Question

By the ration test for series convergence or divergence?

Answer 1

Evaluate the limit:

`lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) (((n+1)!)/((n-3)!))/((n!)/((n-4)!))`

`lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) ((n+1)!)/(n!) ((n-4)!)/((n-3)!)`

`lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) (n+1)/(n-3) =1`

so the ratio test for this series is inconclusive.

However we can see that:

`a_n = (n!)/((n-4)!) =n(n-1)(n-2)(n-3)`

so that:

`a_n > 0` for `n > 3`

`lim_(n->oo) a_n = oo`

and therefore the series is divergent.