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 $0.5378378… = 0.5 + \frac{7}{{185}} = \frac{{199}}{{370}}$