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Non-Negative, Decreasing Sequence: If series converges then lim n*a_n = 0

  1. Mar 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Let {a_n} be positive, decreasing. Show that if a_1 + a_2 + a_3 + ... converges then lim n * a_n = 0.

    2. Relevant equations

    None.

    3. The attempt at a solution

    Consider the harmonic series 1 + 1/2 + 1/3 + ... . Observe that

    [a_n] / [1 / n] = n * a_n .

    Since 1 + 1/2 + 1/3 + ... diverges, by the Limit Comparison Test we know that, if it exists, lim n * a_n = 0, for otherwise a_1 + a_2 + a_3 + ... would also diverge, a contradiction.

    My question is, how do I prove that the limit necessarily exists? Considering f(x) = 1/x, I imagine, for example, a function g(x) in the shape of a step function such that liminf g(x) / f(x) = 0 and limsup g(x) / f(x) = 1. In this case the limit would not exist so I guess (in light of the problem statement) that the series g(1) + g(2) + g(3) + ... must diverge, but I don't know how to show this.
     
    Last edited: Mar 19, 2009
  2. jcsd
  3. Mar 19, 2009 #2

    Dick

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    Ok. If limit n*a_n is not zero, then there is an e>0 such that there exists an infinite number of integers N such that N*a_N>e. Right? If you are onboard with that, then i) there is an N0 such that N0*a_N0>e. There are N0 terms a_n less than N0 and each one is greater than e/N0. So the sum of those terms is greater than e. Now, there is an N1>2*N0 such that N1*a_N1>e. That means each of the a_n for n between N0 and N1 is greater than N1/e. There are more than N1-N0>N1/2 such terms. So the sum of all of those terms is greater than e/2. Now there is an N2>2*N1 such that N2*a*N2>e. The sum of all of the a_n for n between N2 and N1 is again greater than e/2. I can keep this up forever. And get an infinite number of e/2's. But the sum of the a_n is supposed to be finite and e>0. How can that be? I've been a little sketchy about exactly how many terms are between the integers, but that's the spirit of the proof.
     
  4. Mar 19, 2009 #3
    Got it, thank you. I see now that it is much the same as showing that 1 + 1/2 + 1/3 + ... diverges.
     
  5. Mar 19, 2009 #4

    Dick

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    It's exactly the same. You catch on fast.
     
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