How does one prove by induction that
$n! > n^2$
for $n \geq 4$
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This is not true when n=1 – Amr Dec 16 '12 at 20:06
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4How does one prove anything by induction? Can you say which part of the general method is problematic in this case? – Trevor Wilson Dec 16 '12 at 20:08
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Should there be some restriction on the value of $n$, because the statement is false for $n = 1, 2, 3$? – Chris Leary Dec 16 '12 at 20:09
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3The OP probably intended for $n \geq 4$. – bzprules Dec 16 '12 at 20:10
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1@cardinal yes, of course! – amWhy Dec 16 '12 at 20:12
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1I edited it sorry it deserved a -1.... – Olivia Irving Dec 16 '12 at 20:14
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Related question: How to prove $a^n < n!$ for all $n$ sufficiently large, and $n! \leq n^n$ for all $n$, by induction? – Martin Sleziak Dec 16 '12 at 20:21
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See also: Prove by induction that $n^2<n!$ and Hint in Proving that $n^2\le n!$ – Martin Sleziak Feb 09 '15 at 11:59
2 Answers
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To prove that this inequality holds for $n \geq 4$, first we verify the base case, which is trivial, as $24 >16$.
Now assume for some $k$ that $k! > k^2$. Then $(k+1)! > (k+1)k^2 = k^3+k^2 > (k+1)^2$. We can verify the right hand inequality, as this implies that $k^2 > k+1 \implies k^2-k-1>0$, which is clearly true for $k \geq 4$; the inductive step has thus been proven and we're done.
bzprules
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I meant $k^2-k-1>0$; I got that expression by dividing the right hand inequality by $k+1$. – bzprules Dec 16 '12 at 20:32
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Hint:
If $n! > n^2$ holds, can you show that $$ (n+1)! = n! (n+1) > (n+1)^2 = n^2 +2 n+1?$$
Fabian
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