sum_(n=0)^oo (-1)^n/((2n+1)!) Determine the convergence or divergence of the series.

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.