prove by induction that (2n)!≥2^n(n!)^2

Asked Feb 29 '20 at 13:39

Active Feb 29 '20 at 14:04

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I know that the following maths and factorials properties are useful in my problem but I don't really know how to actually get them into it a rigorous way:

$2\times 2^n=2^{n+1}$ $((n+1)!)/(n!)=n+1 $

So I did

$2((2n)!)≥2\times 2^n(n!)^2$

$2((2n)!)≥2^(n+1)(n!)^2$

But then I don't really know what to do for the $2((2n)!)$ and the $(n!)^2$; how to get them to have n+1

edited Feb 29 '20 at 14:04

sirous

10,751

asked Feb 29 '20 at 13:39

EDD2013

Please consider using MathJax for formatting. – blat Feb 29 '20 at 13:44
You should have $2(2n+1)! ≥ 2^{n+1} \left((n+1)! \right)^2$ instead. – Toby Mak Feb 29 '20 at 13:49

prove by induction that (2n)!≥2^n(n!)^2

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