A geometric series is an infinite series whose terms are in a geometric progression, or whose successive terms have a common ratio. If the terms of a geometric series approach zero, the sum of its terms will be finite. As the numbers near zero, they become insignificantly small, allowing a sum to be calculated despite the series being infinite. 

A geometric series with a finite sum is said to converge. A series converges if and only if the absolute value of the common ratio is less than one:

$\left | r \right | < 1$

What follows in an example of an infinite series with a finite sum. We will calculate the sum $s$ of the following series:

$\displaystyle{s = 1+\frac { 2 }{ 3 } +\frac { 4 }{ 9 } +\frac { 8 }{ 27 } + \cdots}$

This series has common ratio $\displaystyle{\frac{2}{3}}$. If we multiply through by this common ratio, then the initial term $1$ becomes $\displaystyle{\frac{2}{3}}$, the $\displaystyle{\frac{2}{3}}$ becomes $\displaystyle{\frac{4}{9}}$, and so on:

$\displaystyle{\frac{2}{3} s = \frac { 2 }{ 3 } +\frac { 4 }{ 9 } +\frac { 8 }{ 27 } +\frac { 16 }{ 81 } + \cdots}$

This new series is the same as the original, except that the first term is missing. Subtracting the new series $\displaystyle{\frac{2}{3}s}$ from the original series, $s$ cancels every term in the original but the first:

$\displaystyle{ \begin{aligned} s-\frac{2}{3}s &=1 \\ \therefore s &= 3 \end{aligned} }$

A similar technique can be used to evaluate any self-similar expression.

A formula can be derived to calculate the sum of the terms of a convergent series. The formula used to sum the first $n$ terms of any geometric series where $r\neq 1$ is:

 $\displaystyle s= a\frac { 1-{ r }^{ n } }{ 1-r } $

If a series converges, we want to find the sum of not only a finite number of terms, but all of them. If $\left | r \right | <1$, then we see that as $n$ becomes very large, $r^n$ becomes very small. We express this by writing that as $n\rightarrow \infty$ (as $n$ approaches infinity), $r^n\rightarrow 0$. 

Applying $r^n\rightarrow 0$, we can find a new formula for the sum of an infinitely long geometric series: 

$\displaystyle{ \begin{aligned} s &= a\frac{1-r^n}{1-r} \\ &\rightarrow a\frac{1}{1-r} \quad \text{as } r^n\rightarrow 0\\ &= \frac{a}{1-r} \end{aligned} }$

Therefore, for $|r|<1$, we can write the infinite sum as: 

$\displaystyle{s = \frac{a}{1-r}}$

Example

Find the sum of the infinite geometric series $64+ 32 + 16 + 8 + \cdots$

First, find $r$, or the constant ratio between each term and the one that precedes it:

$\displaystyle{ \begin{aligned} r &= \frac{32}{64} \\ &= \frac{1}{2} \end{aligned} }$

Substitute $a=64$ and $\displaystyle r= \frac{1}{2}$ into the formula for the sum of an infinite geometric series:

$\displaystyle{ \begin{aligned} s &= \frac{64}{1-\frac{1}{2}} \\ &= \frac{64}{\frac{1}{2}} \\ &= 128 \end{aligned} }$