# Thread: Proof by induction, involving factorials

1. ## Proof by induction, involving factorials

Prove that $\displaystyle 2! \cdot 4! \cdot 6!.....(2n)! \geq \left\[(n+1)!]^n$
I've already proved that P(1) is a true statement.

My inductive hypothesis is that $\displaystyle 2! \cdot 4! \cdot 6!.....(2k)! \geq \left\[(k+1)!]^k$ is true.

I'm trying to show that P(k+1) is true: $\displaystyle 2! \cdot 4! \cdot 6!.....(2k+2)! \geq \left\[(k+2)!]^{k+1}$

$\displaystyle 2! \cdot 4! \cdot 6!.....(2k+2)!$
$\displaystyle \Rightarrow 2! \cdot 4! \cdot 6!.....(2k+2)(2k+1)(2k)!$
$\displaystyle \Rightarrow 2! \cdot 4! \cdot 6!.....(2k)!(2k+2)(2k+1)$

I'm stuck here though. How can this lead to being $\displaystyle \geq \left\[(k+2)!]^{k+1}$?

2. ## Re: Proof by induction, involving factorials

$\displaystyle 2!\cdot 4! \cdot 6! \cdots (2k)!\cdot (2k+2)! \ge [(k+1)!]^k(2k+2)!$ by the induction hypothesis.

\displaystyle \begin{align*}[(k+1)!]^k(2k+2)! & = [(k+1)!]^k(k+1)!(k+2)\cdots (2k+2) \\ & = [(k+1)!]^{k+1}\underbrace{(k+2)\cdots (2k+2)}_{k+1\text{ terms}}\\ & \ge [(k+1)!]^{k+1}\underbrace{(k+2)\cdots (k+2)}_{k+1\text{ times}} \\ & = [(k+2)!]^{k+1}\end{align*}

3. ## Re: Proof by induction, involving factorials

Originally Posted by SlipEternal
$\displaystyle 2!\cdot 4! \cdot 6! \cdots (2k)!\cdot (2k+2)! \ge [(k+1)!]^k(2k+2)!$ by the induction hypothesis.

\displaystyle \begin{align*}[(k+1)!]^k(2k+2)! & = [(k+1)!]^k(k+1)!(k+2)\cdots (2k+2) \\ & = [(k+1)!]^{k+1}\underbrace{(k+2)\cdots (2k+2)}_{k+1\text{ terms}}\\ & \ge [(k+1)!]^{k+1}\underbrace{(k+2)\cdots (k+2)}_{k+1\text{ times}} \\ & = [(k+2)!]^{k+1}\end{align*}
I'm assuming my $\displaystyle (2k+1)$ isn't correct because that would give an odd number.

Are you pulling out $\displaystyle (k+1)!$ from $\displaystyle (2k+2)!$? In the second line.