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Recently, I've been trying to reason about different factorial values.

For example, intuitively, it seems to me that:

$$(n-1)! < \frac{\left(n+\frac{2n}{3}\right)!}{n!}$$

What would be the right approach to demonstrate this?

If I go with induction, the base case is straight forward. This is clearly true for $n=\{3,4,5\}$

For the inductive step, I can prove it holds if $n$ is of form $3i$ or form $3i+2$. I am not able to show it for the case where $n$ is of the form $3i+1$.

If my intuition is correct, what is the easiest way to complete the argument? Would the gamma function, for example, help me here?

If my intuition is not correct, could you show me how to analyze the conditions where it does not hold.

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  • $\begingroup$ Do you take $(n+\frac{2n}{3})!=\Gamma(n+\frac{2n}{3}+1)$ when $n$ is not divisible by $3$, or do you round $n+\frac{2n}{3}$? $\endgroup$
    – Servaes
    Commented May 2, 2017 at 7:34
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    $\begingroup$ $\frac{ (n-1)! }{ n! } = 1/n$ $\endgroup$
    – Hugh Entwistle
    Commented May 2, 2017 at 7:35
  • $\begingroup$ @Servaes, I was rounding. It was not clear to me how to use the gamma function to show that $n+\frac{2n}{3}+1$ increases faster than $(n-1)!$ $\endgroup$
    – Larry Freeman
    Commented May 2, 2017 at 7:37
  • $\begingroup$ @Hugh, I am not clear how that finishes the argument. How would I use your observation to complete the argument or to counter the argument? $\endgroup$
    – Larry Freeman
    Commented May 2, 2017 at 7:43
  • $\begingroup$ A mistake on my part - I did a wonky mental calc ! $\endgroup$
    – Hugh Entwistle
    Commented May 2, 2017 at 23:31

1 Answer 1

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It's not true. By Stirling's formula, $n!\sim \sqrt{2\pi n}\big(\frac{n}{e}\big)^n$, and so $n!(n-1)!=n!^2/n\sim2\pi\big(\frac{n}{e}\big)^{2n}$. But $(5n/3)!\sim\sqrt{10\pi n/3}\big(\frac{5n}{3e}\big)^{5n/3}$. So the ratio of these is about $\big(\frac{n}C\big)^{n/3}$, where $C$ is a constant. This eventually gets much bigger than $1$, so the LHS is eventually larger.

[edit] The first counterexample is $n=31$, where $$30!=265252859812191058636308480000000;$$ $$\frac{51!}{31!}=188635435829605873227261542400000.$$

(The above example rounds down; for rounding to the nearest integer the first counterexample is $n=35$, and for always rounding up it is $n=39$.)

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  • $\begingroup$ Thanks very much! This is exactly the answer that I was looking for! $\endgroup$
    – Larry Freeman
    Commented May 2, 2017 at 7:45

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