Sum of a Geometric series and its Application


The sum of a geometric series is finite as long as the terms approach zero; as the numbers near zero, they become insignificantly small, allowing a sum to be calculated despite the series being infinite. The sum can be computed using the self-similarity of the series.


Consider the sum of the following geometric series:
s \;=\; 1 \,+\, \frac{2}{3} \,+\, \frac{4}{9} \,+\, \frac{8}{27} \,+\, \cdots
This series has common ratio 2/3. If we multiply through by this common ratio, then the initial 1 becomes a 2/3, the 2/3 becomes a 4/9, and so on:
\frac{2}{3}s \;=\; \frac{2}{3} \,+\, \frac{4}{9} \,+\, \frac{8}{27} \,+\, \frac{16}{81} \,+\, \cdots
This new series is the same as the original, except that the first term is missing. Subtracting the new series (2/3)s from the original series s cancels every term in the original but the first:
s \,-\, \frac{2}{3}s \;=\; 1,\;\;\;\mbox{so }s=3.
A similar technique can be used to evaluate any self-similar expression.


For r\neq 1, the sum of the first n terms of a geometric series is:
a + ar + a r^2 + a r^3 + \cdots + a r^{n-1} = \sum_{k=0}^{n-1} ar^k= a \, \frac{1-r^{n}}{1-r},
where a is the first term of the series, and r is the common ratio. We can derive this formula as follows:

&\text{Let }s = a + ar + ar^2 + ar^3 + \cdots + ar^{n-1}. \\[4pt]
&\text{Then }rs = ar + ar^2 + ar^3 + ar^4 + \cdots + ar^{n}  \\[4pt]
&\text{Then }s - rs = a-ar^{n}  \\[4pt]
&\text{Then }s(1-r) = a(1-r^{n}),\text{ so }s = a \frac{1-r^{n}}{1-r} \quad \text{(if } r \neq 1 \text{)}.
As n goes to infinity, the absolute value of r must be less than one for the series to converge. The sum then becomes
a+ar+ar^2+ar^3+ar^4+\cdots = \sum_{k=0}^\infty ar^k = \frac{a}{1-r} \Leftrightarrow |r|<1
When a = 1, this simplifies to:
1 \,+\, r \,+\, r^2 \,+\, r^3 \,+\, \cdots \;=\; \frac{1}{1-r},
the left-hand side being a geometric series with common ratio r. We can derive this formula:

&\text{Let }s = 1 + r + r^2 + r^3 + \cdots. \\[4pt]
&\text{Then }rs = r + r^2 + r^3 + \cdots. \\[4pt]
&\text{Then }s - rs = 1,\text{ so }s(1 - r) = 1,\text{ and thus }s = \frac{1}{1-r}.
The general formula follows if we multiply through by a.
This formula is only valid for convergent series (i.e., when the magnitude of r is less than one). For example, the sum is undefined when r = 10, even though the formula gives s = −1/9.
This reasoning is also valid, with the same restrictions, for the complex case.

Proof of convergence

We can prove that the geometric series converges using the sum formula for a geometric progression:
&1 \,+\, r \,+\, r^2 \,+\, r^3 \,+\, \cdots \\[3pt]
&=\; \lim_{n\rightarrow\infty} \left(1 \,+\, r \,+\, r^2 \,+\, \cdots \,+\, r^n\right) \\
&=\; \lim_{n\rightarrow\infty} \frac{1-r^{n+1}}{1-r}
Since (1 + r + r2 + ... + rn)(1−r) = 1−rn+1 and rn+1 → 0 for | r | < 1.
Convergence of geometric series can also be demonstrated by rewriting the series as an equivalent telescoping series. Consider the function g(K) = (r^(K+1))/(1-r). Note that: r = g(0) - g(1), r^2 = g(1) - g(2), r^3 = g(2) - g(3), . . . Thus: S = r + r^2 + r^3 + . . . = (g(0) - g(1)) + (g(1) - g(2)) + (g(2) - g(3)) + . . . If |r|<1, then g(K) -> 0 as K -> infinity, and so S converges to g(0) = r/(1-r).


Repeating decimals

A repeating decimal can be thought of as a geometric series whose common ratio is a power of 1/10. For example:
0.7777\ldots \;=\; \frac{7}{10} \,+\, \frac{7}{100} \,+\, \frac{7}{1000} \,+\, \frac{7}{10000} \,+\, \cdots.
The formula for the sum of a geometric series can be used to convert the decimal to a fraction:
0.7777\ldots \;=\; \frac{a}{1-r} \;=\; \frac{7/10}{1-1/10} \;=\; \frac{7}{9}.
The formula works not only for a single repeating figure, but also for a repeating group of figures. For example:
0.123412341234\ldots \;=\; \frac{a}{1-r} \;=\; \frac{1234/10000}{1-1/10000} \;=\; \frac{1234}{9999}.
Note that every series of repeating consecutive decimals can be conveniently simplified with the following:
0.09090909\ldots \;=\; \frac{09}{99} \;=\; \frac{1}{11}.
0.143814381438\ldots \;=\; \frac{1438}{9999}.
0.9999\ldots \;=\; \frac{9}{9} \;=\; 1.
That is, a repeating decimal with repeat length n is equal to the quotient of the repeating part (as an integer) and 10n - 1.

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