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**.