Question

How do you apply the ratio test to determine if `sum_(n=1)^oo (e^n(n!))/n^n` is convergent or divergent?

Answer 1

Please see the explanation.

Explanation:

Given: sum_(n=1)^oo (e^n(n!))/n^n

Now for the ratio test :

`L = lim_(nrarroo)|A_(n+1)/A_n|`

`A_n = (e^n(n!))/n^n`

`A_(n+1) = (e^(n+1)((n+1)!))/(n+1)^(n+1)`

`L = lim_(nrarroo)|( (e^(n+1)((n+1)!))/(n+1)^(n+1))/((e^n(n!))/n^n)|`

`L = lim_(nrarroo)|(e^(n+1)((n+1)!))/(n+1)^(n+1)(n^n)/(e^n(n!))|`

`L = lim_(nrarroo)|((e)e^n(n!(n+1)))/(n+1)^(n+1)(n^n)/(e^n(n!))|`

`L = lim_(nrarroo)|((e)cancel(e^n)(cancel(n!)cancel((n+1))))/(n+1)^(ncancel(+1))(n^n)/(cancel(e^n)(cancel(n!)))|`

`L = lim_(nrarroo)|e(n/(n+1))^n|`

`L = 1`

The ratio test is inconclusive.

However, if you do the limit test:

`L = lim_(nrarroo)(e^n(n!))/n^n = oo`

This shows that the sum diverges.