Convergent/divergent series problem

[PaleRider]

Homework Statement

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.

Homework Equations

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 a_n = 0 (after L'Hospital's rule and whatnot). Since a_n is convergent, I now must find the sum of the series...

 

[vela]

How do you know the series converges?

 

[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?

 

[blazeatron]

Ʃ (2^n)/(3^n) = Ʃ (2/3)^n = 2

 

[HallsofIvy]

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.

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.

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?

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