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Easiest way to calculate the Sum of Harmonic serie?

  1. May 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Calculate the sum of the serie.

    2. Relevant equations

    [tex]\sum^{30}_{i=1} 1/i[/tex]

    3. The attempt at a solution

    I broke the problem into parts. I calculated the sum of the geometric series such as:

    [tex]\sum^{4}_{i=1} 2^{-i}[/tex]
    [tex]\sum^{3}_{i=1} 3^{-i}[/tex]
    [tex]\sum^{2}_{i=1} 5^{-i}[/tex]
    [tex]\sum^{1}_{i=1} 7^{-i}[/tex]
    and so on for each base with a prime number. Then, I need to sum them to other terms. However, it can be very error-some.

    The reason to the question is that I sense that there may be an easier way to calculate the sum. What are the ways to calculate harmonic series? Is my way wrong?
     
  2. jcsd
  3. May 9, 2009 #2
    You can use integration by integrating 1/x and making the upper limit 30 and the lower limit 1.
     
  4. May 10, 2009 #3
    It is a serie, not a continuous function. I have never used integration for series. How does its integration differ from the intregation over a continuous function? Is it an approximated value?
     
  5. May 10, 2009 #4
    Jesus Christ!!! Where did you learn this? Are you sure about your method..

    Cause i think that its wrong
     
  6. May 10, 2009 #5
    well this is quite a good sum.

    Express 1/2 as 1-1/2, 1/3 as 1-2/3 and so on...

    So you will get like

    [tex]30 -\sum_{1}^{29}\frac{r}{r+1} = \sum_{1}^{30}\frac{1}{r}[/tex]



    Now covert the sigma term on the left from r=1 to r=30( right now its 1 to 29) Once done with that. Take the sigma term on RHS so you will get rsquare upon r+1.... and some other term...
    Now convert rsqare to rsquare-1+1 and simplify... So on rhs you will r and 1 upon r+1 under sigma and some other term...

    Now take all the terms from rhs to lhs except the 1 upon r+1... Thats it...
     
  7. May 10, 2009 #6

    http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula" [Broken]
     
    Last edited by a moderator: May 4, 2017
  8. May 10, 2009 #7
    You can apply the Euler-Maclaurin summation formula by using that:

    [tex]\lim_{L\rightarrow \infty}\sum_{n=1}^{L}\frac{1}{n} - \log(L)=\gamma[/tex]

    where [itex]\gamma=0.577215664902\cdots[/itex] is Euler's constant.

    What we need to do is avoid using the lower limit of n = 1 in the Euler-Maclaurin summation formula as the series will not converge well. We can write:

    [tex]\sum_{n=1}^{R}\frac{1}{n} = \sum_{n=1}^{L}\frac{1}{n}-\sum_{n=R+1}^{L}\frac{1}{n}[/tex]

    Where L is larger than R+1 and is arbitrary. Applying the Euler-Maclaurin summation formula to the summation from R+1 to L gives:

    [tex]\sum_{n=1}^{R}\frac{1}{n}=\sum_{n=1}^{L}\frac{1}{n}-\log(L) +\mathcal{O}\left(L^{{-1}}\right) + \log(R+1) - \frac{1}{2(R+1)}-\sum_{k=1}^{\infty}\frac{B_{2k}}{2k}\frac{1}{(R+1)^{2k}}[/tex]

    Talking the limit for L to infinity gives:

    [tex]\sum_{n=1}^{R}\frac{1}{n}=\log(R+1) + \gamma - \frac{1}{2(R+1)}-\sum_{k=1}^{\infty}\frac{B_{2k}}{2k}\frac{1}{(R+1)^{2k}}[/tex]

    If we write:

    [tex]\sum_{n=1}^{R}\frac{1}{n}= \frac{1}{R} + \sum_{n=1}^{R-1}\frac{1}{n}[/tex]

    and apply the summation formula to the summation to R-1, we get a nicer looking formula:

    [tex]\sum_{n=1}^{R}\frac{1}{n}=\log(R) + \gamma + \frac{1}{2R}-\sum_{k=1}^{\infty}\frac{B_{2k}}{2k}\frac{1}{R^{2k}}[/tex]


    For R = 30, we only need the first 3 terms of the summation over the Bernoulli numbers to get a result accurate to within 10^(-14). Writing out these first few terms explicitly gives the formula:

    [tex]\sum_{n=1}^{R}\frac{1}{n}=\log(R) + \gamma + \frac{1}{2R}-\frac{1}{12 R^2}+\frac{1}{120 R^4}-\frac{1}{252 R^6}+\cdots[/tex]
     
    Last edited: May 10, 2009
  9. May 10, 2009 #8
    well there is no need to have such complicated solution... Ofcourse its damn right but the method which i have shown is equally correct(thats what i feel).... And we may be able to do it with more ease
     
    Last edited by a moderator: May 4, 2017
  10. May 10, 2009 #9
    [tex]30 + 30/31 -\sum_{1}^{30}\frac{r}{r+1} = \sum_{1}^{30}\frac{1}{r}[/tex]

    [tex]30 + 30/31 = \sum_{1}^{30}\frac{1}{r} + \frac{r}{r+1}[/tex]

    [tex]30 + 30/31 = \sum_{1}^{30}\frac{1}{r} + \frac{r+1-1}{r+1}[/tex]

    [tex]30 + 30/31 = \sum_{1}^{30}\frac{1}{r} + 1 - \frac{1}{r+1}[/tex]

    [tex]30/31 = \sum_{1}^{30}\frac{1}{r(r+1)}[/tex]

    It is hard to follow. Am I on the right track?
     
  11. May 10, 2009 #10
    [tex]
    30 + 30/31 = \sum_{1}^{30}\frac{1}{r} + \frac{r}{r+1}[/tex]

    [tex]
    30 + 30/31 = \sum_{1}^{30}\frac{r+1+r^2}{(r)(r+1)}[/tex]

    [tex]
    30 + 30/31 = \sum_{1}^{30}\frac{1}{r} + \frac{1}{(r)(r+1)} + \frac{r^2 - 1}{(r)(r+1)}[/tex]

    Now convert the middle term to 1 upon r+1 minus 1 upon r. And the last term would get converted to r-1 upon r.

    Now try doing the sum
     
  12. May 11, 2009 #11
    [tex]
    30 + 30/31 = \sum_{1}^{30}\frac{1}{r} - (\frac{1}{r+1} - \frac{1}{r}) + \frac{r - 1}{r}[/tex]

    [tex]
    30 + 30/31 = \sum_{1}^{30}\frac{2}{r} + \frac{r - 1}{r} - \frac{1}{r + 1}[/tex]

    [tex]
    30 + 30/31 = \sum_{1}^{30}\frac{r + 1}{r} - \frac{1}{r + 1}[/tex]

    [tex]
    30 + 30/31 = \sum_{1}^{30}\frac{(r + 1)^2 - r}{r(r+1)}[/tex]

    It is hard to see where the thing is aiming at.
     
  13. May 11, 2009 #12
    seems that i have made some mistake... I am working on it though
     
  14. May 11, 2009 #13
    http://ecademy.agnesscott.edu/~lriddle/series/rear.htm

    Well i think i have bitten of more than what i can chew.... I have tried every way which is taught to a 12th grader... but in vain... Just have a look at the link

    Count Iblis has rather given a very ingenious solution
     
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