Question

How do you apply the ratio test to determine if `Sigma (n!)/n^n` from `n=[1,oo)` is convergent to divergent?

Answer 1

The series:

`sum_(n=1)^oo (n!)/n^n`

is convergent.

Explanation:

Evaluate the ratio:

`abs(a_(n+1)/a_n) = ( ((n+1)!)/(n+1)^(n+1))/ ((n!)/n^n) = n^n/(n+1)^(n+1) ((n+1)!)/(n!) = 1/(n+1)(n/(n+1))^n (n+1)=(n/(n+1))^n`

We can write this expression differently in order to find the limit:

`(n/(n+1))^n = (1/((n+1)/n))^n = 1/(1+1/n)^n`

So:

`lim_(n->oo) abs(a_(n+1)/a_n) = 1/(1+1/n)^n = 1/e <1`

which proves the series to be convergent.