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!

What does ( 0.25)^(3n) converge towards? (EASY Q)

  1. Sep 30, 2009 #1
    Really quick question here:

    I want to find the sommation of f(x) = 0.253n for x being a Natural ( N) number going from 0 to infinity.

    k1k2n converges towards what, as a general rule?

    Where k1, k2 are constants and n= 0,1,2,3 ...

    EDIT: NO, not what it converges to, but what the sommation is equal to... as in f(0)+f(1) +f(2) + ...
     
    Last edited: Sep 30, 2009
  2. jcsd
  3. Sep 30, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It's a geometric series isn't it? It's ((1/4)^3)^n. What do you know about the sum of geometric series?
     
  4. Sep 30, 2009 #3
    Well, I'm looking for = ∑rn-1 for n going from 1 to infinity... But isn't there a formula for where n starts at 0?

    I also know that you can play around sith Sn= (1-rn) / (1-r) , but I don't remember/know where this leads to.
     
    Last edited: Sep 30, 2009
  5. Sep 30, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I'll help you out. The sum for n=0 to infinity of r^n is 1/(1-r) for r<1. That's the n->infinity limit of your Sn.
     
  6. Sep 30, 2009 #5
    All right, thanks a lot
     
    Last edited: Sep 30, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: What does ( 0.25)^(3n) converge towards? (EASY Q)
Loading...