Problem: Prove for natural numbers $n > 2$, $$(\sqrt{2!}-1)((3!)^{\frac{1}{3}}-\sqrt{2!})\cdots(((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}}) < \frac{n!}{(n+1)^n}.$$

I am unable to do this one. Please help.

My attempts: By AM-GM we get,

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

So, now its enough to show, $$\left( \frac{((n+1)!)^{\frac{1}{n+1}}-1}{n}\right)^{n} < \frac{n!}{(n+1)^n}$$ But I don't know how to do that.

I also tried with induction, but it did not work


2 Answers 2


Jack D'Aurizio's answer doesn't have any explicit bound for what $n$ is large enough to cut off the case work. Here's a possible way around this.

I managed to prove this by elementary means. Below is my old approach using properties of the $\Gamma$ function.

I'll start with the form $(n+1)((n+1)!)^{1/(n+1)}-n(n!)^{1/n}<n+1$.

Using AG, it's simple to see that $n!^{1/n}<(n+1)/2$, so we can deal with the following inequalities instead: $$ (n+1)((n+1)!)^{1/(n+1)}-n(n!)^{1/n}<2(n!)^{1/n}\\ (n+1)((n+1)!)^{1/(n+1)}<(n+2)(n!)^{1/n}\\ (n+1)^{n+1}(n+1)n!<(n+2)^{n+1}n!(n!)^{1/n}\\ n+1<\left(1+\frac{1}{n+1}\right)^{n+1}(n!)^{1/n}\\ n+1<\left(1+\frac{1}{n}\right)^n(n!)^{1/n} $$ In the last line I used the well-known inequality $\left(1+\frac{1}{n+1}\right)^{n+1}\geq\left(1+\frac{1}{n}\right)^n$. So now we have: $$ \left(\frac{n^n}{(n+1)^{n-1}}\right)^n<n! $$ We can use induction to get rid of $n!$. The inequality cleary holds for $n=2$. Then: $$ (n+1)!=(n+1)n! >(n+1)\left(\frac{n^n}{(n+1)^{n-1}}\right)^n $$ We could complete the induction step if we knew that $$ (n+1)\left(\frac{n^n}{(n+1)^{n-1}}\right)^n\geq\left(\frac{(n+1)^{n+1}}{(n+2)^n}\right)^{n+1}\\ 1\geq\left(\frac{n+1}{n+2}\frac{(n+1)^{2n}}{n^n(n+2)^n}\right)^n\\ 1\geq\frac{n+1}{n+2}\frac{(n+1)^{2n}}{n^n(n+2)^n} $$ But rearranging a bit, we find: $$ 1\geq\left(\frac{n+1}{n+2}\right)^{n+1}\left(\frac{n+1}{n}\right)^n\\ \left(1+\frac{1}{n+1}\right)^{n+1}\geq\left(1+\frac{1}{n}\right)^n $$ This is the well-known inequality from before, so we may now conclude our induction step: $$ (n+1)!>(n+1)\left(\frac{n^n}{(n+1)^{n-1}}\right)^n\geq\left(\frac{(n+1)^{n+1}}{(n+2)^n}\right)^{n+1} $$ And we are done.

