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Find $\lim_{n\to\infty} ((n!)^{1/n})$. The question seemed rather simple at first, and then I realized I was not sure how to properly deal with this at all. My attempt: take the logarithm, $$\lim_{n\to\infty} \ln((n!)^{1/n}) = \lim_{n\to\infty} (1/n)\ln(n!) = \lim_{n\to\infty} (\ln(n!)/n)$$ Applying L'hopital's rule: $$\lim_{n\to\infty} [n! (-\gamma + \sum(1/k))]/n! = \lim_{n\to\infty} (-\gamma + \sum(1/k))= \lim_{n\to\infty} (-(\lim(\sum(1/k) - \ln(n)) + \sum(1/k)) = \lim_{n\to\infty} (\ln(n) + \sum(1/k)-\sum(1/k) = \lim_{n\to\infty} (\ln(n))$$ I proceeded to expand the $\ln(n)$ out into Maclaurin form $$\lim_{n\to\infty} (n + (n^2/2)+...) = \infty$$ Since I $\ln$'ed in the beginning, I proceeded to e the infinity $$= e^\infty = \infty$$

So am I write in how I approached this or am I just not on the right track? I know it diverges, I was just wanted to try my best to explicitly show it.

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  • $\begingroup$ Better use Stirling. $\endgroup$ Mar 10 '14 at 8:17
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    $\begingroup$ 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. Apply this to $a_n=n!$ $\endgroup$ Mar 10 '14 at 8:19
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    $\begingroup$ This is utterly illegible. Please use latex mode to write formulas, or at the very least fix those linebreaks... $\endgroup$
    – fgp
    Mar 10 '14 at 8:20
  • $\begingroup$ See also $\lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite and other posts linked there. $\endgroup$ Feb 25 '17 at 1:52
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    $\begingroup$ @MaximilianJanisch I have mentioned your suggestion in CRUDE chatroom, perhaps somebody responds there. $\endgroup$ Dec 23 '19 at 10:20
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By rearranging terms, we can see that $$(n!)^2=[1\cdot n][2\cdot (n-1)][3\cdot (n-2)] \cdots [(n-1)\cdot 2][n\cdot 1].$$ Each of the $n$ products $(k+1)\cdot (n-k)$, for $0\le k<n$, is $\ge n$. Thus $$(n!)^2 \ge n^{n} \quad\text{and therefore}\quad (n!)^{1/n}\ge \sqrt{n}.$$

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    $\begingroup$ this is by far the simplest answer +1 $\endgroup$
    – Paramanand Singh
    Jun 13 '15 at 10:22
  • $\begingroup$ @AndréNicolas With $n/2$ no justification is required as one of the factors will be at least $n/2$. It's actually true that $0\le k<n$ implies $(k+1)(n-k)>n$, but why not using an obvious inequality instead? $\endgroup$
    – egreg
    Mar 3 '16 at 13:24
  • $\begingroup$ Because the original inequality is stronger (though far short of the truth) and prettier. $\endgroup$ Mar 3 '16 at 13:27
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    $\begingroup$ @AndréNicolas can you please show me how to prove $\ (k+1)\cdot (n-k)\ge n$? $\endgroup$
    – Manjoy Das
    Apr 22 '20 at 8:12
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    $\begingroup$ @ManjoyDas If $k=0$, it's obvious. Now for $1\leq k\leq\frac n2$, you have $(k+1)\cdot (n-k)\geq (k+1)\frac n2\geq n.$ Similarly, complete the other case. $\endgroup$
    – cqfd
    Apr 22 '20 at 14:35
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$$ \begin{aligned} \lim_{n\to\infty} (n!)^{1/n} &=\lim_{n\to\infty} \exp(\tfrac{1}{n} \ln n!)\\ &= \lim_{n\to\infty} \exp[\tfrac{1}{n} (\ln 1+\ln 2+\cdots + \ln n)]\\ &\ge \lim_{n\to\infty}\exp \left[ \frac{1}{n} \int_1 ^n \ln x dx\right]\\ &=\lim_{n\to\infty} \exp \frac{n\ln n -n+1}{n} \end{aligned} $$

and last side of above inequality diverges.

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    $\begingroup$ How the sum of logarithms becomes greater than the integration? $\endgroup$
    – BijanDatta
    Jan 3 '21 at 15:22
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    $\begingroup$ @BijanDatta Replace the sum to the integration by using a step function and compare the step function with $\ln x$. $\endgroup$
    – Hanul Jeon
    Jan 3 '21 at 15:38
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Taking $\log$ and using Stolz-Cesaro: $$ \log\lim_{n\to\infty}n!^{1/n}= \lim_{n\to\infty}\log n!^{1/n}= \lim_{n\to\infty}{\log 1+\cdots+\log n\over n}= \lim_{n\to\infty}{\log(n+1)\over(n+1)-n}= \lim_{n\to\infty}\log(n+1)=\infty,$$ so $\lim n!^{1/n}=\infty$.

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Assume WLOG that $n$ is even.

Clearly, $$n!=1\cdot2\cdots n > \left(\frac{n}{2}+1\right)\left(\frac{n}{2}+2\right)\cdots n> \left(\frac{n}{2}\right)^{\frac{n}{2}}$$

Can you take it from here?

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  • $\begingroup$ how does the 1st inequality hold? $\endgroup$
    – Manjoy Das
    Apr 22 '20 at 8:08
  • $\begingroup$ all of the factors exist in the LHS product plus more $\endgroup$ Apr 22 '20 at 8:58
  • $\begingroup$ oh yess. I missed that n is even $\endgroup$
    – Manjoy Das
    Apr 22 '20 at 9:34
  • $\begingroup$ The theorem is correct for all $n$ of course, but writing it would just involve some floors or ceils - not sure... $\endgroup$ Apr 22 '20 at 10:34
  • $\begingroup$ I think $1\cdot2\cdots n < \left(\frac{n}{2}+1\right)\left(\frac{n}{2}+2\right)\cdots \left(\frac{n}{2}+n\right)$ would be right $\endgroup$
    – Manjoy Das
    Jun 3 '21 at 16:41
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As suggested by Martín-Blas Pérez Pinilla, using Stirling approximation $$n!\simeq\sqrt{2 \pi } e^{-n} n^{n+\frac{1}{2}}$$ helps a lot. Raising to power $\frac{1}{n}$ then leads to $$(n!)^{\frac{1}{n}}\simeq\ (2 \pi n)^{\frac{1}{n}} \frac{n}{e}$$ For large values of $n$, the first term goes to $1$ and so $(n!)^{\frac{1}{n}}$ behaves as $\frac{n}{e}$

A better approximation can be obtained using Taylor series; writing the beginning of the expansion as $$(n!)^{\frac{1}{n}}\simeq\frac{n}{e}+\frac{\log (2 \pi n)}{2 e}$$ which shows how would behave $$\frac{(n!)^{\frac{1}{n}}}{n}$$ for large values of $n$.

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The inequalities I often find useful because of their simplicity are $(n/e)^n < n! < (n/e)^{n+1} $. These give $\frac{(n!)^{1/n}}{n} \to 1/e $.

These follow from $(1+1/n)^n < e < (1+1/n)^{n+1} $ (which have been proven here many times).

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