Before the closed-form discovery of the area under a hyperbola, Jacob Bernoulli had proven that the following series, named for the harmonic constributions from a strung bridge, diverges, in 1689 A.D. To wit, the divergance of the series is most readily understood to be a function of how far along in the sequence, and then include in each subsequent grouping enough terms to form its own sum of a constant or minimum amount. The harmonic series sum is composed of terms with one for each number of half-wavelengths (which is one for every two fixed point nodes) which fit in the unit (bridge length). The sum of interest is that simple counting sequence used as the denominator, or size of the half-wavelength in the sequence as a function of the unit bridge length. The first such subsequence is as follows.

$$ \sum\limits_1^m n^{-1} = 1 + 2^{-1} + 3^{-1} + \cdots + m^{-1} $$

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, any 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.

In order 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 Harmonic 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 above is 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 $x=1$). The term $n^{-1}$, can also be thought of geometrically as the height of the $n$’th rectangle with width of one. Overlaid are the rectangles of the Harmonic series, with the ten points the hyperbola function is equal to a term of the series marked.

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.