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Partial sum of the harmonic series

  1. May 24, 2014 #1
    1. The problem statement, all variables and given/known data
    I have to find a natural number N that satisfies this equation:

    [itex]\sum^{N}_{i=1} \frac{1}{i} > 100 [/itex]

    2. Relevant equations
    I tried finding a close form of the sum but couldn't find anything useful.

    3. The attempt at a solution
    Well after trying some numbers in maple I found a few very large numbers satisfying the inequality.

    Any hints are much appreciated.
  2. jcsd
  3. May 24, 2014 #2


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    Think of the sum as representing a Riemann sum approximating the value of an integral.
  4. May 24, 2014 #3

    Thanks for the tip, I have only been doing analysis for around 3 months and we haven't started using integrals yet. Until now we have only looked at sequences and series, are there any other ways of doing it?
    Anyways, here is my attempt.
    [itex]\sum[/itex][itex]^{N}_{i}\frac{1}{i} > \int ^{N+1}_{1} \frac{1}{i}di = ln(N+1) = 100[/itex]
    (From wikipedia)
    Then I can easily solve for N and get [itex]N=e^{100}-1\approx 2.6881\cdot10^{43}[/itex]
    Thanks for the help :)
  5. May 24, 2014 #4


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    Yes, I think that's it. But your N there isn't an integer - you want to pick any integer greater than that number. Just a detail.
  6. May 24, 2014 #5


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    Have you seen the proof that ##\sum_1^\infty \frac 1 i## diverges? You can use that to get a value of N without doing any calculations with logs and exponentials.

    The basic idea of the proof is
    1/3 + 1/4 > 2(1/4) = 1/2
    1/5 + 1/6 + 1/7 + 1/8 > 4(1/8) = 1/2
    1/9 .... + 1/16 > 8(1/16) = 1/2

    So you can find N = a power of 2, that makes the sum > 200(1/2).
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