1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Finding sum to convergent series?

  1. Feb 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Decide whether convergent or divergent, if convergent, find sum.

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

    2. Relevant equations


    3. 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.
  2. jcsd
  3. Feb 5, 2012 #2


    User Avatar
    Science Advisor

    Try breaking it into two sums,
    [tex]\sum \frac{3^n}{6^n}+ \sum \frac{2^n}{6^n}[/tex]
    If those converge, separately, then the combination converges to the sum of those two.
  4. Feb 5, 2012 #3
    I respect that. Thank you. However, how do i find what the sum is?
  5. Feb 5, 2012 #4


    User Avatar
    Science Advisor
    Homework Helper

    Write 3^n/6^n as (3/6)^n=(1/2)^n. It's a geometric series. You've probably covered those.
  6. Feb 5, 2012 #5
    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?
  7. Feb 5, 2012 #6


    User Avatar
    Science Advisor
    Homework Helper

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook