Finding sum to convergent series?

[MySecretAlias]

Homework Statement

Decide whether convergent or divergent, if convergent, find sum.


Ʃ as n = 1 and goes to infinity > (3^n + 2^n)/6^n

Homework Equations

a/1-r

The Attempt at a Solution

I'm just confused where to find the "r" to this without actually plugging in values for n, then dividing. For example, to get the answer (at least I think it's correct), I plugged 1 in for n, got 5/6, then i plugged in 2 for n, and got 11/36. I did (11/36) / (5/6), and used that for my r value, plugged that into a/1-r, and got about 1.31

How exactly do I find the r value? Thanks.

 

[HallsofIvy]

Try breaking it into two sums,
\sum \frac{3^n}{6^n}+ \sum \frac{2^n}{6^n}
If those converge, separately, then the combination converges to the sum of those two.

 

[MySecretAlias]

I respect that. Thank you. However, how do i find what the sum is?

 

[Dick]

I respect that. Thank you. However, how do i find what the sum is?

Write 3^n/6^n as (3/6)^n=(1/2)^n. It's a geometric series. You've probably covered those.

 

[MySecretAlias]

Does this mean that r value is simply 1/2? the 2^n does not really matter, since in the long run, it becomes so miniscule?

 

[Dick]

Does this mean that r value is simply 1/2? the 2^n does not really matter, since in the long run, it becomes so miniscule?

Of course the 2 matters, otherwise it wouldn't be (1/2)^n. What's the sum for n=1 to infinity of (1/2)^n? Review geometric series if you have to.