The Harmonic Series

Before the closed-form discovery of the area under a hyperbola, Jacob Bernoulli had proven that the following series (Harmonic) diverges, in 1689 A.D. Diverging as a function of counting numbers ($n$) means the sum of the sequence of terms is larger than any number, provided the count of terms ($m$) is sufficiently large. $$ \sum\limits_1^m n^{-1} = 1 + 2^{-1} + 3^{-1} + \cdots + m^{-1} $$ Where the function, $n^{-1}$, being summed over can be thought of geometrically as the height of a rectangle with width of one. That is without changing the terms (except by putting a factor of one out front, for the width of each rectangle) or the quantity which is the sum, so that the sum is equivalent to the geometric area of all the rectangular area terms up to the $m$'th term/rectangle.

Bernoulli proved the divergence by constructing the series from partitions of subsequences: $$ \sum\limits_{n=1}^\infty n^{-1} = \sum\limits_{n_1=1}^{m_1=1} n_1^{-1} + \sum\limits_{n_2=m_1+1}^{m_2=n_2^2} n_2^{-1} + \sum\limits_{n_3=m_2+1}^{m_2^2} n_3^{-1} + \cdots + \sum\limits_{n_q=m_{q-1}+1}^{m_q^2} n_q^{-1} + \cdots $$ Where each sub-summation, over $n_q$, is greater than or equal to unity thus surpassing, in summation, atny given number. To illustrate this, below it will be elucidated indirectly through the following subsequences, where first instead of going from $m$ to $m^2$ I went as far as necessary to exceed one, starting with $m_2=4$: $$ \sum\limits_2^4 n^{-1} = 2^{-1} + 3^{-1} + 4^{-1} = 1.08\overline{3} $$ $$ \sum\limits_5^{12\lt 25} n^{-1} = 5^{-1} + 6^{-1} +\cdots+ 12^{-1} \approx 1.02 $$ $$ \sum\limits_{13}^{33\lt 169} n^{-1} = 13^{-1} + 14^{-1} +\cdots+ 33^{-1} \approx 1.01 $$ So that in $33$ terms of the series we see the sum is about $4.11$, where I included the first term/subsequence being equal to one.

to add credibility to such a partitioning of the series being a summation of numbers each greater than unity, we use the formula for the area under a smooth hyperbola: For an arbitrary subsequence starting at $(m_{q-1}+1)=1,000$, we use the solution to the continuous function which goes over the counting numbers but instead of each $n^{-1}$-height term getting a unit-width rectangle (same height) it gets a width of some fraction $b=1/s$ where $s$ stands for steps between counting-number terms of the H. series, and for the rectangle adjacent to the $n$'th one, the $(n+1)$'th one of height $(n+1)^{-1}$ also $$ \sum\limits_{1E+3}^{1E+6} n^{-1} = (1E-3) + (1001^{-1}) + (1002^{-1}) +\cdots+ (1E-6) \approx \int\limits_{1E+3}^{1E+6} dx/x = \ln(1E+6)-\ln(1E+3) = 6.9 $$ In use was the scientific notation to indicate a thousand and its square (a million), and to be clear $(1E+3)^{-1}=1000^{-1}=1E-3=0.001$ which is only that simple when the number is $1$ times a power of $10$.

So, while the hyperbola is arbitrarily close to zero for sufficiently large $x$, its rate of getting there is slow enough that the infinite sum of the series is also infinite.

If you were wondering how close an approximation the Natural log is for a summation over the same interval, the error is constrained to be less than that for $m=1$ (difference being one), while being less than that over the entire interval $m \to \infty$. The difference over the entire interval is defined as the Euler-Mascheroni constant: $$ 0.57721\ldots=\lim_{m \to \infty} \sum\limits_1^m n^{-1} - \ln(m) $$ The deviation for finite-$m$ is a saw-tooth function geometrically understood as the area under the hyperbola $x^{-1}$ starting at $x=1$ subtracted from the area of the union of the unit-width-by-$n^{-1}$-high contiguous rectangles, with the first unit-width rectangle being positioned adjacent to the ordinate axis (spanning from zero to one in horizontal and vertical sides), so that the first rectangle isn't under the hyperbola part under evaluation (so that the $m=1$ case works).

Which we can put into perspective by comparing it with the difference between the two quantities for $m=33$: $\ln(33)-\ln(1) = 3.53$, and for the partial-sum of the Harmonic we have, $1 + 1.08 + 1.02 + 1.01 = 4.11$ so the error is $4.11 - 3.53 = 0.613$, which is between the famous limit and one.

Below is a plot of the hyperbola over the (abscissa) interval $[0.8, 10]$, which produces an ordinate interval $[0.1, 1.25]$. The abscissa is drawn from $0$ to $10$, but of course we have to exclude the hyperbola values for some $x$ around zero (because the hyperbola at $x\approx 0$ is much greater than at $x=1$). Overlaid are the rectangles of the Harmonic series, with the ten points the hyperbola function is equal to a term of the series marked.

Harmonic series versus hyperbola.

A plot of the hyperbolic function, on the interval $[1, 10]$. Also the terms of the Harmonic series are represented as rectangles of unit-width.

To compare the eventually discovered integral of the hyperbola

In summary, the integral of the hyperbola was used to approximate the Harmonic series, elucidating the subtlety of the divergance of the series.

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