Name: Triangle Inequality Theorem
Uploaded: 2019-07-19
Duration: 5 min 58 s
Description: Measure the sum of lengths of two sides of a triangle and you'll find it to be greater than the length of the other side. Learn more about triangle inequalit...

Question

Show that

(n!)^{2} > n^{n}

and

2 \cdot 4 \cdot 6 \dots . 2 n < (n + 1)^{n}

.

Accepted Answer

To show that $(n!)^{2} > n^{n}$

We know that , $(n - r) (r - 1) \geq 0$ , $r \leq n$ , where $r$ is a natural number

$\Rightarrow r (n - r + 1) \geq n$

For $r = 1$ & $n,$ the equality will hold

$∴ r (n - r + 1) = n$

Hence, for different values to $r,$ we have

$1 (n) = n$

$2 (n - 1) > n$

$3 (n - 2) > n$

$4 (n - 3) > n$

$⋮ ⋮ ⋮$

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

$n (1) = n$

On multiplication of all these cases, we get

$1 (n) \times 2 (n - 1) \times 3 (n - 2) \times . . . \times n (1) > n \times n \times n \times . . . \times n$ ( $n$ times)

$\Rightarrow (1 \times 2 \times 3 \times . . . \times n) (n \times n - 1 \times n - 2 \times . . . \times 1) > n^{n}$

$\Rightarrow (n!) (n!) > n^{n}$

$\Rightarrow (n!)^{2} > n^{n}$

To show that $2 \cdot 4 \cdot 6 \cdot \cdot \cdot 2 n < {(n + 1)}^{n}$

We know, for different positive real numbers, $A . M . > G . M .$

Applying the inequality on $n & 1, n - 1 & 2, \dots, 1 & n$ , we get

$\frac{n + 1}{2} > {(1 \cdot n)}^{\frac{1}{2}}$ $\Rightarrow (n + 1)^{2} > 2 \cdot 2 n$

$\frac{(n - 1) + 2}{2} > {\{2 \cdot (n - 1)\}}^{\frac{1}{2}}$ $\Rightarrow (n + 1)^{2} > 4 \cdot (n - 1)$

$\frac{(n - 2) + 3}{2} > {3 \cdot (n - 2)}^{\frac{1}{2}}$ $\Rightarrow {(n + 1)}^{2} > 6 \cdot (2 n - 4)$

$⋮ ⋮ ⋮$

$\frac{(n - (n - 1)) + n}{2} > {(n \cdot 1)}^{\frac{1}{2}}$ $\Rightarrow (n + 1)^{2} > 2 n \cdot 2$

On multiplying all the inequalities, we get

$(n + 1)^{2 n} > (2.4.6 \dots . 2 n)^{2}$

Or, $(n + 1)^{n} > 2.4.6 \dots . 2 n$

Or, $2.4.6 \dots . 2 n < (n + 1)^{n}$

Show that $(n!)^{2} > n^{n}$ and $2 \cdot 4 \cdot 6 \dots . 2 n < (n + 1)^{n}$ .

Important Questions on Inequalities

Learn and Prepare for any exam you want !

Show that (n!)2>nn and 2·4·6….2n<(n+1)n.

Important Questions on Inequalities

Learn and Prepare for any exam you want !

Show that $(n!)^{2} > n^{n}$ and $2 \cdot 4 \cdot 6 \dots . 2 n < (n + 1)^{n}$ .