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!

Convergence of r^n

  1. Sep 25, 2011 #1
    I know this is like very basic, but my brain just somehow couldn't accept it!!

    1. The problem statement, all variables and given/known data
    I don't understand why does the sequence (rn) converges to 0 as n -> infinity when -1<|r|<1

    3. The attempt at a solution
    i did quite a few ways to convince myself.
    Firstly, we know that (1/r)n<(1/r) only if 0<r<1. So by squeeze theorm, (1/r)n converges to 0. Then it also holds for -1<r<0 cz |(1/r)n| = (1/r)n.

    this way seems to be correct but it doesn't seems to be convincing enough.
     
  2. jcsd
  3. Sep 25, 2011 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    If you want to make it more tangible, you could try it with some numbers:

    If r = 0.1, then you get { 0.1, 0.01, 0.001, 0.0001, ... } and you see this quickly tends to 0.
    If r = 0.5, then again { 0.5, 0.25, 0.125, 0.0625, ...}
    Even for r = 0.99, { 0.99, 0.9801, ... } goes much more slowly, but 0.99100 is already of the order of 5 x 10-5.

    Clearly, the boundary case is r = 1, as { 1, 1, 1, .... } never converges to 0.

    For the rigorous proof, refer to your own post :-) What you did there is correct.
     
  4. Sep 25, 2011 #3
    |r| < 1 so r = 1/x where |x|>1. r^n = 1/(x^n). If |x| >1 then clearly x^n becomes infinitely large as n goes to infinity. So 1/(big number) goes to zero.
     
  5. Sep 25, 2011 #4
    Ok. Thanks so much
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Convergence of r^n
  1. Is C^n or R^n compact? (Replies: 11)

  2. Convergence of n^(1/n) (Replies: 1)

Loading...