# What is the geometric average of 2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10...?

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 $$\lim_{x\to \infty} \frac{f(n)}{g(n)} = 1 \text{ ?}$$

• en.wikipedia.org/wiki/Stirling%27s_approximation Jul 12, 2015 at 2:01
• 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. ${}\qquad{}$ Jul 12, 2015 at 2:04

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 $n\to\infty$.

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