The Harmonic Series
Before the closed-form discovery of the area under a hyperbola, Johann Bernoulli had proven that the Harmonic Series diverges, in 1689 AD. The Series is the infinite sum of fractions, similar in that way to the Geometric Series, but the kernel for this series is $(1/n)$, not $(r^n)$. It was given its name for the harmonic overtone contributions to the primary vibration (note, standing wave) of a stringed musical instrument. First, the divergence of the series is established to be a function of its being unbounded, on route to comparing the series to the area under the hyperbola, $y=1/x$, as described by the natural logarithm.
The partial sum, called a Harmonic Number, symobolized as $H_m$, is the sum from term one up through the $m$'th term as follows,
Bernoulli proved the divergence of the series by constructing the series from partitions of the sequence, each difference of harmonic numbers was chosen for its approximation to unity, so that the sum of the unbounded series is in fact unbounded, in a logarithmic way.
$$ \sum\limits_{n=1}^\infty n^{-1} = \sum\limits_{n_1=1}^{m_1=1} n_1^{-1} + \sum\limits_{n_2=m_1+1}^{m_2=n_2^2} n_2^{-1} + \sum\limits_{n_3=m_2+1}^{m_3^2} n_3^{-1} + \cdots + \sum\limits_{n_q=m_{q-1}+1}^{m_q^2} n_q^{-1} + \cdots $$Where each partial series, over $n_q$, is greater than or equal to unity thus surpassing, any given number proposed as the upper bound. To illustrate this, it is given enumeratively in the following subsequences, where first instead of going from $m$ to $m^2$ the upperbound is placed as far as necessary for the partial series to exceed one, starting with $m_1=1, m_2=4=2^2$:
$$ \sum\limits_2^{4=2^2} n^{-1} = 2^{-1} + 3^{-1} + 4^{-1} = 1.08\overline{3} $$$$ \sum\limits_5^{12= 0.48*5^2} n^{-1} = 5^{-1} + 6^{-1} +\cdots+ 12^{-1} \approx 1.02 $$
$$ \sum\limits_{13}^{33=0.19*13^2} n^{-1} = 13^{-1} + 14^{-1} +\cdots+ 33^{-1} \approx 1.01 $$
So that in $33$ terms of the series we see the sum is about $4.11$, compare to $\log_e(33)=3.50$, with a difference of $0.61$, clearly divergant since there is no upperbound to $x^2$.
For a few values of $m\in\{1,2,3,10,40,100\}$ the value of $H_m$ is compared to the natural log of $m$, $f(m)=\log_e(m)=\ln(m)$, with the difference, $H_m-\ln(m)$, labelled as $\Delta$. Note the asymptotic error as $m$ increases, $0.577 \lt \Delta \le 1$, indicating the natural logarithm is a good and well-behaved approximation for the Harmonic Numbers.
| $m$ | $H_m$ | $\ln(m)$ | $\Delta$ |
|---|---|---|---|
| 1 | 1 | 0 | 1 |
| 2 | 1.5 | 0.69 | 0.81 |
| 3 | 1.83 | 1.1 | 0.73 |
| 10 | 2.93 | 2.30 | 0.63 |
| 40 | 4.28 | 3.69 | 0.59 |
| 100 | 5.19 | 4.60 | 0.59 |
| 1000 | 7.49 | 6.91 | 0.58 |
The difference between the Harmonic Number and the natural logarithm has most of its error at the beginning, so that the difference between two Harmonic Numbers is less than the difference between the logarithm of the two numbers, as shown in the following table, where the first column is the difference between subsequent Harmonic Numbers, the second column is the difference of natural logarithms, and the third column is the difference between the
| $(m_1,m_2)$ | $H_{m_2}-H_{m_1}$ | $\ln(m_2)-\ln(m_1)$ | $\Delta=\text{col.}3-\text{col.}2$ |
|---|---|---|---|
| $(1,2)$ | $2^{-1}=0.5$ | 0.69 | 0.19 |
| $(2,3)$ | $3^{-1}=0.33$ | 0.41 | 0.08 |
| $(3,4)$ | $4^{-1}=0.25$ | 0.28 | 0.03 |
| $(4,5)$ | $5^{-1}=0.2$ | 0.22 | 0.02 |
| $(5,6)$ | $6^{-1}=0.16$ | 0.18 | 0.02 |
| $(99,100)$ | $100^{-1}=0.01$ | $0.01005$ | $5E-5$ |
| $(100,1000)$ | $7.49-5.19=2.30$ | $2.30$ | $0.00$ |
The error in the approximation of the Harmonic series by the natural logarithm is decreasing as $m$ increases, so the error in differences is also decreasing. Clearly, as the smaller of the two $m$'s increases (the lower-bound of the sequence), the difference between those two Harmonic numbers and the difference between their two corresponding natural logarithms converges. This is because the distance in abscissa is staying the same (one unit for successive Harmonic numbers), while the height of the rectangles is decreasing, so that the area under the hyperbola (exactly the difference of natural logs) is approaching the area of the rectangle as $m$ increases (difference between the hyperbola and the floored denominator-height becomes smaller relative to one, the distance in abscissa).
So, back to the partitioning of the H.S. by sequences of the form, $(m,m^2)$, amounting to a summation of numbers each greater than unity, or $\ln(x^2)-\ln(x)=\ln(x)$, we use the formula for the area under a smooth hyperbola: For an arbitrary subsequence starting at $(m_{q-1}+1)=1,000$ and ending at $m_q=1,000,000$, the area under the hyperbola from $1,000$ to $1,000,000$ is:
$$ H_{1E-6} -H_{1E-3} \approxeq \int\limits_{1E+3}^{1E+6} dx/x = \ln(1E+6)-\ln(1E+3) = 6.9 $$In use above is the scientific notation to indicate a thousand and its square (a million), and to be clear $(1E+3)^{-1}=1000^{-1}=0.001=1E-3$ which is only that slick 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 Constant (published in 1734):
$$ 0.57721\ldots=\lim_{m \to \infty} \sum\limits_1^m n^{-1} - \ln(m) $$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, square in square coordinates), so that the first rectangle isn't under the hyperbola part under evaluation (so that the $m=1$ case of maximum deviation works as expected, with $\ln(1)=0$).
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.
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.
In summary, the integral of the hyperbola was used by Euler to approximate the Harmonic series, elucidating the subtlety of the divergence of the series, as the logarithm is slowly changing, but unbounded.