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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!

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    $\begingroup$ I assume you mean $p\le 1$ vs $p> 1$. $\endgroup$ – J.G. Mar 26 at 14:25
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    $\begingroup$ 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... $\endgroup$ – 5xum Mar 26 at 14:29
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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 $\ln x\in o(x)$.

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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.

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