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

Expert Answers
marizi eNotes educator| Certified Educator

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 |`


`| (-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.