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

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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**.