For $n \ge 0$, is $\dfrac{\left(n+\left\lfloor\frac{2n}{3}\right\rfloor\right)!}{n!} > \dfrac{(2n)!}{\left(2n-\left\lfloor\frac{n}{3}\right\rfloor\right)!}$

I will consider an answer using the gamma function! I have used the floor function in my question only because I was not clear how to make the argument using the gamma function.

I believe the answer is yes. Here's my argument:

  • For all $n$, $\left\lfloor\frac{2n}{3}\right\rfloor \ge 2\left\lfloor\frac{n}{3}\right\rfloor$

  • So, there are $2$ terms $(n+1)(n+2)\dots(n+\frac{2n}{3})$ for every $1$ term of $(2n)(2n-1)\dots(2n-\frac{n}{3}+1)$ and since $(n+1)(n+2)=n^2+3n+2 > 2n$ where $(n+1)*(n+2)$ is the least product and $2n$ is the greatest value.

Edit: I am open to an answer with the gamma function.

  • $\begingroup$ How are you interpreting the factorial of a non-integer? $\endgroup$ – Umberto P. May 2 '17 at 15:34
  • $\begingroup$ I am applying a floor function. I will add that detail to the question. Will that make a difference here? $\endgroup$ – Larry Freeman May 2 '17 at 15:38
  • 1
    $\begingroup$ It makes a difference in that your terms are otherwise undefined. It is hard to prove anything about undefined terms. $\endgroup$ – Umberto P. May 2 '17 at 15:41
  • $\begingroup$ Great point. I agree about the importance of being exact. I just meant that once I define it as I should, will it change the result. Apologies if my question was unclear. I will edit the question to correct the ambiguity. :-) $\endgroup$ – Larry Freeman May 2 '17 at 15:44
  • $\begingroup$ The previous version of the question was fine, it is enough to state that $m!$ means $\Gamma(m+1)$ for a non-integer $m$. I would propose a rollback, this version is unnecessarily uglier and more difficult to prove. $\endgroup$ – Jack D'Aurizio May 2 '17 at 15:58

We are stating that $$ 2\log\Gamma\left(\frac{5}{3}n+1\right)>\log\Gamma(2n+1)+\log\Gamma(n+1)\tag{1}$$ holds for any $n\geq 0$. In order to prove it is enough to show that

$$ 2\sum_{m\geq 1}\left(\frac{1}{m}-\frac{1}{m+\frac{5}{3}n}\right)> \sum_{m\geq 1}\left(\frac{1}{m}-\frac{1}{m+2n}\right)+\sum_{m\geq 1}\left(\frac{1}{m}-\frac{1}{m+n}\right)\tag{2}$$ then integrate both sides of $(2)$ with respect to the $n$ variable. $(2)$ is equivalent to $$ \sum_{m\geq 1}\left(\frac{1}{m+2n}+\frac{1}{m+n}-\frac{2}{m+\frac{5}{3}n}\right)>0 \tag{3} $$ but that is trivial since the main term of $(3)$ can be represented as $\frac{n (m+3 n)}{(m+n) (m+2 n) (3 m+5 n)}>0$.

  • $\begingroup$ Awesome! I reallly appreciate the analysis. $\endgroup$ – Larry Freeman May 2 '17 at 15:57
  • $\begingroup$ @LarryFreeman: thank you, but in order to prevent downvotes, I have to say this answer replies to the previous version of the question, not involving any $\left\lfloor \cdot\right\rfloor$. $\endgroup$ – Jack D'Aurizio May 2 '17 at 16:00
  • $\begingroup$ Got it. I will add a note to the question so that your answer is 100% . I thought that the floor version was easier. I luke tour analysis much better! $\endgroup$ – Larry Freeman May 2 '17 at 16:02

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