Prove the inequality $$\sqrt[n]{n!}\lt \sqrt[n+1]{(n+1)!}$$


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As $n!$ is only defined if $n\in \mathbb{N}\cup \{ 0\}$, use logarithms. You can make:

$$\frac{1}{n} \log(n!)< \frac{1}{n+1}\log((n+1)!)$$

$$\Leftrightarrow (n+1)\log(n!)< n\log((n+1)!)$$

$$\Leftrightarrow (n+1)\log(n!)< n\log(n!)+ n\log(n+1)$$

$$\Leftrightarrow n\log(n!)+\log(n!)< n\log(n!)+ n\log(n+1)$$ $$\log(n!)< n\log(n+1)$$

$$\Leftrightarrow \log(n!)<log((n+1)^n)$$

And that is true because $n!<(n+1)^n$ noting that you have $n$ times $n+1$, which is greater than the product of the numbers from 1 to $n$.


This is only an explanation rather than a rigorous proof, but it might be more convincing than a rigorous proof (and can be converted to one, if desired).

Note that

$$(n!)^{1/n}\lt((n+1)!)^{1/(n+1)}\ \iff{\ln1+\ln2+\cdots+\ln n\over n}\lt{\ln1+\ln2+\cdots+\ln n+\ln(n+1)\over n+1}$$

Each expression in the equivalent inequality can be interpreted as an average. The second can be seen as an updated average, with one additional value, $\ln(n+1)$, that is larger than all the other values. It's thus clear that this average exceeds the first. (E.g., if your grades go up from year to year, your grade point average should not go down.)


$$\sqrt[n]{n!}\lt \sqrt[n+1]{(n+1)!}$$ Raise this to the $nm$-th power: $$(n!)^{n+1} \lt ((n+1)!)^n$$ $$(n!)^nn!<(n!(n+1))^n$$ $$(n!)^nn!<(n!)^n(n+1)^n$$ $$n! \le (n+1)^n$$ Each of the $n$ factors of $n!$ is smaller than $n+1$.

  • $\begingroup$ I see now this is similar to the corrected proof of @MonsieurGalois but without logarithm $\endgroup$ – miracle173 Aug 15 '16 at 2:58
  • $\begingroup$ Sorry, I was not at the computer, so I didn't corrected properly. $\endgroup$ – MonsieurGalois Aug 15 '16 at 3:01

Start with the pretty easy fact that the average of the logarithms of the first $n$ integers is less than the logarithm of $n+1$: $$ \frac1n\sum_{k=1}^n\log(k)\lt\log(n+1)\tag{1} $$ Divide by $n+1$ $$ \frac1{n(n+1)}\sum_{k=1}^n\log(k)\lt\frac1{n+1}\log(n+1)\tag{2} $$ Add $\frac1{n+1}\sum\limits_{k=1}^n\log(k)$ to both sides $$ \frac1n\sum_{k=1}^n\log(k)\lt\frac1{n+1}\sum_{k=1}^{n+1}\log(k)\tag{3} $$ And $(3)$ is the logarithm of the desired inequality.

  • $\begingroup$ This is very nice! $(1)$ seems kind of weak, it's pretty surprising that you get the final result from it $\endgroup$ – Yuriy S Aug 15 '16 at 8:52
  • $\begingroup$ @YuriyS: $(1)$ is pretty weak. However, I worked this backwards from the way it was presented, and $(1)$ was the final result. $\endgroup$ – robjohn Aug 15 '16 at 15:47

Since $n > 0$, proving that $$\sqrt[n]{n!}\lt \sqrt[n+1]{(n+1)!}$$ is the same as proving that $$\log\left(\sqrt[n]{n!}\right)\lt \log\left(\sqrt[n+1]{(n+1)!} \right)\tag 1$$that is to say that $$\frac 1n\log\left( n!\right)\lt \frac 1{n+1}\log\left( (n+1)!\right)$$ Using Stirling approximation $$m! \sim \sqrt{2 \pi m}\left(\frac{m}{e}\right)^m$$ which translates to $$\log(m!)=\log\sqrt{2 \pi}+\frac 12 \log(m)+m \log(m)-m$$ the left hand side of $(1)$ becomes $$(\log (n)-1)+\frac{\log (2 \pi )+\log (n)}{2 n}$$ while the right hand side of $(1)$ becomes $$(\log (n+1)-1)+\frac{\log (2 \pi )+\log (n+1)}{2 (n+1)}$$ Now, writing $n+1=n\left(1+\frac 1n\right)$ and using truncated Taylor series or equivalents, we should arrive to $$ \log\left(\sqrt[n+1]{(n+1)!}\right)-\log\left(\sqrt[n]{n!}\right) \sim \frac 1n >0$$

If you add one more terms from Stirling series, you should get $$\log\left(\sqrt[n+1]{(n+1)!}\right)-\log\left(\sqrt[n]{n!}\right)=\frac 1n-\frac{\log(2\pi n)}{2n^2}+O\left(\frac{1}{n^{5/2}}\right)$$ where the rhs is always positive.


Ad Did commented, the above assumes that $n$ is sufficiently large enough in order that Stirling approximation be very good.

  • 2
    $\begingroup$ Sorry but this approach needs much more caution, to produce anything meaningful. 1. Stirling formula is not an identity but an expansion with an error term hence one should follow some $o$ term starting at "which translates to". 2. If this succeeds, the most it can produce is that the inequality holds for n large enough, not for every n as asked. $\endgroup$ – Did Aug 15 '16 at 7:16

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