# `sum_(n=1)^oo (-1)^n/3^n` Determine the convergence or divergence of the series.

To determine the convergence or divergence of the series `sum_(n=1)^oo (-1)^n/3^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/3^n` , we have `a_n =(-1)^n/3^n` .

Then, `a_(n+1) =(-1)^(n+1)/3^(n+1)` .

We set up the limit as:

`lim_(n-gtoo) | [(-1)^(n+1)/3^(n+1)]/[(-1)^n/3^n]|`

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

`| [(-1)^(n+1)/3^(n+1)]/[(-1)^n/3^n]| =| (-1)^(n+1)/3^(n+1) *3^n/(-1)^n|`

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

`| ((-1)^n (-1)^1)/(3^n *3^1)*3^n/(-1)^n|`

Cancel out common factors `(-1)^n` and `(3^n)` .

`| (-1)^1/ 3^1 |`

Simplify:

`| (-1)^1/ 3^1 | =| (-1)/ 3 |`

` = |-1/3| `

` =1/3`

Applying ` |[(-1)^(n+1)/3^(n+1)]/[(-1)^n/3^n]|= 1/3` , we get:

`lim_(n-gtoo) | [(-1)^(n+1)/3^(n+1)]/[(-1)^n/3^n]|=lim_(n-gtoo) 1/3 `

`lim_(n-gtoo) 1/3 = 1/3 `

The limit value `L=1/3` satisfies the condition: `L lt1` .

Therefore, the series converges absolutely.

Approved by eNotes Editorial Team