To determine the convergence or divergence of the series `sum_(n=0)^oo (-1)^n/((2n+1)!)` , we may apply ratio test.

In Ratio test, we determine the limit as:

`lim_(n-gtoo)|a_(n+1)/a_n| = L`

or

`lim_(n-gtoo)|a_(n+1)*1/a_n| = L`

 Then ,we follow the conditions:

a) `L lt1` then the series converges absolutely.

b) `Lgt1` then the series diverges.

c)` L=1` or does not exist  then the test is inconclusive.The series may be divergent, conditionally convergent, or absolutely convergent.

For the series `sum_(n=0)^oo (-1)^n/((2n+1)!)` , we have:

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

Then,

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

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

            `=(-1)^(n+1)/((2n+2+1)!)`

            `=(-1)^(n+1)/((2n+3)!)`

            `=((-1)^n*(-1))/((2n+3)(2n+2)((2n+1)!))`

Applying the Ratio test on the power series, we set-up the limit as:

`lim_(n-gtoo) |((-1)^n*(-1))/((2n+3)(2n+2)((2n+1)!)) *((2n+1)!)/(-1)^n|`

Cancel out common factors: `(-1)^n` and `(2n+1)!` .

`lim_(n-gtoo) |(-1)/((2n+3)(2n+2))|`

Evaluate the limit.

`lim_(n-gtoo) |(-1)/((2n+3)(2n+2))| =|-1| lim_(n-gtoo) |1/((2n+3)(2n+2))|`

                                         `=1* 1/oo`

                                         `=1*0`

                                         `=0`

The `L=0` satisfies ratio test condition: `Llt1`  since `0lt1` .

Thus, the series `sum_(n=0)^oo (-1)^n/((2n+1)!)` is absolutely convergent.