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!

If Ʃa_n converges, with a_n > 0, then Ʃ(a_n)^2 always converges

  1. Mar 15, 2013 #1
    1. The problem statement, all variables and given/known data
    if Ʃa_n converges, with a_n > 0, then Ʃ(a_n)^2 always converges


    2. Relevant equations

    n/a

    3. The attempt at a solution

    I am at a complete loss. I have tried using partial sums, cauchy criterion, and I tried using ratio test which seems to work but I am not sure.

    Since Ʃa_n converges then by ratio test

    lim n->∞ a_n+1 / a_n < 1

    Now we apply ratio test to Ʃ(a_n)^2

    lim n→∞ (a_n+1)^2 / (a_n)^2 = (lim n→∞ a_n+1 / a_n)^2 < 1^2 = 1

    Thus by ratio test Ʃ(a_n)^2 converges.

    This working did not utilize the condition a_n > 0, so it seems suspect to me.
     
  2. jcsd
  3. Mar 15, 2013 #2

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hint: if ##0 < a_n < 1##, then ##a_n^2 < a_n##.
     
  4. Mar 15, 2013 #3
    Ok so you use comparison test right? since (a_n)^2 < a_n and a_n converges then (a_n)^2 converges as well, but is it correct to assume 0<a_n<1? Can you assume that because since
    Ʃa_n converges that implies a_n → 0 and for some n≥N a_n is in (0,1)?
     
  5. Mar 15, 2013 #4

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, that's right, or putting it more precisely, there exists an ##N## such that ##a_n \in (0,1)## for all ##n \geq N##. And then use the comparison test.
     
  6. Mar 15, 2013 #5
    thanks a bunch J
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted