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? Inverse Series method?

  1. Nov 8, 2007 #1

    im trying to find out how to get the inverse of the series 1/1+1/2+1/3+1/4....
    i know the output of 30 is approximately 3.99 and 450 is around 6.6, but I'd like to be able to find the fraction range for outputs of numbers such as 10, 20, and 72. The online series calculators tend to die around output 14. My knowledge of integrals and series is weak at best.
  2. jcsd
  3. Nov 9, 2007 #2

    Gib Z

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    Well the series you stated is known as the Harmonic series, and is famously known to diverge because if it did not, it would imply the number of elements in the set of the Prime numbers is finite.

    It seems like even though it diverges, you want to approximate its values for a certain amount of terms, and I can help you there. There is a well known asymptotic approximation for the Harmonic series, which diverges as well, but even though it does, its a good approximation because it diverges at a similar rate.

    Basically [tex]\lim_{n\to\infty} \frac{ \sum_{k=1}^{n} (1/k) }{\log_e (n) +\gamma} = 1[/tex] where lower case gamma denotes the Euler-Mascheroni constant, which from memory is about 0.577 though there are better numerical approximations.

    What that means is that as n becomes larger, [itex]\log_e n + \gamma[/itex] will approximate the n-th partial sum of the harmonic series will less and less error. So say I wanted to know how many terms I would need for an output of 10, an approximate solution can be obtained by setting [itex]\log_e n + \gamma = 10[/itex], so n is approximately 12,370.

    As a side note, we can see why the asymptotic expression for the harmonic series is a very plausible thing if we plot y=1/x on the Cartesian plane. Looking at the region for values of x onwards, we know the area bounded by x=1, x=k, y=1/x and y=0 (the area under the curve between x=1 and k) is given by [itex]\log_e k[/itex]. Now draw rectangles, first one between x=1 and 2, with height of 1, then between x=2 and 3, height of 1/2, and general term; rectangle between x=a and a+1, height of 1/a.

    We can see that the difference in the area of the rectangles and the area under the curve is only a finite amount, the rectangles approximate the curve better and better. Now it is merely a numerical computation to evaluate the finite amount of difference.
  4. Nov 9, 2007 #3


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    My interpretation of "inverse" was the other way around: For what n is the series equal to (or larger than) M? That problem only makes sense because the series diverges and so is, for some n, larger than or equal to any M. Unforunately, I also do not see any reasonable way of finding that.
  5. Nov 10, 2007 #4

    Gib Z

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    I believe I did answer that =] In fact I evaluated an example, M=10, then n= 12, 370.
  6. Nov 10, 2007 #5
    Here is a simple approximation.

    Consider [tex]f(x) = \frac{1}{x}[/tex] on [tex][1,\infty)[/tex] this is a decreasing,continous,positive function. So that means, [tex]\sum_{k=1}^{n-1}\frac{1}{k} \geq \int_1^n \frac{dx}{x} \implies \sum_{k=1}^n \frac{1}{k} \geq \log n + \frac{1}{n}[/tex] and [tex]\sum_{k=2}^n \frac{1}{k} \leq \int_1^n \frac{dx}{x} \implies \sum_{k=1}^n \frac{1}{k} \leq \log n + 1[/tex]. Together these inequalities say [tex]\log n + \frac{1}{n} \leq \sum_{k=1}^n \frac{1}{k} \leq \log n + 1[/tex]
  7. Jun 12, 2008 #6
    Update: I fiddled with the series for a while and discovered (probably to no one's surprise) that the difference between the sum of any two integers such as ~11 and ~12, or between ~15 and ~16 is a ratio of total fractions approaching e (2.7182818...)

    Here's a small list of the final "k"
    (sigma[a,b,1/k] a = first number, b = last number, equation sums all ranges a to b, thus +1/1, +1/2, +1/3… etc. Example: sigma[1,11,1/k] = 3.0198… Rounds down to 3.)

    note: im using ~ for "approximately equals)

    1, 835,426
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