`sum_(n=1)^oo (-1)^n/2^n` Determine whether the series converges absolutely or conditionally, or diverges.

Expert Answers
marizi eNotes educator| Certified Educator

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 |`


`| (-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.