Integrals 2: Continuous Geometric Series

Geometric Series Approximation

As simple and easy to remember as the GS formula is, its continuous limit is also very accessible, which we will derive directly.

The Continuous Limit of the Series

The integral of $r^x$ is the continuous summation, or Riemann Integral, of the exponential with a proper fraction as base, $r=e^{\ln(r)}$.

$$ \sum_{n=0}^\infty r^n \to \int_0^\infty r^x \, dx $$

With $r^x=e^{\ln (r) x}$, so we have:

$$ \int_0^\infty r^x \, dx = \frac{1}{\ln (r)} r^x |_0^\infty = \frac{-1}{\ln (r)} $$

$$ F=\int_0^{x}0.5^t \, dt \frac{1}{\ln (0.5)}(0.5^x-1) \approx 1.44(1-0.5^x) $$

Figure 1. Graph of the integral of the plain argument exponential with a base of one half: $F = 1.44(1-0.5^x)$ (dashed) versus the integrand with the endpoint as abscissa, $x$ (solid line, with area below it depicting the constribution to $F$).

Figure 1 (above) shows the continously evaluated Geometric Series, with the value of the integral being a cumulative sum of the area swept out by the values of the integrand, $f=0.5^x$.

This can be compared to the series for a range of proper fractions, which has some similarity to the approximation of the Harmonic Numbers to the integral of the Hyperbola in that the result is asymptotically accurate.

$r$	$\frac{1}{1-r}$	$\frac{-1}{\ln (r)}$	$\Delta$
$0.25$	$\frac{4}{3}$	$0.72$	$0.61$
$\frac{1}{3}$	$1.5$	$0.91$	$0.59$
$0.5$	$2$	$1.44$	$0.56$
$0.8$	$5$	$4.48$	$0.52$
$0.9$	$10$	$9.49$	$0.51$
$0.9_n$	$1EE{+}n$	$1EE{+}n \, - 0.50_{n}83$	$0.50_{n}83$

Table 1. The Geometric Series for ratios listed are compared to the integral of ratio as the base of an exponential function. The pattern demonstrated is that there is a decreasing error with increasing ratio, and that the approximation is always less by an amount in the set $[0.5, 1.0]$. In the last line, for a ratio of $n$-nines, the subscript of $n$ on the digit indicates an $n$-times repetition of that digit in the number.

The formula for $\Delta$ in the last line of Table 1 (above), demonstrating a lowerbound of one half, was obtained with a desktop calculator.

$$\begin{equation}\label{delta_geo} \Delta(0.9_n) = \frac{1}{1-0.9_n} - \frac{-1}{\ln (0.9_n)} = 0.50_{n}83 \end{equation} $$

The approximation is more accurate the closer $r$ is to one, since the deviation is almost a fixed amount of undershot for a given series' value. Compare this to the Harmonic Numbers approximation, Euler Constant, which undershoots the Harmonic Numbers series for the same reason $-1/\ln(r)$ undershoots $1/(1-r)$, which is the continuous approximation is continuously decreasing while the corresponding series is constant for each unit of abscissa.

Let's look at the set of proper fractions as domain interval for the two functions, to compare the Geometric Series with our continuous function approximation graphically.

Figure 2. The Geometric Series, $f(r)=\frac{1}{1-r}$, is drawn here in red , and its continuous approximation, $f(r) = \frac{-1}{\ln(r)}$, which is drawn in cobalt, the strictly lower ordinate value.

Figure 3. Error Function: The difference between the Geometric Series and its continuous approximation, $f(r) = \frac{1}{1-r} + \frac{1}{\ln(r)}$, is drawn here to emphasize the pattern of error has an upper bound of one, and a lower bound of one half.

Figure 3 (above) demonstrates the formula for $\Delta$ with an $n$-nines ratio, \eqref{delta_geo}. While the error is in a fixed range, the fraction of the underlying value is decreasing with Geometric Series sum. For a $n$-nines fraction, the error is effectively vanishing.

$$\begin{equation}\label{delta_percent} \frac{\Delta(0.9_n)}{\frac{1}{1-0.9_n}} \approx (0.5) (1EE{-}n) = 5EE{-}(n+1) \end{equation} $$

[1] Scriba, Christoph J. (1963). The inverse method of tangents: A dialogue between Leibniz and Newton. Archive for History of Exact Sciences 2 (2):113-137.

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