# Limit of a sequence involving root of a factorial: $\lim_{n \to \infty} \frac{n}{ \sqrt [n]{n!}}$ [duplicate]

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I need to check if $$\lim_{n \to \infty} \frac{n}{ \sqrt [n]{n!}}$$ converges or not. Additionally, I wanted to show that the sequence is monotonically increasing in n and so limit exists. Any help is appreciated. I had tried taking log and manipulating the sequence but I could not prove monotonicity this way.

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## marked as duplicate by Martin Sleziak, Yiorgos S. Smyrlis, drhab, Claude Leibovici, studiosusSep 19 '14 at 9:02

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Since you are new, I want to give you some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. If this is homework, please add the [homework] tag; people will still help, so don't worry. –  Zev Chonoles Jul 17 '12 at 12:21
Use Stirling's Approximation: $n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n.$ –  draks ... Jul 17 '12 at 12:25
Thanks for your help. @Zev: I had tried taking log and manipulating the sequence but I could not prove monotonicity this way. –  Gautam Shenoy Jul 17 '12 at 12:38
The limit is computed in this question: Finding the limit of $\frac{n}{\sqrt$n${n!}}$. I don't know whether the sequence is monotone (starting from some $n_0$). –  Martin Sleziak Jul 17 '12 at 14:05

## 3 Answers

Use Stirling's approximation: $n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n$ and you'll get $$\lim_{n \rightarrow \infty} \frac{n}{(n!)^{1/n}} =\lim_{n \rightarrow \infty} \frac{n}{(\sqrt{2 \pi n} \left(\frac{n}{e}\right)^n)^{1/n}} =\lim_{n \rightarrow \infty} \frac{n}{({2 \pi n})^{1/2n} \left(\frac{n}{e}\right)} =\lim_{n \rightarrow \infty} \frac{e}{({2 \pi n})^{1/2n} }=e,$$ because $\lim_{n\to \infty} ({2 \pi n})^{1/2n}= \lim_{n\to \infty} n^{1/n}=1$.

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Using Stirling's Approximation like that is incorrect as n! is not precisely equal to the aforementioned expression. But you did give me a valuable idea in the process. Thanks, I shall post the solution once I type it. –  Gautam Shenoy Jul 17 '12 at 12:42
$n!$ is asymptotic to that expression, which is all you need to justify draks' argument. –  Gerry Myerson Jul 17 '12 at 12:54
@Gerry: You're right. –  Gautam Shenoy Jul 17 '12 at 16:14

What you have is actually an indefinite integral in disguise. Let's first consider the reciprocal of what you have: \begin{eqnarray*} \lim_{n\to\infty}\frac{(n!)^{1/n}}{n} & = & e^{{\displaystyle \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\ln\left(\frac{k}{n}\right)}}\\ & = & e^{{\displaystyle \int_{0}^{1}\ln xdx}}\\ & = & e^{-1}. \end{eqnarray*} Thus we get that $$\lim_{n\to\infty}\frac{n}{(n!)^{1/n}}=e.$$

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Alternatively, you could use the fact that for a sequence $(a_n)$ of positive terms, if $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}$ exists, then so does $\lim\limits_{n\rightarrow\infty}\root n\of{a_n}$ and the two limits are equal.

For your problem, consider $a_n={n^n\over n!}$. Then $${a_{n+1}\over a_n}={(n+1)^{n+1}\over (n+1)!}\cdot {n!\over n^n}= {1\over n+1}\cdot\Bigl({n+1\over n}\Bigr)^n\cdot(n+1)=\Bigl(1+{1\over n}\Bigr)^n \ \ \buildrel{n\rightarrow\infty}\over\longrightarrow\ \ e.$$

Thus $\lim\limits_{n\rightarrow\infty}{a_{n+1}\over a_n}=e$. As $\root n\of {a_n}={n\over(n!)^{1/n}}$, we have $\lim\limits_{n\rightarrow\infty}{n\over(n!)^{1/n}}=e$ as well.

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Can you explain why $\lim\limits_{n\rightarrow\infty}\root n\of{a_n}$ follows from $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}$? –  draks ... Jul 17 '12 at 14:26
@draks See Lemma 3 in these notes of Pete L. Clark. You can also find this in Theorem 3.37 in Walter Rudin's Principles of Mathematical Analysis. –  David Mitra Jul 17 '12 at 14:30
Thanks David. Your method was correct. I wish I had thought of it earlier. –  Gautam Shenoy Jul 17 '12 at 17:57