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How to evaluate $$ \lim_{n \to \infty} \sqrt[n]{\frac{n^n}{n!}}? $$

My Attempt:

For each $n = 1, 2, 3, \ldots$, let us put $$ s_n := \sqrt[n]{\frac{n^n}{n!}}. $$

We note that $$ \begin{align} \sqrt[n]{\frac{n^n}{n!}} &= \frac{\sqrt[n]{n^n}}{\sqrt[n]{n!}} \\ &= \frac{n}{\sqrt[n]{n!}} \\ &\geq \frac{n}{\sqrt[n]{n \times \cdots \times n}} \\ &= \frac{n}{\sqrt[n]{n^n}} \\ &= \frac{n}{n} = 1 . \end{align} \tag{1} $$ Moreover, we also have $$ \begin{align} \frac{s_{n+1}}{s_n} &= \frac{ \frac{n+1}{\sqrt[n+1]{(n+1)!}} }{ \frac{n}{\sqrt[n]{n!}} } \\ &= \frac{n+1}{n} \frac{\sqrt[n]{n!} }{\sqrt[n+1]{(n+1)!} } \\ &= \frac{n+1}{n} \frac{1}{\sqrt[n+1]{n+1} } \big( n! \big)^{\frac{1}{n} - \frac{1}{n+1} } \\ &= \frac{1}{n} \big( n+1 \big)^{1 - \frac{1}{n+1} } \big( n! \big)^{ \frac{1}{n(n+1)} } \\ &= \end{align} $$

How to proceed from here?

Or, can we show that this sequence is (or is not) a Cauchy sequence?

Saaqib Mahmood
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    It is a well known result that $\liminf_n\frac{|a_{n+1}|}{|a_n|}\leq\liminf_n\sqrt[n]{|a_n|}\leq\limsup_n\sqrt[n]{|a_n|}\leq\limsup_n\frac{|a_{n+1}|}{|a_n|}$. In your case, set $a_n=\frac{n^n}{n!}$. Then $\frac{a_{n+1}}{a_n}=\big(1+\frac{1}{n}\big)^n$ – Mittens May 24 '23 at 03:13
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    $n! \sim \sqrt{2\pi n}\left( \frac{n}{e} \right)^n$ https://en.wikipedia.org/wiki/Stirling%27s_approximation – zkutch May 24 '23 at 03:21
  • https://math.stackexchange.com/q/3300681/42969 – Martin R May 24 '23 at 03:25
  • You can also check Stolz–Cesàro Theorem there is exactly a same example in the wiki https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem – Nanayajitzuki May 24 '23 at 03:37

1 Answers1

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I am going to complete two ideas suggested in the comments. Metamorphy notified that the second paragraph was faulty, hopefully this fixes it!

We can use the corollary of the Stols-Cesàro Theorem for limits of a geometric mean: Let $(y_n)$ be any sequence of real numbers such that the limit $\lim_{n \rightarrow \infty}\frac{y_{n+1}}{y_n}$ exists. Then $\lim_{n \rightarrow \infty}\sqrt[n]{y_n} = \lim_{n \rightarrow \infty}\frac{y_{n+1}}{y_n}$. Now we can solve the limit: $$\lim_{n \rightarrow \infty} \sqrt[n]{\frac{n^n}{n!}} = \lim_{n \rightarrow \infty}\frac{(n+1)^{n+1}n!}{(n+1)!n^n} = \lim_{n \rightarrow \infty}\frac{(n+1)^{n+1}}{(n+1)n^n} = \lim_{n \rightarrow \infty}(1+1/n)^n=e.$$

We can also use Stirling's approximation: $n! \sim \sqrt{2\pi n}(\frac{n}{e})^n$. $$\lim_{n \rightarrow \infty} \sqrt[n]{\frac{n^n}{n!}} \sim \lim_{n \rightarrow \infty} \sqrt[n]{\frac{n^n}{\sqrt{2\pi n}(\frac{n}{e})^n}} = \lim_{n \rightarrow \infty} e(\sqrt{2\pi n})^{-\frac{1}{n}} = e.$$ Now we can use the asymptotic bounds of Stirling's approximation to show that this equals our original limit. For all $n\geq 1$, $$\sqrt{2\pi n}(\frac{n}{e})^ne^{\frac{1}{12n+1}}< n! < \sqrt{2\pi n}(\frac{n}{e})^ne^{\frac{1}{12n}},$$ and so we can use the squeeze theorem to show $$\lim_{n \rightarrow \infty} e(e^{\frac{1}{12n}}\sqrt{2\pi n})^{-\frac{1}{n}}\leq\lim_{n \rightarrow \infty} \sqrt[n]{\frac{n^n}{n!}}\leq\lim_{n \rightarrow \infty} e(e^{\frac{1}{12n+1}}\sqrt{2\pi n})^{-\frac{1}{n}},$$ which simplifies to $$e\leq\lim_{n \rightarrow \infty} \sqrt[n]{\frac{n^n}{n!}}\leq e \implies \lim_{n \rightarrow \infty} \sqrt[n]{\frac{n^n}{n!}}=e.$$

Gary
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Eleanor Blake
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  • I think what metamorphy is getting at is that the limit in the last sentence as written doesn't exist. Instead, $n! - \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n \in O(n^{n - \frac{1}{2}} e^{-n})$, which grows rapidly. That's probably enough for our purposes here, though. A more immediately useful way of writing down that asymptotic behavior is that $n! = \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n (1 + R(n))$ for some function $R(n) \in O\left(\frac{1}{n}\right)$. – Travis Willse May 24 '23 at 06:50