I want to prove this,

$$2^23^34^45^5\cdots n^n \leq \Big(n+1-(n!)^{\frac{1}{n}}\Big)^{(n(n+1))0.5}$$

I try induction but we can't multiply by $(n+1)^{n+1}$ it's too big .On the other hand i try to study the following function : $$ f(n)=\Big(n+1-(n!)^{\frac{1}{n}}\Big)^{(n(n+1))0.5} )$$

But it reveals nothing.

I try also to approximate the factorial but it's not enought

I prefer hints.

Thanks a lot for sharing your time.

  • $\begingroup$ Note: Same setup as mathoverflow.net/questions/344504/an-olympiad-like-inequality with $f(x) = x$, though the general case isn't true. $\endgroup$ – Calvin Lin Nov 13 '19 at 7:17
  • $\begingroup$ $n\ge 3$ is needed. Use bounds for $\sum k\ln k$ and $(n!)^{1/n}$. $\endgroup$ – River Li Nov 13 '19 at 14:43
  • $\begingroup$ With Stirling, you can show $n+1-(n!)^{1/n}>n\left(1-\frac{1}{e}\right)-\varepsilon$ for large enough $n$. Numerically checking for some values of $\varepsilon$, it appears that for large enough $n$, the stronger bound $2^23^34^45^5\cdots n^n \leq \left(n\left(1-\frac{1}{e}\right)-\varepsilon\right)^{n(n+1)/2}$ holds. $\endgroup$ – alex.jordan Nov 13 '19 at 15:46
  • 1
    $\begingroup$ Still any question left ? ;) $\endgroup$ – user90369 Nov 13 '19 at 20:46

This is a long chain of symbol-pushing, with repeated use of Stirling. Maybe it has mistakes. If it does not, it surely could be cleaned up and streamlined. But at the moment, I think it works. The strategy is to use these Stirling bounds: $$\sqrt{2\pi n}\frac{n^n}{e^n}<n!<\sqrt{2\pi n}\frac{n^n}{e^n}\left(1+\frac{1}{11n}\right)$$ using the appropriate one of these two inequalities to keep the argument going. Eventually there are two sides that have simplified to each having a finite limit, and the lef-side limit is less than the right-side limit.

The left side is: $$P(n)=\frac{(n!)^n}{\prod_{k=1}^{n-1}k!}$$ By Stirling: $$ \begin{align} P(n) &<\frac{\left(n!\right)^n}{\prod_{k=1}^{n-1}\left(\sqrt{2\pi k}\frac{k^k}{e^k}\right)}\\ &=\frac{\left(n!\right)^n}{(2\pi)^{(n-1)/2}\sqrt{(n-1)!}e^{-n(n-1)/2}\prod_{k=1}^{n-1}\left(k^k\right)}\\ &=\frac{n^n\left(n!\right)^n}{(2\pi)^{(n-1)/2}\sqrt{(n-1)!}e^{-n(n-1)/2}\prod_{k=1}^{n}\left(k^k\right)}\\ &=\frac{n^n\left(n!\right)^n}{(2\pi)^{(n-1)/2}\sqrt{(n-1)!}e^{-n(n-1)/2}P(n)}\end{align} $$

Multiply by $P(n)$ and take the square root:

$$ P(n)<\sqrt{\frac{n^n\left(n!\right)^n}{(2\pi)^{(n-1)/2}\sqrt{(n-1)!}e^{-n(n-1)/2}}} $$

So we want to show $$\sqrt{\frac{n^n\left(n!\right)^n}{(2\pi)^{(n-1)/2}\sqrt{(n-1)!}e^{-n(n-1)/2}}}<\left(n+1-(n!)^{1/n}\right)^{n(n+1)/2}$$ which is equivalent to: $$\frac{n\left(n!\right)e^{(n-1)/2}\sqrt[2n]{2\pi n}}{\sqrt{2\pi}\sqrt[2n]{n!}}<\left(n+1-(n!)^{1/n}\right)^{n+1}$$

But since $n!>\sqrt{2\pi n}\frac{n^n}{e^n}$, it suffices to show: $$\frac{n\left(n!\right)e^{(n-1)/2}\sqrt[2n]{2\pi n}}{\sqrt{2\pi}\sqrt[2n]{\sqrt{2\pi n}\frac{n^n}{e^n}}}<\left(n+1-(n!)^{1/n}\right)^{n+1}$$

which is equivalent to: $$\frac{\sqrt{n}\left(n!\right)e^{n/2}\sqrt[4n]{2\pi n}}{\sqrt{2\pi}}<\left(n+1-(n!)^{1/n}\right)^{n+1}$$

Using Stirling to more precision, for large enough $n$, $n!<\sqrt{2\pi n}\frac{n^n}{e^n}\left(1+\frac{1}{11n}\right)$. And it suffices to show:

$$\frac{\sqrt{n}\sqrt{2\pi n}\frac{n^n}{e^n}\left(1+\frac{1}{11n}\right)e^{n/2}\sqrt[4n]{2\pi n}}{\sqrt{2\pi}}<\left(n+1-\left(\sqrt{2\pi n}\frac{n^n}{e^n}\left(1+\frac{1}{11n}\right)\right)^{1/n}\right)^{n+1}$$

which is equivalent to $$n\sqrt[n+1]{\frac{\left(1+\frac{1}{11n}\right)\sqrt[4n]{2\pi n}}{e^{n/2}}}<n+1-\frac{n}{e}\left(\sqrt{2\pi n}\left(1+\frac{1}{11n}\right)\right)^{1/n}$$

It suffices to show $$n\sqrt[n+1]{\frac{\left(1+\frac{1}{11n}\right)\sqrt[4n]{2\pi n}}{e^{n/2}}}<n-\frac{n}{e}\left(\sqrt{2\pi n}\left(1+\frac{1}{11n}\right)\right)^{1/n}$$ which is equivalent to $$\sqrt[n+1]{e^{n/2+1}\left(1+\frac{1}{11n}\right)\sqrt[4n]{2\pi n}}<e-\sqrt[n]{\sqrt{2\pi n}\left(1+\frac{1}{11n}\right)}$$

The left side converges to $\sqrt{e}\approx1.648$ while the right side converges to $e-1\approx1.718$. So this inequality holds for large enough $n$. (It appears from a spreadsheet that it holds for $n\geq56$.)

It remains to check the original inequality for small values of $n$.


For e.g. $~n=2~$ it's not correct, but from a certain value $~n_0~$ already.

Comparing asymptotic approximations.

$\displaystyle \left(n^{-\left({\frac{n^2}{2}+\frac{n}{2}}\right)}\prod\limits_{k=1}^n k^k\right)^\frac{2}{n^2}\approx \left(A e^{-\frac{n^2}{4}}n^{\frac{1}{12}} \right)^\frac{2}{n^2} \approx \frac{1}{\sqrt{e}}$

where $~A~$ is called the Glaisher–Kinkelin constant

$\displaystyle \left(n^{-\left({\frac{n^2}{2}+\frac{n}{2}}\right)}\left(n+1-\sqrt[n]{n!}\right)^{\frac{n(n+1)}{2}}\right)^\frac{2}{n^2}\approx 1-\frac{1}{n}-\frac{\sqrt[n]{n!}}{n}\approx 1-\frac{1}{e}$

It's $~\displaystyle \frac{1}{\sqrt{e}} < 1-\frac{1}{e}~$ .

This means that $~n_0~$ exists so that the claim is correct for $~n\geq n_0~$ .


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