# Prove that $n^{n+1} \leq (n+1)^{n} \sqrt[n]{n!}$

Let $n$ be a positive integer. I conjectured that the following inequality is true $$n^{n+1} \leq (n+1)^{n} \sqrt[n]{n!} .$$ Anyhow I could neither prove nor disprove it. I could only check, by using Stirling's Formula, that the ratio of the right and left members tends to 1 as $n \rightarrow \infty$. Any help is welcome.

\begin{align} \log\left(n\left(1+\frac1n\right)^{-n}\right) &=\log(n)-n\log\left(1+\frac1n\right)\\ &=\log(n)+n\log\left(1-\frac1{n+1}\right)\\ &\le\log(n)-\frac{n}{n+1}\\ &\le\log(n)-\frac{n-1}n\\ &=\frac1n\int_1^n\log(x)\,\mathrm{d}x\\ &\le\frac1n\sum_{k=2}^n\log(k)\\ &=\log\left(\sqrt[n]{n!}\right) \end{align} Exponentiate and rearrange to get $$n^{n+1}\le(n+1)^n\sqrt[n]{n!}$$

Taking logarithms the inequality is equivalent to $$(n+1)\log n\le n\log(n+1)+\frac1n\sum_{k=1}^n\log k.$$ We estimate now the sum: $$\sum_{k=1}^n\log k\ge\int_1^n\log x\,dx=n\log n-n+1.$$ It will be enough to prove that $$n\log n\le n\log(n+1)-1+\frac1n,$$ or equivalently $$1-\frac1n\le n\log\Bigl(1+\frac1n\Bigr).$$ This follows from the inequality $\log(1+x)\ge x-x^2/2$, $0<x<1$.

• A very elegant proof. I could never have found it by myself! Thank you very much, Julian. – Maurizio Barbato Mar 13 '16 at 16:44

From $$e^{-1} = \left(e^{-\frac{1}{n+1}}\right)^{n+1} > \left(1-\frac{1}{n+1}\right)^{n+1}(\because e^x \ge x+1)$$

and

$$\left(1-\frac{1}{n+1}\right)^{n+1} \cdot \left(1 + \frac{1}{n}\right)^{n+1} = 1$$

Note

$$e \le (1+\frac{1}{n})^{n+1}=\frac{(1+n)^{n+1}}{n^{n+1}} \Leftrightarrow n^{n+1} \le \frac{(n+1)^{n+1}}{e}$$

Also note $$n!e^n=\sum_{k=0}^{\infty}{n^k\frac{n!}{k!}} \ge \sum_{k=0}^{n}{n^k\frac{n!}{k!}} \ge \sum_{k=0}^{n}{n^k\binom{n}{k}}= ({n+1})^n \Leftrightarrow \sqrt[n]{n!} \ge \frac{n+1}{e}$$

This gives us that $$(n+1)^{n} \sqrt[n]{n!} \ge \frac{(n+1)^{n+1}}{e} \ge n^{n+1}$$.

• – S.C.B. Mar 13 '16 at 14:53
• To prove the second inequality, note that $n! e^{n} = \sum_{k=0}^{\infty} \frac{n!}{k!} n^{k} > \sum_{k=0}^{n} \frac{n!}{k!} n^{k} > \sum_{k=0}^{n} {n \choose k} n^{k} = (n+1)^n$. Now by using the two inequalities quoted in the answer we get: $n^{n+1} < (n+1)^{n+1} / e = (n+1)^{n} (n+1)/e < (n+1)^{n} \sqrt[n]{n!}$. Great proof! Thank you very much. – Maurizio Barbato Mar 13 '16 at 16:39
• @MauryBarbato Thank you for your kind complements! – S.C.B. Mar 13 '16 at 16:40
• Fixed small typo. – S.C.B. Nov 30 '16 at 7:39

If you feel inclined to do "nice" (erm) calculus, you can always choose to study one of these two functions: $$f\colon x\in [1,\infty) \mapsto \Gamma(x+1) (x+1)^{x^2} - x^{x^2+x}$$ or $$g\colon x\in [1,\infty) \mapsto \frac{\Gamma(x+1) (x+1)^{x^2}}{x^{x^2+x}}$$ and show respectively that $f\geq 0$ or $g \leq 1$ on their domain. (for instance, as $g$ is increasing, if I'm not mistaken, so after showing this you would only have to compute $\lim_\infty g$).

This will not be very enjoyable, but should work: whichever of the two functions you choose, it'll be differentiable and nicely behaved. (If you want, you can even define them on $[2,\infty)$ instead if technicalities at $1$ pop up.)

• Clement C., are my statements true/helpful? I'm not experienced so I am not sure... – S.C.B. Mar 13 '16 at 14:36
• Regarding my answer: the idea being "if it's true on $\mathbb{R}$ (or here on $[1,\infty)$), then it'll be true on $\mathbb{N}$." But on the reals, one has all the machinery of real analysis and calculus to draw upon... – Clement C. Mar 13 '16 at 14:38
• @Clement C. Apart from the fact that the $2x$ should be replaced by $x^2$ in the definitions of $f$ and $g$, I don't see how to prove that $f \geq 0$ or $g \geq 1$. These two functions seem quite complicated to study. By the way, what does it mean "erm"? – Maurizio Barbato Mar 13 '16 at 17:04
• @MauryBarbato Thanks for catching that -- fixed. "Erm" means that the "nice" was sarcastic: this is a systematic method, but not the most elegant and systematic one. As for the functions: they may be complicated to study, but at the end of the day brute-force and differentiation should lead to the proof -- it'll just be a bit painful. – Clement C. Mar 13 '16 at 17:08
• @Clement C. Why technicalities should pop up at 1? I know very little about the game function, and it seems to me quite cumbersome to deal with it and its derivatives. – Maurizio Barbato Mar 13 '16 at 19:51