Question

prove the inequalities ${\left(n!\right)}^2\le n^n\left(n!\right)<\left(2n\right)!$ for all positive integers n .

Answer 1

$consider$

$(n!{)}^2\le n^n(n!)part$

$n\le n$

$(n-1)\le n$

$(n-2)\le n$

$.....$

$.....$

$.....$

$1\le n$

$\therefore n(n-1)(n-2)......1\le n.n.n......ntimes$

$n!\le n^n$

multiply both sides by n!, we get:

$(n!{)}^2\le n^n(n!)$

Now consider the second part $n^n(n!)\le (2n!)$

$as$

$n\le 2n$

$n\le (2n-1)$

$n\le (2n-2)$

$.....$

$.....$

$.....$

$\le [2n-(n-1\left)\right]$

$\therefore n.n.....ntimes\le 2n(2n-1)(2n-2)........[2n-(n-1\left)\right]$

$n^n\le 2n(2n-1)(2n-2)........[n+1]$

multiply both sides by n!, we get

$n^{n}n!\le 2n(2n-1)(2n-2)........[n+1](n!)$

$n^{n}n!\le (2n!)$

Hence Proved!