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.



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

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.



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


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