I'm trying to prove that $n! > n^2$ for $n\geq 4$ by use of mathematical induction, but I get to the inductive step and get lost. But I'm struggling with the inductive step as expected.
Asked
Active
Viewed 139 times
2 Answers
0
$4!=24>16=4^2$, so the basic step holds. Then:
$$ (n+1)! = (n+1) n! \color{red}{>} (n+1) n^2 > (n+1)(n+1) = (n+1)^2 $$ and we are fine. We used the inductive hypothesis in $\color{red}{>}$.
Jack D'Aurizio
- 353,855
0
HINT: multiplying $$n!>n^2$$ by $n+1>0$ we get $$(n+1)!>n^2(n+1)$$ and now you have to show that $$n^2(n+1)>(n+1)^2$$ which is easy.
Dr. Sonnhard Graubner
- 95,283
-
You probably start from $$n!>n^2,$$ not $$(n+1)!>(n+1)^2.$$ – Did Oct 03 '15 at 14:35
-
this step is trivial i have it not forgotten, what is the matter with you? – Dr. Sonnhard Graubner Oct 03 '15 at 14:36
-
1The matter with you is that you cannot even set correctly a proof by induction (and we will not mention initialization so that kids won't be afraid...). – Did Oct 03 '15 at 14:40
-
yes you have ritgh, i have made a typo, sorry for it! – Dr. Sonnhard Graubner Oct 03 '15 at 14:40