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Series convergence proof

  1. Oct 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Assume that the series(an) is convergent and that an >= 0 for all n in N. Prove that the series((a^2)n) converges.


    2. Relevant equations



    3. The attempt at a solution

    Alright, this is what I've got so far:
    Assume that the series of an is convergent and that an>=0 for all n in N. In order for the series to be convergent, that would mean that the sequence (an) converges to 0. By definition of convergence, that would mean for epsilon greater than 0, there exists an N in N so that for n>=N:
    |an|<epsilon and furthermore:
    -epsilon<an<epsilon

    This is where I get stuck... am I allowed to just multiply through by an to show that
    -e(an)<((a^2)n)<e(an)?

    And since series of (an) converges, and e is a constant, that would mean that series(epsilon*an) also converges, and by the comparison test that would mean that:
    series((a^2)n) converges as well.

    I don't know if what I'm doing is right, if it isn't then any tips would be great!
     
  2. jcsd
  3. Oct 12, 2009 #2
    why must [itex]a_n[/itex] converge to 0?

    consider the sequence [itex]a_n=\{1,2,3,4,5,5,5,5,5,5,5,5,5,5,5,..........\}[/itex]. here you have [itex]a_n \geq 0[/itex] and it converges to [itex]5 \neq 0[/itex]

    i think your sort of on the right track though.
    im not very good at sequences so i would wait for someone else's opinion but surely you could say that since [itex]a_n[/itex] converges, [itex] \exists N_0 \in \mathbb{N}[/itex] such that [itex]\forall n \geq N_0[/itex] we have [itex]-\epsilon < a_n < \epsilon[/itex]
    now multiply through by [itex]a_n[/itex]
    getting [itex]-\epsilon a_n < (a_n)^2 < \epsilon a_n[/itex]
    now the sequence [itex]-\epsilon a_n[/itex] will be bounded as follows from the convergence of [itex]a_n[/itex]:
    [itex]-\epsilon ( -\epsilon) < -\epsilon a_n < -\epsilon (\epsilon)[/itex]
    and similarly for the sequence [itex]\epsilon a_n[/itex]:
    [itex]\epsilon (-\epsilon) < \epsilon a_n < \epsilon(\epsilon)[/itex]
    so [itex](a_n)^2[/itex] is bounded between
    [itex]-\epsilon(-\epsilon) < (a_n)^2 < \epsilon( \epsilon) \Rightarrow \epsilon^2 < (a_n)^2 < \epsilon^2[/itex]
    then by the sandwich theorem [itex]a_n[/itex] converges.

    however, as i said, im not very good at analysis so this could be completely wrong but its my shot at the answer. wait for someone better to give their 2 cents....
     
  4. Oct 12, 2009 #3

    Dick

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    Since you know a_n->0 since the series a_n converges, then there is an N such that |a_n|<1 for all n>N. That means |a_n|^2<|a_n|. Think comparison test.
     
  5. Oct 12, 2009 #4
    so basically what you did was the same as me but you assumed that epsilon was less than or equal to 1, but you still multiplied through by a_n? If that's the case, then I understand, if not, please correct me where I'm wrong...
     
  6. Oct 12, 2009 #5
    but i do understand for all values between 0 and 1, the square of that value is less than the original value, would it be better to just explain that rather than multiplying through by a_n?
     
  7. Oct 12, 2009 #6

    Dick

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    You can do it either way, sure. Showing |a_n|^2<e|a_n| also shows |a_n|^2 converges. Just seemed nicer to pick e=1.
     
  8. Oct 12, 2009 #7
    alright, thanks so much for your help!!
     
  9. Oct 12, 2009 #8
    why does this ahve to converge to 0?
     
  10. Oct 12, 2009 #9
  11. Oct 12, 2009 #10
    Looking at your example, saying that an converges to 5 would mean that the series of an, would be increasing by 5 each time. The only way for the series to approach a number is if the values of an you keep adding on get smaller and smaller, and eventually approach zero, that way the sum can approach a specific number. Meaning if an doesn't approach zero that means that the sum of an diverges, because it would be unbounded.
     
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