# Convergent/divergent series problem

1. Feb 21, 2012

### PaleRider

1. The problem statement, all variables and given/known data
Determine whether the series is convergent of divergent. If it is convergent find its sum.

Ʃ (1+2^n)/(3^n) as n=1 to infinite.

2. Relevant equations

3. The attempt at a solution
I'm not exactly sure where to start with this. I tried to test for divergence but I don't know how to solve the limit. I tried to find a pattern with partial sums, but did not find anything relevant. Am I approaching this the wrong way? I kinda need a kick in the right direction.

EDIT: Was looking at the limit for test of divergence incorrectly. So now I know that the limit of an = 0 (after L'Hospital's rule and whatnot). Since an is convergent, I now must find the sum of the series...

Last edited: Feb 21, 2012
2. Feb 21, 2012

### vela

Staff Emeritus
How do you know the series converges?

3. Feb 21, 2012

### PaleRider

I know the series is convergent because the limit of the series = 0. Test for divergence says that the series is divergent if the limit of the series does not exist, or if the series does not equal 0.

I made some more progress:

Ʃ (1 + 2^n)/(3^n)
= Ʃ 1/(3^n) + (2^n)/(3^n)
= Ʃ 1/(3^n) + Ʃ (2^n)/(3^n)
= 1/2 + Ʃ (2^n)/(3^n)
=?

I'm not quite sure how to solve the remaining summation. Is it a geometric series?

4. Feb 21, 2012

### blazeatron

Ʃ (2^n)/(3^n) = Ʃ (2/3)^n = 2

5. Feb 21, 2012

### HallsofIvy

Staff Emeritus
You may be misunderstanding. It is true that if the limit of the sequence $\lim_{n\to 0} a_n$ is not 0 then the series $\sum_{n=0}^\infty a_n$ cannot converge. But if the limit of the sequence is 0, then the series may or may not \onverge. $\lim_{n\to 0} 1/n = 0$ but $\sum 1/n$ does NOT converge.

$$\frac{2^n}{3^n}= \left(\frac{2}{3}\right)$$.