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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, Moishe Kohan Sep 19 '14 at 9:02

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    $\begingroup$ Use Stirling's Approximation: $ n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n. $ $\endgroup$ – draks ... Jul 17 '12 at 12:25
  • $\begingroup$ 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 $n_0$). $\endgroup$ – Martin Sleziak Jul 17 '12 at 14:05
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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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  • $\begingroup$ 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. $\endgroup$ – Gautam Shenoy Jul 17 '12 at 12:42
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    $\begingroup$ $n!$ is asymptotic to that expression, which is all you need to justify draks' argument. $\endgroup$ – Gerry Myerson Jul 17 '12 at 12:54
  • $\begingroup$ @Gerry: You're right. $\endgroup$ – Gautam Shenoy Jul 17 '12 at 16:14
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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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  • $\begingroup$ (+1) However, it is a definite integral. $\endgroup$ – robjohn May 26 at 0:35
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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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  • $\begingroup$ Can you explain why $\lim\limits_{n\rightarrow\infty}\root n\of{a_n}$ follows from $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}$? $\endgroup$ – draks ... Jul 17 '12 at 14:26
  • $\begingroup$ @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. $\endgroup$ – David Mitra Jul 17 '12 at 14:30
  • $\begingroup$ Thanks David. Your method was correct. I wish I had thought of it earlier. $\endgroup$ – Gautam Shenoy Jul 17 '12 at 17:57

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