6

So far I have:

Base case: $$ n = 1 : (2(1))! > (1!)^2$$ $$ 2! > 1!^2$$ $$2 > 1$$ Induction step: Assume this is true for some $n > 1$ Let n = p + 1 $$(2(p+1))! > ((p+1)!)^2$$ $$(2p)!(2p+1)(2p+2) > ((p+1)!)^2$$ $$(2p)!>\frac{((p+1)!)^2}{(2p+1)(2p+2)}$$ Now I add 1 to the smaller term in the demoninator, for cancellation purposes which makes the RHS smaller so inequality still holds.

$$(2p)!>\frac{(p!)^2(p+1)^2}{(2p+2)(2p+2)}$$ $$(2p)!>\frac{(p!)^2}{4}$$ Now I am stumped on how to get rid of the extra 4 in the denominator.

Martin Sleziak
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Ryan Joseph
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  • Inductively? Would be easier to note that $\binom{2n}{n} > 1 $, as long as $n\ge 1$. – hmakholm left over Monica Nov 21 '17 at 22:16
  • @HenningMakholm: Would you say ${2n \choose n} \gt 1$ since there are at least two ways of choosing $n$ objects from $2n$ (e.g. choose the first $n$ or the last $n$, and these are different choices when $n \ge 1$)? – Henry Nov 22 '17 at 01:42
  • Regardless of which answer best suits your purposes, please note the point in Mark Bennet's. Your calculation is backwards from the direction it needs to go. Proofs do not follow the order "conclusion $\implies$ known". They follow the order "known $\implies$ conclusion". In the induction step, you know that $(2p)! > (p!)^2$. You need to start with that, and proceed to $(2(p+1))! > ((p+1)!)^2$. – Paul Sinclair Nov 22 '17 at 03:49
  • Here is a stronger inequality: Prove that $2^n(n!)^2 \leq (2n)!$ – Martin Sleziak Nov 22 '17 at 06:48

4 Answers4

14

Use $$(2n)! = n! (n+1)(n+2)\cdots(2n)> n! \cdot 1\cdot 2 \cdots n = (n!)^2.$$

Alternatively, if you know binomial coefficients, $$ (2n)! = \binom{2n}{n} (n!)^2 > (n!)^2, $$ since $\binom{2n}{n}$ is an integer $> 1.$

Reiner Martin
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7

By the inductive hypothesis we have $(2n)! \geq (n!)^2$, we need to show that $(2n+2)! \geq ((n+1)!)^2$ so we need \begin{eqnarray*} (2n+2)(2n+1) \geq (n+1)^2 \\ 3n^2+4n+1 \geq 0 \end{eqnarray*} which is obviously true.

Donald Splutterwit
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6

You are arguing the wrong way. If you start with $$(2n)!\gt (n!)^2$$ and multiply this by $(2n+2)(2n+1)$ you get $$(2(n+1))!\gt(2n+2)(2n+1)(n!)^2\gt((n+1)!)^2$$leaving you some details to fill in.

The way you are arguing at the moment attempts to deduce the result for $n$ from the result for $n+1$.

Note also that if induction is not required, the fact that the binomial coefficient $\binom {2n}n$ is a positive integer does the job. There are also various ways of proving that $\binom nr$ is a positive integer for $0\le r \le n$ including some quite simple induction methods, which would then give the result you want as a special case.

Mark Bennet
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5

$(2(p+1))!=(2p+2)(2p+1)(2p)!>2(p+1)(2p+1)(p!)^{2}$, now \begin{align*} 2(p+1)(2p+1)(p!)^{2}-((p+1)!)^{2}&=[2(2p+1)p!-(p+1)!](p+1)!\\ &=[2(2p+1)-(p+1)]p!(p+1)!\\ &=(3p+1)p!(p+1)!>0 \end{align*}

user284331
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