1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

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

    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    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
    etc

    So you can find N = a power of 2, that makes the sum > 200(1/2).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Partial sum of the harmonic series
  1. Series sum (Replies: 9)

  2. Sum of series (Replies: 2)

  3. Sum of series (Replies: 15)

  4. Sum this series (Replies: 3)

  5. Summing a series (Replies: 1)

Loading...