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How can one find this limit: $$\lim_{n \to \infty} \frac{n^3}{(3n)!^\frac{1}{n}}$$

Hospitals is out of the question, in this case, because n! is not a differentiable function.

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    $\begingroup$ Are you familiar with Stirling's approximation? $\endgroup$ – vadim123 Oct 10 '13 at 16:30
  • $\begingroup$ No, I am not familiar with it. $\endgroup$ – John Oct 10 '13 at 16:31
  • $\begingroup$ Mathematica returns $e^3/27$ $\endgroup$ – Fred Kline Oct 10 '13 at 16:41
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Consider the power series $$ \sum_{n=0}^\infty a_nz^n,\qquad\mbox{where}\quad a_n:=\frac{n^{3n}}{(3n)!}, $$ and let $R$ be the radius of convergence. Then $$ \frac{1}{R}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\sqrt[n]{a_n}=\lim_{n\to\infty}\frac{n^3}{(3n)!^\frac{1}{n}}. $$ Now, the first limit it easy to calculate. We have $$ \frac{a_{n+1}}{a_n}=\frac{(n+1)^{3n+3}(3n)!}{(3n+3)!n^{3n}}=\left(1+\frac{1}{n}\right)^{3n}\cdot\frac{(n+1)^3}{(3n+1)(3n+2)(3n+3)}, $$ and the first factor tends to $e^3$ and the second to $\frac{1}{27}$ as $n\to\infty$, what leads to $$ \lim_{n\to\infty}\frac{n^3}{(3n)!^\frac{1}{n}}=\frac{e^3}{27}. $$

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    $\begingroup$ I just thought of that. Great!! $\endgroup$ – John Oct 10 '13 at 16:53
  • $\begingroup$ you're welcome :) $\endgroup$ – sranthrop Oct 10 '13 at 16:53
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$$\lim_{n \to \infty} \frac{n^3}{(3n)!^\frac{1}{n}}$$ $$=exp[\frac{3n\ln(3)-\ln(3n)-\ln(3n-1)-\cdots-\ln(1)}{n}]$$ $$=exp[\frac{-1}{n}\left(\sum\limits_{i=1}^{n} \ln(\frac{i}{n})\right)]$$ $$=exp[-\int\limits_{0}^{3}ln(x)\,dx]=exp[-3\ln(3)+\int\limits_{0}^{ln(3)}e^{x}\, dx]$$ $$=exp[-3\ln(3)+3]=\frac{e^{3}}{27}$$

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$A=\lim_{n\to\infty}\frac{n^3}{(3n)!^{1/n}}$ is the cube of $B=\lim_{n\to\infty}\frac{n}{(3n)!^{1/3n}}$, which is a third of $C=\lim_{n\to\infty}\frac{3n}{(3n)!^{1/3n}}$, which is equal to $D=\lim_{n\to\infty}\frac n{n!^{1/n}}$ if the latter exists.

The logarithm of the latter is $E=\lim_{n\to\infty}\log n-\frac1n\Sigma_{i=1}^n\log i$.

Now the derivative of $x\log x-x$ is $\log x$, so $n\log n-n+1=\int_1^n\log x\ge\int_1^n\log\lfloor x\rfloor=\Sigma_{i=1}^{n-1}\log i$, and similarly $n\log n-n+1=\int_1^n\log x\le\int_1^n\log\lceil x\rceil=\Sigma_{i=1}^n\log i$.

Thus $n\log n-n+1\le\Sigma_{i=1}^n\log i\le(n+1)\log n-n+1$, so $1-\frac1n-\frac{\log n}n\le\log n-\frac1n\Sigma_{i=1}^n\log i\le 1-\frac 1n$.

This shows that $E=1$. Since $E$ was defined as $\log D$, $C$ was equal to $D$, $B$ was a third of $C$ and $A$ was the cube of $B$, it follows that $A=\frac{e^3}{27}$.

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