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I have to compute $\lim_{n\rightarrow\infty}\frac{n^n}{(n!)^2}$.

I tried say that this limit exists and it's l, so we have $\lim_{n\rightarrow\infty}\frac{n^n}{(n!)^2} = L$ then I rewrited it as: $\lim_{n\rightarrow\infty}(\frac{\sqrt n}{\sqrt[n]{n!}})^{2n}$ then I used natural log over the whole expresion but didn't got into a nice place.

I don't know about Pi function or gamma function so therefore can't really use L'Hospital's rule.

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By ratio test

$$\frac{(n+1)^{n+1}}{((n+1)!)^2}\frac{(n!)^2}{n^n}=\frac1{n+1}\left(1+\frac1n\right)^n\to 0$$

then

$$\lim_{n\rightarrow\infty}\frac{n^n}{(n!)^2}=0$$

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Another way. Note that as $n\to +\infty$, $$\frac{n^n}{(n!)^2}=\prod_{k=1}^n\frac{n}{ k(n+1-k)}\leq \frac{n}{ \lfloor\frac{n+1}{2}\rfloor (n+1-\lfloor\frac{n+1}{2}\rfloor)}\leq \frac{4}{n}\to 0$$ because each factor $\frac{n}{ k(n+1-k)}\leq 1$, and the smallest one is in the middle.

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By the root test, the limit is $n/(n!)^{2/n}$.

By looking at the last 2/3 of 1 to n, $n! > (n/3)^{2n/3}$ so $(n!)^{2/n} > (n/3)^{4/3}$ so the ratio is less than $3^{4/3}/n^{1/3}$ which goes to zero.

This method csn be used to shiw that $n^n/(n!)^a \to 0$ for any $a > 1$.

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{n!}$-Stirling Asymptotic Expansion:

\begin{align} \lim_{n \to \infty}{n^{n} \over \pars{n!}^{2}} & = \lim_{n \to \infty}{n^{n} \over \bracks{\root{2\pi}n^{n + 1/2}\expo{-n}}^{2}} = {1 \over 2\pi}\lim_{n \to \infty}\bracks{{1 \over n}\pars{\expo{2} \over n}^{n}} \\[5mm] & = {1 \over 2\pi}\lim_{n \to \infty}{\exp\pars{n\bracks{2 - \ln\pars{n}}} \over n} = \bbx{\large 0} \end{align}

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