Question

Show that ${\left(|\underset{–}{n}\right)}^2>n^n$, and $\mathrm{2.4.6...2}n<{(n+1)}^n$

Answer 1

Let $n^n=n\times n\times n\times ...$ upto $n$ times.

$\underset{–}{|n}=n\times (n-1)\times ...\times 3\times 2\times 1$

$\underset{–}{|n}=1\times 2\times 3\times ...\times (n-1)\times n$

Upon multiplying the above two equation we get,

$\underset{–}{|n^2}=(n\times 1)((n-1)\times 2)...(1\times n)$

$(n\times 1)\ge n$

$(n-1)\times 2>n$

$(n-2)\times 3>n$

Thus in every such a way is greater than $n$

Therfore upon multiplying each term we get ,

$\underset{–}{|n^2}\ge n^n$

According to AM -GM inequality , Arithematic mean between the numbers is always $\ge$ geometric mean .

$\frac{(2+4+6+\mathrm{8...}+2n)}{n}\ge {(2\times 4\times 6....2n)}^{\frac{1}{n}}$

$\frac{2(1+2+3+4+...+n)}{n}\ge {(2\times 4\times 6....2n)}^{\frac{1}{n}}$

Sum of $n$ natural number will be $\frac{n(n+1)}{2}$

Therfore,

$\frac{2n(n+1)}{2n}\ge {(2\times 4\times 6....2n)}^{\frac{1}{n}}$

${(n+1)}^n\ge (2\times 4\times 6....2n)$