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It is known that the Khinchin constant is not the geometric mean of the first $n$ coeffecients, as $n$ approaches infinity, of the continued fraction of e, which is $$[2; 1, 2, 1, 1, 4, 1,1, 6, 1, 1, 8, 1, 1, 10, 1, 1, 12,\ldots]$$ Seeing the graph of $f(n)$, the geometric mean of the first $n$ numbers, it seems like there would be a fairly elementary function $g(n)$ which follows and approximates it. Since $2\cdot4\cdot6\cdot8\cdot10\cdots = 2^n\cdot n!$, we can say $$f(n)=\sqrt[n]{\left\lfloor \frac{n}{3} \right\rfloor !\cdot2^{\lfloor n/3 \rfloor + 1}}$$ And since the floor function doesn't matter much on a larger scale, and to be more consistent with the $\frac{n}3+1$, we can essentially rewrite the function as $$f(n)=\sqrt[n]{\Gamma\left(\frac{n}{3}+1\right)\cdot2^{n/3+ 1}}$$After graphing this function and reading a list of identities of the gamma function I could not simplify this, even though the graph looks fairly straightfoward. My question is, is there a simpler, or perhaps constant $g(n)$ where $$\lim_{x\to \infty} \frac{f(n)}{g(n)} = 1 \text{ ?}$$

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    $\begingroup$ en.wikipedia.org/wiki/Stirling%27s_approximation $\endgroup$
    – Will Jagy
    Jul 12 '15 at 2:01
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    $\begingroup$ You can write $3\cdot 5$ or $3\times 5$. Using an asterisk for ordinary multiplication is a workaround for occasions when one is limited to the characters on the keyboard. I edited accordingly. ${}\qquad{}$ $\endgroup$ Jul 12 '15 at 2:04
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By Stirling's approximation, we know that $(n!)^{1/n} \sim n/e$ as $n\to\infty$, so $$ \left(\left\lfloor \frac{n}{3} \right\rfloor!\right)^{1/n} \;=\; \left( \left(\left\lfloor \frac{n}{3} \right\rfloor!\right)^{3/n}\right)^{1/3} \;\sim\; \left(\frac{n}{3e}\right)^{1/3}, $$ and hence $$ f(n) \;\sim\; \left(\frac{2n}{3e}\right)^{1/3} $$ as $n\to\infty$.

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Stirling's formula states that $n! \sim \sqrt{2\pi n} (n/e)^n$, and in particular $n!^{1/n} \sim n/e$. Therefore $$ f(n) = (n/3)!^{1/n} (2^{n/3+1})^{1/n} \sim (n/3e)^{1/3} 2^{1/3} = \sqrt[3]{\frac{2n}{3e}}. $$

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