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Convergence of a series

  1. Apr 25, 2008 #1
    [SOLVED] convergence of a series

    1. The problem statement, all variables and given/known data
    Prove that the convergence of [itex]\sum a_n[/itex] implies the convergence of [itex]\sum \frac{a_n}{n}[/itex] if [itex]a_n \geq 0[/itex].

    2. Relevant equations

    3. The attempt at a solution
    I want to use the comparison test. So, I want to find [itex]N_0[/itex] so that [itex]n \geq N_0[/itex] implies [tex]\frac{\sqrt{a_n}}{n} \leq a_n[/tex] which is clearly not in general possible. So I am stuck. Maybe I need to replace a_n by a subsequence of a_n or something?
  2. jcsd
  3. Apr 25, 2008 #2
    consider any partial sum A_m = Sum(a_k/k, k = 1,..., m).

    |A_m| = |a_1/1 + ... + a_k/k| = |(1/1) * a_1 + ... + (1/k)*a_k| <= ...
  4. Apr 25, 2008 #3
    Sorry. I am trying to prove
    \sum \frac{\sqrt{a_n}}{n}

    converges not

    \sum \frac{a_n}{n}

    For some reason, I cannot edit the opening post.
  5. Apr 25, 2008 #4
    ok even easier, just means you won't have to show something else I had in mine,

    Let A_m = sum(sqrt(a_k)/k, k = 1,..., m) be any partial sum, then

    |A_m| = |sqrt(a_1)/1 + ... + sqrt(a_k)/k| = |(1/1)*sqrt(a_1) + ... + (1/k)sqrt(a_k)| <= ...

    use some famous inequality for the next step(not the triangle inequality), what's the other one I'm sure you know!
  6. Apr 25, 2008 #5
    Cauchy-Schwarz. Yay!

    BTW you're k's and m's are mixed up.
  7. Apr 25, 2008 #6
    woops yea, good to see you got it! I'll look at your other question now
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