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can anyone help me with this limit? $$\lim_{n\to \infty} n!\frac{e^n}{n^n}$$ I know that the result is infinity but I can't find a way to prove it.

What I tried was to "split" the expression in half to have a pruduct of n/2 and n/2 fractions and then made it smaller by taking first n/2 fractions as $\frac{e}{n}\frac{n}{2}$ and the other n/2 fractions as $\frac{e}{n}$ leading to $\lim_{n\to \infty}(\frac{e}{n}\frac{n}{2})^\frac{n}{2}(\frac{e}{n})^\frac{n}{2}$ which equals zero so this way of using squeeze theorem doesn't work. L'Hospital rule didn't help me either.

Thank you for your advice or hints

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3 Answers 3

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Suppose that $n$ is a positive integer. By the concavity of $\log$, we have \begin{align*} \log n! = & \sum\limits_{k = 1}^n {\log k} \ge \sum\limits_{k = 1}^n {\int_{k - 1/2}^{k + 1/2} {\log xdx} } = \int_{1/2}^{n + 1/2} {\log xdx} \\ & = \left( {n + \frac{1}{2}} \right)\log \left( {n + \frac{1}{2}} \right) - \left( {n + \frac{1}{2}} \right) + \frac{{1 + \log 2}}{2} \\ & \ge \left( {n + \frac{1}{2}} \right)\log n - n . \end{align*} Thus, $$ \log \left( {\frac{{n!e^n }}{{n^n }}} \right) \ge \frac{1}{2}\log n, $$ i.e., $$ \frac{{n!e^n }}{{n^n }} \geq \sqrt n. $$ This shows that the LHS tends to $+\infty$ as $n\to+\infty$.

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Using Stirling's approximation we have that

$$\sqrt{2\pi n} \leq \frac{n!e^n}{n^n}.$$

And since you lower bound diverges you have that limit diverges.

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\begin{equation}\displaystyle\lim_{n\rightarrow\infty}\Bigg(\frac{n!e^n}{n^n}\Bigg) = \lim_{n\rightarrow\infty}\Bigg(\prod_{i=0}^{n-2}\Bigg(1-\frac{i+1}{n}\Bigg)\Bigg)e^n = \lim_{n\rightarrow\infty}\Bigg(\underbrace{1\times 1\times..\times1}_\text{n-1 times}\times\frac{1}{n}\times e^n\Bigg)=\lim_{n\rightarrow\infty}\Big(\frac{e^n}{n}\Big)=\lim_{n\rightarrow\infty}e^{n-1}=\infty\end{equation} (by L'Hôpital's rule). To clarify, \begin{equation} \frac{n!}{n^n} = \frac{n(n-1)(n-2)...1}{\underbrace{n\times n\times n... \times n}_\text{n times}} = \frac{n}{n}\times\frac{n-1}{n}\times...\times\frac{1}{n} = 1\times\Bigg(1-\frac{1}{n}\Bigg)\times....\times\frac{1}{n} \end{equation} And, $$\frac{1}{n} = \Bigg(1-\frac{n-1}{n}\Bigg)$$ which is the result when $i=n-2$. Thus, the limit diverges.

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  • $\begingroup$ You are claiming that $$ \frac{{n!}}{{n^{n - 1} }} \to 1 \times 1 \times \cdots \times 1 = 1, $$ whereas, by Stirling's formula, we know that $$ \frac{{n!}}{{n^{n - 1} }} = \sqrt {2\pi } \frac{{n^{3/2} }}{{e^n }}(1 + o(1)) \to 0. $$ Please, revise your answer! $\endgroup$
    – Gary
    Jun 9, 2020 at 7:34

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