Calculus of a Single Variable, Chapter 9, 9.5, Section 9.5, Problem 37

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.