Analytical Proof of the statement: $n! > n^{100}$

Asked Feb 27 '23 at 10:41

Active Feb 27 '23 at 10:41

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I think intuitively $n! > n^{100}$, this does not work for values of $n<126$. It is clear that after $n=126$ the sequence $n!$ grows faster than $n^{100}$.

I am trying to give an analytical proof of the statement.

Attempt: Let $a_n = \frac{n!}{n^{100}}$ then $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = n+1$ and this diverges. I am not sure if this suffices to conclude that $n!>n^{100}$.

I am sure there is a simpler way to prove this but I can't figure how.

asked Feb 27 '23 at 10:41

2

Does this answer your question? How to prove $n! >n^a$ for all $a\in \mathbb{R}$ (for sufficiently large $n$)? – Anne Bauval Feb 27 '23 at 10:43
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"$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = n+1$" means nothing. The RHS should not depend on $n.$ – Anne Bauval Feb 27 '23 at 10:44
@AnneBauval Yes thank you. – Feb 27 '23 at 10:46
You can write that as: $$\frac{a_{n+1}}{a_n}\sim n+1.$$ – Thomas Andrews Feb 27 '23 at 12:06
I was thinking of that but then how do you come to the conclusion that $n! > n^{100}$? – Feb 27 '23 at 20:24

Analytical Proof of the statement: $n! > n^{100}$

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