Proof that if $n\in\mathbb N$, with $n\geq6$ then $$\sqrt[n]{n!}\leq\frac{n}{2}$$

As of now, I can´t use derivatives, functions, and stirlings approximations, as the proofs for them havent been seen in my class or they were seen after this problem

  • $\begingroup$ as of now, i have tried to prove it by bernoulli innequality, induction and using the minimal value of the nth root of n. as of now, i dont know stirlings aproximation $\endgroup$ – user489562 Oct 10 '17 at 3:30
  • $\begingroup$ @hotpocketbitch Use induction. $\endgroup$ – Michael Rozenberg Oct 10 '17 at 3:37
  • $\begingroup$ can this be solved by setting $f(x)=(n/2)^n-n!$ and then taking derivatives? $\endgroup$ – vita nova Oct 10 '17 at 3:51
  • $\begingroup$ See also here: math.stackexchange.com/questions/2446871 $\endgroup$ – Michael Rozenberg Oct 10 '17 at 4:31

By induction let's prove $n!\le\dfrac{n^n}{2^n}$ for $n\ge 6$.

For $n=6$ then $n!=720$ and $\dfrac{n^n}{2^n}=729$ so the inequality is verified.

Now let's assume the inequality true at rank $n$ then


We have by binomial formula: $(n+1)^n=n^n+\binom{n}{1}n^{n-1}+\underbrace{...+1}_{>0}\ge 2n^n$

Thus $(n+1)!\le\dfrac{(n+1)(n+1)^n}{2^{n+1}}\le\dfrac{(n+1)^{n+1}}{2^{n+1}}$ and the induction step is verified.


As Ross Millikan commented, when you have problems involving factorials, using Stirling approximation makes life quite easy.

Consider $$\sqrt[n]{n!}\leq\frac{n}{2}\implies \frac 1n\log(n!) \leq \log(n)-\log(2)\implies\log(n!) \leq n\log(n)-n\log(2)$$ By Stirling approximation $$\log(n!)=n (\log (n)-1)+\frac{1}{2} \left(\log (2 \pi )+\log \left({n}\right)\right)+\frac{1}{12 n}+O\left(\frac{1}{n^3}\right)$$

Just continue.


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