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I am trying to use mathematical induction to prove

$$(2n)!\ge2^n(n!)^2\quad\text{for }n\in\mathbb{N}$$

I am stuck at the $n=k+1$ point.

Barry Cipra
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user156248
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3 Answers3

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$(2(n+1))! = (2n+2)! = (2n)!(2n+1)(2n+2) \geq 2^n\cdot (n!)^2\cdot (2n+1)(2n+2) > 2^n(n!)^2\cdot (n+1)(2n+2) = 2^{n+1}((n+1)!)^2$. Done !

DeepSea
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  • hi! thanks for answering... I'm just wondering if you can explain how you reached the answer? where did (2n)!(2n+1)(2n+2) come from? and where did (2n+1)(2n+2) from 2n⋅(n!)2⋅(2n+1)(2n+2)? – user156248 May 11 '15 at 21:40
  • $(2n+2)!=(2n+2)(2n+1)!=(2n+2)(2n+1)(2n)!$ by the definition of $K!=K.(K-1)!$ – OKPALA MMADUABUCHI May 11 '15 at 21:51
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Induction is not really needed here. Observe that $${(2n!) \over (n!)^2} = {2n \over n}{2n-1 \over n- 1} ...{n \over 1}$$ Each factor is at least $2$ so the product is at least $2^n$.

Hint for an inductive proof: Let $a_n = {(2n)! \over (n!)^2} $. Then $${a_{n+1} \over a_n} = {(2(n+1))! \over (2n)!} { (n!)^2 \over (n+1)!^2 }$$ $$ = { (2n + 2)(2n + 1) \over (n+1)(n+1)}$$ $$= 2{2n + 1 \over n + 1}$$ $$> 2$$ This should be highly helpful for an inductive proof.

Zarrax
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At the $k+1$ step we obtain $(2k+2)!\geq 2^{k+1}(k+1)!^2$, which is equivalent to: $(2k+2)(2k+1)(2k)!\geq 2^k2(k+1)^2(k)!^2$. By our inductive hypothesis the largest that $2^k(k)!^2$ could possibly be is $(2k)!$, so we can replace $2^k(k)!^2$ on the left hand side with $(2k)!$, and cancel the term from both sides to obtain $(2k+2)(2k+1)\geq 2(k+1)!^2$. Doing some basic algebra we obtain $4k^2+6k+2\geq 2k^2+2k+1$, which is true by inspection (note that the coefficients of the polynomial on the left hand side are all larger than those on the right hand side, and so for any $k \in \mathbb{N}$ this will hold).

AnotherPerson
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  • thank you thank you thank you! Your explanation cleared up a lot of things. I'm just wondering why this is possible when you said " By our inductive hypothesis the largest that 2k(k)!2 could possibly be is (2k)!, so we can replace 2k(k)!2 on the left hand side with (2k)!, and cancel the term from both sides to obtain (2k+2)(2k+1)≥2(k+1)!2. " – user156248 May 11 '15 at 23:12
  • oh and did you mean right hand side for "we can replace 2k(k)!2 on the left hand side with (2k)!"? – user156248 May 11 '15 at 23:13
  • It is possible because we assume that $(2k)!\geq 2^n(n!)^2$ after we show the base case at $n=1$, and the reasoning is that if the relation $\geq$ holds when they are equal, then it will also hold for any case where $(2k)!> 2^k(k!)^2$. – AnotherPerson May 11 '15 at 23:16
  • To answer you second question, yes. Sorry about that. – AnotherPerson May 11 '15 at 23:17