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Integral test, basic comparsion test, limit comparsion test

  1. Apr 4, 2007 #1
    1) [​IMG]

    This question comes from a section (infinite series) related to the integral test, basic comparsion test, and limit comparsion test, so I believe that I have to use one of them. However, I seriously have no idea how to prove this...can someone give me some hints or guidelines on how to solve this problem?

    Thanks a lot!
     
  2. jcsd
  3. Apr 4, 2007 #2

    HallsofIvy

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    Compare ak/k with 1/k2!
     
  4. Apr 4, 2007 #3
    But I don't know what a_k is, how can I know which of (a_k)/k and 1/k^2 is bigger?

    And also, how can we make use of the fact that Sigma (a_k)^2 converges?
     
    Last edited: Apr 4, 2007
  5. Apr 4, 2007 #4
    I vaguely recall that the Cauchy-Schwarz inequality was useful in a problem similar to this. Or possibly the same problem. I forget.
     
  6. Apr 4, 2007 #5
    I haven't learnt the Cauchy-Schwarz inequality in this course, is there any way to prove without it?
     
  7. Apr 4, 2007 #6
    I don't know of any off the top of my head. Sorry. But you might look for versions that you could prove relatively easily from what you know. Said inequality comes in many different flavors.
     
  8. Apr 4, 2007 #7
    Any hints? Can someone please help me?
     
  9. Apr 4, 2007 #8
    use the limit comparison test...

    if lim(n->infinity) of (a_k/k)/(a^2_k) = L > 0 (is greater than zero because a_k has all nonnegative terms) and sum of (a^2_k) converges (which it says it does), then sum of (a_k/k) must converge by the limit comparison test

    lim(n->infinity) of (a_k/k)/(a^2_k)= lim of 1/(k*a_k) which is greater than zero
    ...if you didn't get the limit part
     
  10. Apr 5, 2007 #9
    But lim(n->infinity) of (a_k/k)/(a^2_k) might not even exist, so how can you use the limit comparsion test?
     
  11. Apr 5, 2007 #10

    Dick

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    This is SO exactly Cauchy-Schwarz (as Mystic998 pointed out). If A is the infinite dimensional vector with components a_k (under the usual L_2 norm) and B is 1/k then this is the same as A.B<=|A|*|B|. A.B is what we want to prove finite, |A| is given to be finite and |B| is known to be finite. QED. If you want to prove this without CS I do think you need to find a proof of CS and modify it to this special case.
     
  12. Apr 6, 2007 #11
    no the limit has to exist because the only way it wouldn't was if a_k or k were 0, which would make the summations pretty stupid because adding up infinitely many terms of 0 obviously converges to 0, or the limit could equal infinity which does satisfy the test (L>0). i'm telling you man limit comparison is the way to go. almost positive.
     
  13. Apr 6, 2007 #12

    Dick

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    I think if you want to apply the limit comparison test to (a_k/k)/(a^2_k) to conclude a_k/k converges you need to show that the limit is zero. Not greater than zero.
     
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