This was a trivial exercise in induction that I am unable to prove algebraically, or otherwise.
Prove that $n!>n^3\quad\mbox{if}\quad n\gt 5$
This was a trivial exercise in induction that I am unable to prove algebraically, or otherwise.
Prove that $n!>n^3\quad\mbox{if}\quad n\gt 5$
that's probably not very smart, but if you take the limit $\lim_{x \to \infty} \frac{x!}{x^3}$ and differentiate three times (L'Hospital's rule), you get a constant in the denominator and some expression that tends to $\infty$ in the numerator if you take the limit (since all n are integers and (fg)'=f'g+fg' you are guaranteed to get a monotonic sum in the numerator, therefore no $\infty -\infty$ indeterminancy)
I think the following is the simplest approach:
If $n>5$ then $n! \geq n (n-1)(n-2) \cdot3 \cdot 2 \,.$
Now $3(n-2) >n$ and $2(n-1) >n$ for $n >5$, which completes the proof.
You could start with $n!\gt n(n-1)(n-2)(n-3)$, then you just have to prove the quartic beats the cubic.
For $n>5$, $n! \geq n (n-1)(n-2)(3)(2)$.
But $3(n-2)>n$ and $2(n-1)>n$, so $n!>n^3$.
You want $(n-1)! > n^2$ for $n > 5.$ But for $n$ in this range, $(n-1)! \geq 6(n-1)(n-2)$. Hence it suffices to show that $6(1- \frac{1}{n})(1- \frac{2}{n}) > 1$ for $n > 5.$ But $6(1- \frac{1}{n})(1- \frac{2}{n}) \geq \frac{10}{3}$ for $n > 5.$
If you follow the derivation of Stirling's Approximation, you get $n!\gt \sqrt{2\pi n}\left(\frac{n}{\text{e}}\right)^n$, which makes it obvious for $n \gt 3\text{e}$ or $n \gt 8$, then just check 5,6,7 by hand.
Hint: Take logarithms.
Then you only have to show that $\sum_{k=2}^n \log k > 3\log n \ \ \ \text{for} \ n\geq 5.$
Removing the middle terms, and grouping the last two with the first three, for $n\geq 6$ we have: $ \sum_{k=2}^n \log k \geq \log n +(\log(n-1)+ \log (2))+(\log(n-2)+\log(3))$ $\geq \log n+\log 2(n-1)+\log 3(n-2)$ $>\log n+\log n+\log n=3\log n.$
Hope that helps,
The Stolz–Cesàro theorem does for sequences what l"Hopital does for functions, and can be used to show that $n!/n^3$ goes to infinity (which isn't exactly what's wanted here, I know). The theorem says, if $a_1,a_2,\dots$ and $b_1,b_2,\dots$ are real sequences, if $b_n$ is strictly increasing and unbounded, then $\lim{a_n\over b_n}=\lim{a_{n+1}-a_n\over b_{n+1}-b_n}$ if the second limit exists.
Applied three times to $n!/n^3$, you get $(n^3+3n^2+5n+2)n!/6$, which clearly goes to infinity.