sum_(n=1)^oo (-1)^n/e^n Determine the convergence or divergence of the series.

Recall the Ratio test determines the limit as:
lim_(n-gtoo)|a_(n+1)/a_n| = L
 Then ,we follow the conditions:
a) L lt1 then the series is absolutely convergent
b) Lgt1 then the series is divergent.
c) L=1 or does not exist  then the test is inconclusive. The series may be divergent, conditionally convergent, or absolutely convergent.
We may apply the Ratio Test to determine the convergence or divergence of the series sum_(n=1)^oo (-1)^n/e^n .
For the given series sum_(n=1)^oo (-1)^n/e^n , we have a_n =(-1)^n/e^n .
 Then, a_(n+1) =(-1)^(n+1)/e^(n+1) .
We set up the limit as:
lim_(n-gtoo) | [(-1)^(n+1)/e^(n+1)]/[(-1)^n/e^n]|
 To simplify the function, we flip the bottom and proceed to multiplication:
| [(-1)^(n+1)/e^(n+1)]/[(-1)^n/e^n]| =| (-1)^(n+1)/e^(n+1)*e^n/(-1)^n|
Apply Law of Exponent: x^(n+m) = x^n*x^m . It becomes:
| ((-1)^n (-1)^1)/(e^n *e^1)*e^n/(-1)^n|
Cancel out common factors (-1)^n and e^n .
| (-1)^1/ e^1 |
Simplify:
| (-1)^1/ e^1 | =| (-1)/ e |
          = |-1/e|
          =1/e
Applying | [(-1)^(n+1)/e^(n+1)]/[(-1)^n/e^n]|= 1/e , we get:
lim_(n-gtoo) | [(-1)^(n+1)/e^(n+1)]/[(-1)^n/e^n]|=lim_(n-gtoo) 1/e
lim_(n-gtoo) 1/e=1/e or 0.3679 (approximated value)
 The limit value  L~~0.3679 satisfies the condition: L lt1 .
 Therefore, the series sum_(n=1)^oo (-1)^n/e^n is absolutely convergent.