By M.I prove that?

Question

$(2n)! < 2^{2n} (n!)^2$; How to prove this by mathematical induction ?

Answer 1

Start with the base case $n=1$ to get

$$\begin{align} (2\cdot 1)!<2^{2\cdot1}(1)!^2\Leftrightarrow 2<4 \end{align}$$

now assume the two cases $n=m$ and $n=m+1$ as following

$$\begin{align} (2\cdot m)!&<2^{2\cdot m}(m)!^2\\ (2\cdot (m+1))!&<2^{2\cdot (m+1)}(m+1)!^2 \end{align}$$

now observe that $(2m+2)(2m+1)(2m)!=(2m+2)!$ and that $(2m+2)^22^{2m}m!=2^{2\cdot (m+1)}(m+1)!^2$ to get from the first to the second case. Now we have to compare the factors of the RHS and the LHS and show which one is greater for $m\ge1$. So we get

$$\begin{align} (2(m+1))!=(2m+2)(2m+1)(2\cdot m)!&<(2m+2)(2m+1)2^{2\cdot m}(m)!^2\\ &<(2m+2)(2m+2)2^{2\cdot m}(m)!^2\\ &=2^2(m+1)^22^{2m}(m)!^2\\ &=2^{2(m+1)}(m+1)!^2\\ \Leftrightarrow (2(m+1))!&<2^{2(m+1)}(m+1)!^2 \end{align}$$

and we are done. For inductive proof concerning inequalities it is always a good to try to just multiply one side with the missing factor or to add the missing summand to one side. Then start to reshape it in way that the factor/summand of the other side shows up while always staying focused on the relation.

Answer 2

Hint: $(2(n+1))!=(2n+2)!=(2n)!(2n+1)(2n+2)<$

$<(2n)!(2n+2)(2n+2)=(2n)!2^2(n+1)^2$

From here try to use the induction hypothesis.