Calculus of a Single Variable, Chapter 9, 9.5, Section 9.5, Problem 7

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.