sum_(n=1)^oo (-1)^n/(n!) Determine whether the series converges absolutely or conditionally, or diverges.

To determine the convergence or divergence of the series sum_(n=1)^oo (-1)^n/(n!) , we may apply the Ratio Test.
In Ratio test, we determine the limit as:
lim_(n-gtoo)|a_(n+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 given series sum_(n=1)^oo (-1)^n/(n!) ,  we have a_n =(-1)^n/(n!) .
 Then, a_(n+1) =(-1)^(n+1)/((n+1)!) .
We set up the limit as:
lim_(n-gtoo) | [(-1)^(n+1)/((n+1)!)]/[(-1)^n/(n!)]|
 To simplify the function, we flip the bottom and proceed to multiplication:
| [(-1)^(n+1)/((n+1)!)]/[(-1)^n/(n!)]| =| (-1)^(n+1)/((n+1)!)*(n!)/(-1)^n|
Apply Law of Exponent: x^(n+m) = x^n*x^m . It becomes:
| ((-1)^n (-1)^1)/((n+1)!)*(n!)/(-1)^n|
Cancel out common factors (-1)^n and apply (-1)^1 = -1
| -(n!)/((n+1)!) |
Simplify:
| -(n!)/((n+1)!) |=(n!)/((n+1)!)
                =(n!)/(n!(n+1))
                =1/(n+1)
The limit becomes:
lim_(n-gtoo)1/(n+1) =(lim_(n-gtoo) (1))/(lim_(n-gtoo) (n+1) )
                     = 1/(oo+1)
                      =1/oo
                      =0
 The limit value L=0 satisfies the condition: L lt1 .
 Therefore, the series sum_(n=1)^oo (-1)^n/(n!) is absolutely convergent.

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