{2\pi}x^{x+1/2}e^{-x}$$. Thus $$\lim_{x\to\infty}\frac{x!^p}{x^x}=(2\pi)^{p/2}\exp\lim_{x\to\infty}\left[\left((p-1)x+\frac{p}{2}\right)\ln x-px\right].$$If $$p>1$$, this is $$\exp\infty=\infty$$ due to the $$x\ln x$$ term. If $$p<1$$, the same logic obtains a limit of $$\exp-\infty=0$$. If $$p=1$$, we get $$\sqrt{2\pi}\exp\lim_{x\to\infty}(\tfrac12\ln x-x)=\exp-\infty=0$$ because $$\ln x\in o(x)$$.

Solution via Ratio Test:

The ratio between two successive terms is: $${{(n!)^p}\over{n^n}}\over{{((n+1)!)^p}\over{(n+1)^{n+1}}}$$

Simplifying gives: $$\left({{n!}\over{(n+1)!}}\right)^p \cdot {{(n+1)^{n+1}}\over {n^n}}$$

Further simplification yields: $${1\over (n+1)^p} \cdot \left({{n+1}\over {n}}\right)^n\cdot (n+1)$$

With some rearrangement we have: $${1\over (n+1)^{p-1}} \cdot \left({{n+1}\over {n}}\right)^n$$

The second term goes off to $$e$$ as $$n$$ goes to infinity. The first goes to $$0$$ when $$p$$ is larger than $$1$$, $$1$$ when $$p$$ is equal to $$1$$, and $$+\infty$$ when $$p$$ is less than $$1$$. In the first case, the ratio between successive terms is $$0$$. In the second case, it is $$e$$, and in the third case it is unbounded. By the ratio test, it must convege in the first case and diverge in the other two.