I'm told to show that $2^n < n!$ using induction
This is my attempt at it:
BC: $n=4, 2^4 = 16 < 4!$
IH: n = k, $2^k < k!$
IS: try n = k+1
I'm told to only work from one side, so I try the left side:
2^(k+1)
But I'm stuck here, any ideas?
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the base case seems to have failed. $2^1 = 2 > 1 = 1!$. – Future Jan 24 '16 at 04:00
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2$2^{k+1} = 2 \times 2^k < 2 \times k!$. Your base case should be $4$ though. – Jan 24 '16 at 04:00
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I've fixed it, thanks – EDEDE Jan 24 '16 at 04:01
3 Answers
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It is true only for $n\ge 4$.
Hint for the inductive step:
$2^{k+1}=2\cdot2^k<2\cdot k!$, so it is enough to prove $2\cdot k!\le (k+1)!$.
Bernard
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If $2\cdot k!\le (k+1)!$ and $2^{k+1}<2\cdot 2^k$, the inductive step will follow by transitivity. Now $2\cdot k!\le (k+1)!=(k+1) ,k! \iff 2\le k+1$, i.e. $;k\ge 1$. – Bernard Jan 24 '16 at 04:13
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You have $2^k<k!$ by induction hypotheses. Then $2^k(k+1)<(k+1)k!=(k+1)!$, if $k>3$ you have that $2^{k+1}<(k+1)2^k$ and the result follows.
EQJ
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$1 \lt k \implies 2 \lt k+1$. So $2^{k+1} = 2^k \times 2 < k! * (k+1) = (k+1)!$ – Steven Alexis Gregory Jan 24 '16 at 04:07
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By hypothesis $k>3$. It is just a redundant argument, I know. Otherwise you can not use induction. – EQJ Jan 24 '16 at 04:10
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If k>3, then is it true that k>1? Hence is it true that 2 < k+1? – Steven Alexis Gregory Jan 25 '16 at 21:48
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This is true for $n \geq 4$.
Base step: $2^4 < 4!$ i.e. $16 < 24$ true!
Induction Step: suppose $2^{n-1} < (n-1)!$. Since $2<n$ you can multilpy the left side for $2$ and the right side for $n$ keeping the inequality true, so you get
$2 \cdot 2^{n-1} < n(n-1)!$ i.e. $2^n < n!$.
Maffred
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