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A Convergence Question

  1. Jan 18, 2009 #1
    The problem statement, all variables and given/known data
    Suppose {a_k} is a decreasing sequence of real numbers with a_k >= 0 for all k. Show that if a_1 + a_2 + ... converges, then lim k*a_k = 0. Is the converse true?

    Relevant equations
    Fact: If {a_k} is a sequence of real numbers such that a_1 + a_2 + ... converges, then lim a_k = 0.

    The attempt at a solution
    It seems to me that {a_k} would have to decrease faster than {1/k} in order for ka_k to converge to 0. For example, if {a_k} = {1/k}, lim k*a_k = 1. However, if {a_k} = {1/2^k}, then lim k*a_k = 0. This intuition has failed to lead me to the answer though. It has also come to my attention that if a_1 + a_2 + ... converges to A, that 0 <= lim k*a_k <= A, given that the limit exists. This fact however has also been unhelpful. Any tips?
     
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  3. Jan 18, 2009 #2

    Dick

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    Suppose the limit k*a_k is not zero. That means that there is an e>0 such that k*a_k>e for an infinite number of different values of k. In particular you can find a subsequence k_i such that k_{i+1}>2*k_i for all i and k_i*A_{k_i}>e. Use that a_k is decreasing and think about how you can estimate the sum of the a_i for k_i<i<k_{i+1}.
     
  4. Jan 18, 2009 #3
    The sum of the a_i for k_i<i<k_{i+1} is greater than or equal to the sum a_i for k_i < i <= 2k_i and that sum is greater than or equal to k_i*a_{2*k_i}. Where do we go from here?
     
  5. Jan 18, 2009 #4

    Dick

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    There are more than k_{i+1}/2 terms in that group of terms and they are all greater than or equal to e/k_{i+1}. Hence?
     
  6. Jan 18, 2009 #5
    Actually, there are less than k_{i+1}/2 since k_{i+1}/2 > k_i. And where do you get that the are >= e/k_{i+1}?
     
  7. Jan 18, 2009 #6

    Dick

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    If k_{i+1}/2>k_{i} then k_{i+1}-k_{i}>k_{i+1}-k_{i+1}/2. I'm subtracting a LARGER number on the right side. And I picked a subsequence such that a_k_{i}*k_{i}>e and the a_k's are decreasing.
     
  8. Jan 18, 2009 #7
    OK, I understand now. But I don't know what you want me to conclude. For any k, we have that a_1 > a_2 > ... > a_k > e/k. What is so special about k_{i+1}?
     
  9. Jan 18, 2009 #8

    Dick

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    The point is that summing the terms between a_k_{i} and a_k_{i+1} gives you more than e/2. There are an infinite number of such intervals. In such a case the sum of the a_k must diverge. It's a proof by contradiction!
     
  10. Jan 18, 2009 #9
    Wow. I honestly did not see that. Let me ask you: What led you to figure that there would be an infinite number of intervals in the sum of a_k that are greater than e/2? Did you look at a particular example?
     
  11. Jan 18, 2009 #10

    Dick

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    It was the same intuition that made you say a_k must approach 0 faster than 1/k. The proof is roughly the same as the proof that a_k=1/k diverges. I looked at it and scratched my head for quite a while. It's not that easy. But no, particular examples aren't that helpful. Did you find the counterexample for the converse?
     
  12. Jan 18, 2009 #11
    I've never seen the proof that a_k = 1/k diverges. You wrote "for quite a while", which I found funny since I've been scratching my head since yesterday.

    This should work as a counter-example: Let a_k = 1/k^{3/2} so that k*a_k = 1/sqrt(k), which converges to 0, but the sum diverges.
     
  13. Jan 18, 2009 #12

    Dick

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    Look up the proof, you'll see the resemblance. No, the sum of a_k=1/k^(3/2) does converge. It's a power series. You want something that diverges REALLY slowly. Slowly enough that the series is divergent but k*a_k still goes to zero.
     
  14. Jan 18, 2009 #13
    Oops. I was thinking about the sum of 1/sqrt(k) which diverges. My mistake. Let me ponder and tinker with this tonight. Hopefully I'll have something by tomorrow.
     
  15. Jan 18, 2009 #14

    Dick

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    BTW the elementary proof the harmonic series diverges is that the sum of 1/k from 1/2^k to 1/2^(k+1) is greater than 1/2. There are 2^k terms of size greater than 1/2^(k+1). Let me know what you find for the counterexample. There's a pretty easy one. It's just a LITTLE smaller than 1/k.
     
    Last edited: Jan 19, 2009
  16. Jan 19, 2009 #15
    I did some googling and found that the sum of [tex]\sqrt[n]{2} - 1[/tex] diverges slowly and also [tex]\lim_{n \to \infty} (\sqrt[n]{2} - 1) = 0[/tex]. I couldn't think of one as I don't really have a feel for these things yet.
     
  17. Jan 19, 2009 #16
    I just had a flash: Were you thinking about 1/(k ln k)? That should work too, I think.
     
  18. Jan 19, 2009 #17

    Dick

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    That's the one. An integral test shows it diverges, but k*(1/(k ln k)) goes to zero.
     
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