Question

Show that $n^n>\mathrm{1.3.5....}(2n-1)$.,where $nϵN$

Answer 1

Using AM $\ge$ GM

$\frac{(2n-1)+1}{2}\ge \sqrt{(}2n-1\times 1)$

or $n\ge \sqrt{(}(2n-1).1)$

on squaring both sides we get,

$n^2\ge (2n-1).1$

Case-1 $\{n=even\}$ So on repeating above equation $\frac{n}{2}$ times changing RHS by $\{3,(2n-3)\}$ and so on we get ,

$n^n\ge \mathrm{1.3.5}\dots (2n-1)$

Case-2 $\{n=odd\}$ In this case repeat above process $\frac{(n-1)}{2}$ we get

$n^{n-1}\ge 1.3\dots (n-2)(n+2)\dots (2n-1)$

on multiplying both sides by $n$ we get,

$n^n\ge \mathrm{1.3.5}\dots (2n-1)$