What is the Inverse Series Method for Calculating Output Fractions?

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Discussion Overview

The discussion centers around the Inverse Series Method for calculating output fractions related to the Harmonic series. Participants explore how to determine the number of terms needed for the series to reach specific output values, as well as approximations and interpretations of the series' behavior.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks to find the inverse of the Harmonic series for specific outputs, noting that online calculators fail for larger outputs.
  • Another participant explains the divergence of the Harmonic series and provides an asymptotic approximation involving the Euler-Mascheroni constant, suggesting a method to estimate the number of terms needed for an output of 10.
  • Some participants interpret the term "inverse" differently, focusing on determining for which n the series equals or exceeds a given output M, acknowledging the divergence of the series.
  • A participant presents inequalities that bound the Harmonic series, indicating that the series can be approximated using logarithmic functions.
  • One participant shares an observation that the difference between sums of consecutive integers approaches the mathematical constant e, providing a list of approximate values for various outputs.

Areas of Agreement / Disagreement

Participants express differing interpretations of the term "inverse" in relation to the series, and while some provide approximations and methods, there is no consensus on a definitive approach to calculating the outputs for specific values.

Contextual Notes

Participants acknowledge limitations in their understanding of integrals and series, and there are unresolved mathematical steps in determining the exact outputs for the series.

shintashi
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hi,

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.
 
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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.
 
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.
 
HallsofIvy said:
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.

I believe I did answer that =] In fact I evaluated an example, M=10, then n= 12, 370.
 
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]
 
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: I am using ~ for "approximately equals)

1=1
2~4
3~11
4~31
5~83
.
.
.
227
616
1,674
4,550
12,367
33,617
91,380
248,397
675,215
1, 835,426
4,989,205
13,562,065
36,865,516
100,210,865
272,401,372
 

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