Question

How do you apply the ratio test to determine if `Sigma n^n/((2n)!)` from `n=[1,oo)` is convergent to divergent?

Answer 1

The series:

`sum_(n=1)^oo n^n/((2n)!)`

is convergent.

Explanation:

Evaluate the ratio:

`abs (a_(n+1)/a_n) = ((n+1)^(n+1)/((2(n+1)!)) )/ (n^n/((2n)!))`

`abs (a_(n+1)/a_n) = (n+1)^(n+1)/n^n ((2n)!)/((2(n+1)!))`

`abs (a_(n+1)/a_n) = (n+1)((n+1)/n)^n ((2n)!)/((2n+2)!)`

`abs (a_(n+1)/a_n) = (n+1)(1+1/n)^n 1/((2n+2)(2n+1))`

`abs (a_(n+1)/a_n) =(1+1/n)^n cancel(n+1) /(2cancel((n+1))(2n+1))`

`abs (a_(n+1)/a_n) =(1+1/n)^n 1 /(2(2n+1))`

Now we have:

`lim_(n->oo) (1+1/n)^n = e`

`lim_(n->oo) 1/(2n+1) = 0`

so:

`lim_(n->oo) abs (a_(n+1)/a_n) = 0`

which proves the series to be convergent based on the ratio test.