To determine the convergence or divergence of the series `sum_(n=1)^oo (-1)^n/2^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** ...

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To determine the convergence or divergence of the series `sum_(n=1)^oo (-1)^n/2^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/2^n` , we have `a_n =(-1)^n/2^n` .

Then, `a_(n+1) =(-1)^(n+1)/2^(n+1)` .

We set up the limit as:

`lim_(n-gtoo) | [(-1)^(n+1)/2^(n+1)]/[(-1)^n/2^n]|`

To simplify the function, we flip the bottom and proceed to multiplication:

`| [(-1)^(n+1)/2^(n+1)]/[(-1)^n/2^n]| =| (-1)^(n+1)/2^(n+1)*2^n/(-1)^n|`

Apply Law of Exponent: `x^(n+m) = x^n*x^m` . It becomes:

`| ((-1)^n (-1)^1)/(2^n *2^1)*2^n/(-1)^n|`

Cancel out common factors `(-1)^n` and `(2^n)` .

`| (-1)^1/ 2^1 |`

Simplify:

`| (-1)^1/ 2^1 | =| (-1)/ 2 |`

` = |-1/2| `

` =1/2`

Applying `| [(-1)^(n+1)/2^(n+1)]/[(-1)^n/2^n]|= 1/2` , we get:

`lim_(n-gtoo) | [(-1)^(n+1)/2^(n+1)]/[(-1)^n/2^n]|=lim_(n-gtoo) 1/2`

` lim_(n-gtoo) 1/2=1/2`

The limit value `L=1/2 or 0.5` satisfies the condition: `L lt1` since `1/2lt1` or `0.5lt1` .

Therefore, the series `sum_(n=1)^oo (-1)^n/2^n ` **converges absolutely.**