Infinite Geometric Series

A geometric sequence in which the number of terms increases without bounds is called an infinite geometric series.

If the absolute value of the common ratio r is less than 1, {S_n}, the sum of n terms always approaches a definite limit as n increases without bounds. As we have proved, the sum of a finite geometric series is

{S_n} = \frac{{{a_1} - {a_1}{r^n}}}{{1 - r}},{\text{ }}r \ne 1

We rewrite it as

{S_n} = \frac{{{a_1}}}{{1 - r}} - \frac{{{a_1}{r^n}}}{{1 - r}}

If now r is numerically less than 1, i.e., \left| r \right| < 1, the numerical value of {r^n} decreases as n increases. By taking n which is sufficiently large, we can make {r^n} as small as we want. Hence, by taking n large enough, we can make {S_n} differ from \frac{{{a_1}}}{{1 - r}} by as little as we want, i.e., we can make {S_n} approach \frac{{{a_1}}}{{1 - r}} as a limit. Symbolically

{S_\infty } = \mathop {\lim }\limits_{n \to \infty } {S_n} = \frac{{{a_1}}}{{1 - r}}

Here {S_\infty } is the sum of an infinite geometric progression with the first term as {a_1} and common ratio r. We can also call it an infinite series. Accordingly, the expression {a_1} + {a_1}r + {a_1}{r^2} + \cdots is called an infinite geometric series. If the terms continuously decrease as {S_\infty } approaches a limiting value as n becomes infinitely large, it is said to be a convergent infinite series.

 

Example:

Find the sum of the infinite geometric sequence
2,{\text{ }}\frac{4}{3},{\text{ }}\frac{8}{9},{\text{ }}\frac{{16}}{{27}}, \cdots
Solution:

We have
{a_1} = 2, r = \frac{2}{3} < 1, then
{S_\infty } = \frac{{{a_1}}}{{1 - r}} = \frac{2}{{1 - \frac{2}{3}}} = 6

 

Recurring or Periodic Decimals

An interesting application of a geometric progression with infinitely many terms is the evaluation of recurring or periodic decimals.

When we attempt to express a common fraction such as \frac{3}{8} or \frac{4}{{11}} as a decimal fraction, the decimal always either terminates or ultimately repeats in blocks. Thus
\frac{3}{8} = 0.375 (decimal terminates)
\frac{4}{{11}} = 0.363636... (decimal repeats)

In the division process by which we express the fraction \frac{p}{q} as a decimal fraction the remainders can only be the numbers 0,1,2,3,4, \ldots ,q - 1. If at any stage in the division we obtain a remainder of 0, the process terminates. Otherwise, after not more than q divisors, one of the remainders 0,1,2,3,4, \ldots ,q - 1 must recur and the decimal begins to repeat.

 

Example:

Express the recurring decimal fraction 0.5378378378... as a common fraction.

Solution:

The given decimal fraction can be written in the form
0.5378378... = 0.5 + 0.0378 + 0.0000378 + \cdots

 

Hence our number consists of the decimal 0.5 plus the sum of an infinite geometric progression with the first term {a_1} = 0.0378 and common ratio r = 0.001. The sum of the infinite progression is expressible as the fraction
{S_\infty } = \frac{{0.0378}}{{1 - 0.001}} = \frac{{0.0378}}{{0.999}} = \frac{{378}}{{9990}} = \frac{7}{{185}}

Hence 0.5378378... = 0.5 + \frac{7}{{185}} = \frac{{199}}{{370}}