Define $f(x)=x(x!)^{1/x}$. The statement above amounts to proving $f(x+1)-f(x)<x+1$ for $x$ large enough and checking by hand for all smaller $n$. Now suppose we managed to show that beyond a certain $x$, we had $f'(x)<x$. Then by Lagrange's theorem there would be some $\xi\in(x,x+1)$ such that $$ f(x+1)-f(x)=f'(\xi)<\xi<x+1 $$ To prove this, we'll have to make some pretty ugly bounds for $f'$. First: $$ f'(x)=(x!)^{1/x}\left(1-\frac{1}{x}\log(x!)+\psi(x+1)\right) $$ Here $\psi$ is the digamma function. Wikipedia has the following estimates for $x!$: $$ \sqrt{2\pi}x^{x+1/2}e^{-x}\leq x!\leq e^{1-x}x^{x+1/2} $$ Also, the digamma function page has the following inequality: $$ \psi(x)<\log x-\frac{1}{2x} $$ Putting all this together for $x>1$: \begin{align} f'(x)&<(e^{1-x}x^{x+1/2})^{1/x}\left(1+\log (x+1)-\frac{1}{2(x+1)}-\frac{1}{x}\left(\frac{1}{2}\log(2\pi)+\left(x+\frac{1}{2}\right)\log(x)-x\right)\right)\\ &<e^{1/x-1}x^{1+1/2x}(2+\log(x+1)-\log(x))\\ &<e^{1/x-1}x^{1+1/2x}(2+1/x) \end{align} Here I dropped a bunch of negative terms in the second line. I also used $\log(1+1/x)<1/x$ in the third line. Define $g(x)=e^{1/x-1}x^{1/2x}(2+1/x)$. Then we want to prove $g(x)<1$. Apparently (I calculated this with Mathematica because I don't feel like differentiating $g$): $$ g'(x)=-\frac{1}{2}e^{-1+1/x}x^{-3+1/2x}(1+4x+(2x+1)\log(x)) $$ But this is negative for any $x>1$. Now we might numerically calculate that $g(9)\approx0.98$ and because $g$ is decreasing, we have $g<1$ for $x\geq 9$. Now all that remains is checking for all $n<9$.

I don't like this approach because it's too numerical at the end. At the same time, $f(x+1)-f(x)$ seems like a very well-behaved function - indeed, I'm sure there exists some way of going about this more analytically.

  • $\begingroup$ B E A U T I F U L... well actually rather ugly but an actually rigorous answer! I wish that the digamma function and your g(x) were not involved however because the book that I pulled it from hasn't defined either of these... it's from a chapter which covers the AGM inequality and Bernoulli inequality. The text is "More Calculus of a Single Variable". Well.... I think that for now... this is the best we have. If I can derive a more clean proof I'll post it..... until then, I believe your solution is the ONLY real solution (after one checks n=1,...9 which we leave to the reader LOL). $\endgroup$
    – Squirtle
    Jun 5, 2019 at 21:54
  • $\begingroup$ Some thoughts... it might be helpful to split it up into cases. If x(n) equals the product on the LHS, then a good strategy might be the following: Suppose that for all $n\le N$ we have $x(n)<\frac{n!}{(n+1)^n}$. Then we could try proving the result by contradiction. It may help to split it up into cases. Case 1: Assume $x(N+1) = \frac{(N+1)!}{(N+2)^{N+1}}$ and derive a contradiction. Case 2: Repeat with $=$ replaced by $>$. Sometimes splitting this up helps. unfortunately, it hasn't helped me yet. $\endgroup$
    – Squirtle
    Jun 5, 2019 at 21:58
  • $\begingroup$ @Squirtle Also, which book is the one you got this from? $\endgroup$
    – J_P
    Jun 6, 2019 at 1:28
  • $\begingroup$ More calculus of a single variable $\endgroup$
    – Squirtle
    Jun 6, 2019 at 1:35
  • $\begingroup$ Oh, that's the title, I thought it was the chapter... $\endgroup$
    – J_P
    Jun 6, 2019 at 10:52

The last inequality can be written as $$(n+1)\cdot\left((n+1)!^{\frac{1}{n+1}}-1\right)<n\cdot n!^\frac{1}{n}$$ or: $$(n+1)\cdot (n+1)!^{\frac{1}{n+1}} -n\cdot n!^\frac{1}{n}< n+1$$ that is straightforward to prove through Stirling's approximation: $n!\approx\frac{n^n}{e^n}\sqrt{2\pi n}$ gives that the LHS behaves like $\frac{2}{e}\cdot(n+1)$ and $\frac{2}{e}<1$.

  • 3
    $\begingroup$ I recently asked this question as well... and this answer doesn't at all seem to obviously work. If I did something similar and the result failed on the order of 10^-11 for all n greater than 5. This is to say, I feel your answer is too cavalier.... and if it can be made more rigorous would you please do this. I spent several hours yesterday working on this, so it is not as if I haven't tried this approach or seriously considered it. Can you clear it up more, please? $\endgroup$
    – Squirtle
    Jun 4, 2019 at 18:35
  • $\begingroup$ This answer could be completed if this issue gets resolved $\endgroup$
    – user679638
    Jun 5, 2019 at 10:08

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