Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A general formula for the sum of the series 1/k?

  1. Mar 23, 2013 #1
    Is there any general expression for the sum to nth term of the series 1/k?

    I know that for sufficiently large indices, a good approximation can be ln(b/a) where b and a are the upper and lower limits respectively.
    I've tried to do something very simple for the exact sum by constructing unit length rectangles between the curve of the function 1/x and the x axis. This shows clearly that as the values of x gets larger and larger, the integral, being the portion of the rectangle below the curve, gets closer and closer to the actual sum, the total area of the rectangles.
    59vgjn.png
    But I don't seem to be able to get any further than this. I did try to account for the extra (above-curve) portions of the rectangles by treating them approximately as triangles(what better option did I have?) and using the 0.5*b*h formula for their area but later realizing that the heights are changing, I thought it wouldn't be so easy.
    So, anyone willing to give it a try?
     
  2. jcsd
  3. Mar 23, 2013 #2

    pwsnafu

    User Avatar
    Science Advisor

  4. Mar 24, 2013 #3
    I couldn't understand wikipedia pwsnafu. Care to explain?
     
  5. Mar 24, 2013 #4

    pwsnafu

    User Avatar
    Science Advisor

    Sure. The harmonic numbers are defined ##H_n = \sum_{k=1}^n \frac{1}{k}##. Now obviously when you calculate your sum, you'll get a rational number. Fun little fact, the numerators you get (in order) is called the Wolstenholme numbers, and the denominators are called...I don't think they have a name. Hmm.

    Anyway, you want to calculate ##\sum_{k=a}^b \frac{1}{k}## when that is ##\sum_{k=1}^b \frac{1}{k} - \sum_{k=1}^{a-1} \frac{1}{k} = H_{b}-H_{a-1}##. So that's the answer, you just look them up on the internet and we are done.

    But that's not really a satisfactory answer. Now Wikipedia shows two other forms:
    ##H_n = \int_0^1 \frac{1-x^n}{1-x} dx = \sum_{k=1}^{n}(-1)^k \frac{1}{k} {n \choose k}##.
    The first equality is easy: we know
    ##\frac{1-x^n}{1-x} = \frac{(1-x)(1+x+x^2+\ldots+x^{n-1})}{1-x} = 1+x+x^2+\ldots+x^{n-1}##.
    Integrating term by term we get
    ##x+\frac{x^2}{2}+\frac{x^3}{3} + \ldots + \frac{x^n}{n}##. The terms of the integral is just x=1 and x=0, so we obtain ##H_n##.

    Follow?
     
  6. Mar 24, 2013 #5
    I got lost from the integral for Hn.
    Even if I take it as a given, I don't get how it equals the summation part.
     
  7. Mar 24, 2013 #6

    pwsnafu

    User Avatar
    Science Advisor

    But you do follow what I wrote right?

    So moving onto the summation. We want to solve ##\int_0^1 \frac{1-x^n}{1-x}dx##.
    We do the substitution ##u = 1-x##. We get
    ##\int_0^1 \frac{1-(1-u)^n}{u} du##
    if this is not immediately obvious verify it.

    Now we expand. You know from the binomial theorem that
    ##(1-u)^n = 1 - u + {n \choose 2} u^2 - \ldots (-1)^n u^n##
    So
    ##1-(1-u)^n = u - {n \choose 2} u^2 + \ldots (-1)^{n-1} u^n##
    And
    ##\frac{1-(1-u)^n}{u} = 1 - {n \choose 2} u + \ldots (-1)^{n-1} u^{n-1}##

    So we calculate
    ##\int_0^1 1 - {n \choose 2} u + \ldots (-1)^{n-1} u^{n-1} \, du##
    is equal to ##u - {n \choose 2} \frac{u^2}{2} + \ldots (-1)^{n-1} \frac{u^n}{n} |_0^1##
    and we are left with
    ##1 - {n \choose 2} \frac{1}{2} + \ldots (-1)^{n-1} \frac{1}{n}##
    and that's exactly the sum formula.

    Edit: 500th post. Yay!
     
  8. Apr 17, 2013 #7
    Thank you very much!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: A general formula for the sum of the series 1/k?
  1. Sum of 1^2 +2^2 +.k^2 (Replies: 7)

Loading...