Answer to Question #161765 in Calculus for Vishal

Question

Prove that the series ∞n=1 n^n/n! diverges

Answer 1

By the Inequality of arithmetic and geometric means we have:

$\sqrt{1\cdot (n-1)}\leq n/2$ or $1\cdot (n-1)\leq (n/2)^2$

$\sqrt{2\cdot (n-2)}\leq n/2$ , or $2\cdot (n-2)\leq (n/2)^2$

...

if n is even, the last inequality will be $n/2\leq n/2$,

if n is odd, the last inequality will be $(n-1)/2\cdot (n+1)/2\leq (n/2)^2$

Multiplying all this inequalities (without the square roots), we obtain:

$(n-1)!\leq (n/2)^{n-1}$ and

$n!\leq 2(n/2)^n$

Therefore, $\frac{n^n}{n!}\geq\frac{n^n}{2(n/2)^n}=2^{n-1}$

The values $2^{n-1}$ are unbounded from above, therefore the sequence ${n^n}/n!$ is unbounded too and hence, it is divergent.