Prove that $$e^{\binom{n}{2}}>n!$$
$n \in \mathbb{Z_+}$
Sorry, couldn't attempt it.
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1It will have to be $n \in \mathbb{Z}_+$, not $\mathbb{Z}$. – Jeppe Stig Nielsen Aug 11 '13 at 14:10
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Is that the same as $\mathbb{N}$? I changed it :) – please delete me Aug 11 '13 at 14:32
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$\mathbb N$ is an irritating thing, because consensus isn't complete about whether $0\in\mathbb N$. I believe most assume $0\in\mathbb N$, so you want $\mathbb Z_+$. – Thomas Andrews Aug 11 '13 at 17:51
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Ok, fixed. Thank you. – please delete me Aug 11 '13 at 17:55
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Actually I wasn't sure if people use $\mathbb{Z}^+$ or $\mathbb{Z}_+$. – Jeppe Stig Nielsen Aug 11 '13 at 19:23
2 Answers
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Hint: Use $\binom{n}{2} = 0+1+2+\cdots+(n-1)$.
Thomas Andrews
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Does that mean $0+1+2+\cdots+(n+1)>ln(n!)$ Where do I do from there? – please delete me Aug 11 '13 at 14:33
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Second hint: $e^{0+1+2\dots +(n-1)}= e^0e^1e^2\dots e^{n-1}$. Or just try induction on $n$. – Thomas Andrews Aug 11 '13 at 14:41
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Am I allowed to assume that $e^n>n+1,n \in \mathbb{N}$? Otherwise, how can I prove it? – please delete me Aug 11 '13 at 16:11
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Can you prove $e^n>1+n$, perhaps? (It's only true for $n>0$. $n=1$ gives equality.) How do you define $e^n$? – Thomas Andrews Aug 11 '13 at 16:14
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$e^{n}=\frac{n^{1}}{1!}+\frac{n^{2}}{2!}+\frac{n^{3}}{3!}+\cdots$ so the inequality must be true, right? – please delete me Aug 11 '13 at 16:20
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Sorry, I made a mistake, it should be $e^n=1+n+\cdots$ so it is bigger then the other side since we are adding something positive to $1+n$. Thank you a lot for the assistance. I really need to learn to prove things properly :( – please delete me Aug 11 '13 at 16:40
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I quite like Thomas Andrews' approach. Alternatively you can estimate $$ \ln(n!)=\sum_{k=1}^n\ln k <\int_1^{n+1}\ln x\,dx. $$ And calculating that integral gives you a good enough upper bound on the r.h.s. Admittedly this needs more machinery, but it also gives a better approximation to $n!$.
Jyrki Lahtonen
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