Show that ${(n!)^2>n^n}$.
I am trying to use the ${AM \geq GM}$ inequality but I am not getting what I should get. Please help! Thank you!! :))
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1I guess $$n!<n^n$$ – avz2611 May 09 '14 at 06:49
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1are you mad ??? this is totally wrong question...see for n=3.. – soumajit das May 09 '14 at 06:50
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1yaa @ user142634 – soumajit das May 09 '14 at 06:50
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1i guess it is (n!)^2 – soumajit das May 09 '14 at 06:51
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1as for (n!)^2>n^n @ abir mukherjee – soumajit das May 09 '14 at 06:53
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Sorry for the typo!! I have edited the question! :)) – Abir Mukherjee May 09 '14 at 08:57
2 Answers
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$$(n!)^2=n1\ (n-1)2\ (n-2)3\cdots 2(n-1)\ 1n > n^n$$ because for $1\le k\le n-1$ each factor $(n-k)(k+1)=(n-k-1)k+n>n$.
Martín-Blas Pérez Pinilla
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Using the induction:
- For $n=1$ the result is clear.
- The inductive step $$(n+1)!=(n+1)n!\le (n+1)n^n\le (n+1)(n+1)^n=(n+1)^{n+1}$$ so $$n!\le n^n,\qquad \forall n\ge1$$