Prove ${a_n} = \root n \of {n!} $ is monotonically increasing to $\infty$

I already showed that $a_n$ diverges to infinity like this:
I used to the lemma which says that if $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=L$ then $\root n \of {a_n} = L$


  1. How to show that $a_n$ is strictly monotone?
  2. The lemma above is very useful, is there an alternative to show $a_n$ diverges?

I was wrong, the lemma doesn't fit here at all.

  • 1
    $\begingroup$ See this for 2.. $\endgroup$ – David Mitra Jul 19 '14 at 16:01

If we can show that $(n!)^{n+1}\lt ((n+1)!)^n$, we will be finished.

Equivalently,we want to show that $n!\lt (n+1)^n$. This is clear.

  • $\begingroup$ I thought about taking exponent from both sides, but that's getting a bit complicated from there. $\endgroup$ – Elimination Jul 19 '14 at 16:07
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    $\begingroup$ The equivalence of the first and second forms are clear, the two sides of the first have a common factor $(n!)^n$. Then we take each term of the first to the power $\frac{1}{n(n+1)}$. (I wanted to give an argument that uses only arithmetic.) $\endgroup$ – André Nicolas Jul 19 '14 at 16:12
  • $\begingroup$ Thank you, I also like the fact that it's pure arithmetic proof. $\endgroup$ – Elimination Jul 19 '14 at 16:24
  • $\begingroup$ You are welcome. For proving divergence, which as been done many times on MSE, I like pairing, as in $(n!)(n!)=(1\cdot n)(2\cdot (n-1))\cdots (n\cdot 1)\ge n^n$. Now take the $\frac{1}{2n}$-th root. $\endgroup$ – André Nicolas Jul 19 '14 at 16:35

Hint: Using GM-AM inequality. one have that $$ (n!)^{\frac{1}{n}}=(\prod_i i)^{\frac{1}{n}}\le \frac{\sum_i i}{n} = \frac{n+1}{2}. $$

Proof: Then you will have $$ (n!)^{\frac{n+1}{n}}=(n!)^{\frac{1}{n}}(n!)\le \frac{n+1}{2}(n!) = \frac{1}{2}((n+1)!)\le(n+1)!, $$ and consequently $$ (n!)^{\frac{1}{n}}\le ((n+1)!)^{\frac{1}{n+1}}, $$


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