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Prove that $$e^{\binom{n}{2}}>n!$$

$n \in \mathbb{Z_+}$

Sorry, couldn't attempt it.

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    $\begingroup$ It will have to be $n \in \mathbb{Z}_+$, not $\mathbb{Z}$. $\endgroup$ – Jeppe Stig Nielsen Aug 11 '13 at 14:10
  • $\begingroup$ Is that the same as $\mathbb{N}$? I changed it :) $\endgroup$ – please delete me Aug 11 '13 at 14:32
  • $\begingroup$ $\mathbb N$ 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 $\mathbb Z_+$. $\endgroup$ – Thomas Andrews Aug 11 '13 at 17:51
  • $\begingroup$ Ok, fixed. Thank you. $\endgroup$ – please delete me Aug 11 '13 at 17:55
  • $\begingroup$ Actually I wasn't sure if people use $\mathbb{Z}^+$ or $\mathbb{Z}_+$. $\endgroup$ – Jeppe Stig Nielsen Aug 11 '13 at 19:23
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Hint: Use $\binom{n}{2} = 0+1+2+\cdots+(n-1)$.

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  • $\begingroup$ Does that mean $0+1+2+\cdots+(n+1)>ln(n!)$ Where do I do from there? $\endgroup$ – please delete me Aug 11 '13 at 14:33
  • $\begingroup$ Second hint: $e^{0+1+2\dots +(n-1)}= e^0e^1e^2\dots e^{n-1}$. Or just try induction on $n$. $\endgroup$ – Thomas Andrews Aug 11 '13 at 14:41
  • $\begingroup$ Am I allowed to assume that $e^n>n+1,n \in \mathbb{N}$? Otherwise, how can I prove it? $\endgroup$ – please delete me Aug 11 '13 at 16:11
  • $\begingroup$ Can you prove $e^n>1+n$, perhaps? (It's only true for $n>0$. $n=1$ gives equality.) How do you define $e^n$? $\endgroup$ – Thomas Andrews Aug 11 '13 at 16:14
  • $\begingroup$ $e^{n}=\frac{n^{1}}{1!}+\frac{n^{2}}{2!}+\frac{n^{3}}{3!}+\cdots$ so the inequality must be true, right? $\endgroup$ – please delete me Aug 11 '13 at 16:20
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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 $n!$.

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