24.2: Infinite Geometric Series

In some cases, it makes sense to add not only finitely many terms of a geometric sequence, but all infinitely many terms of the sequence! An informal and very intuitive infinite geometric series is exhibited in the next example.

Example \(\PageIndex{1}\)

Consider the geometric sequence \[1, \dfrac 1 2, \dfrac 1 4, \dfrac 1 8, \dfrac 1 {16}, \dots \nonumber \]

Solution

Here, the common ratio is \(r=\dfrac 1 2\), and the first term is \(a_1=1\), so that the formula for \(a_n\) is \(a_n=\left(\dfrac 1 2\right)^{n-1}\). We are interested in summing all infinitely many terms of this sequence:

\[1+ \dfrac 1 2+ \dfrac 1 4+ \dfrac 1 8+\dfrac 1 {16}+ \dots \nonumber \]

We add these terms one by one, and picture these sums on the number line:

\[\begin{aligned} 1&= 1\\ 1+\dfrac 1 2&= 1.5\\ 1+\dfrac 1 2+\dfrac 1 4&= 1.75\\ 1+\dfrac 1 2+\dfrac 1 4+\dfrac 1 8 &= 1.875\\ 1+\dfrac 1 2+\dfrac 1 4+\dfrac 1 8+\dfrac 1 {16} &= 1.9375\end{aligned} \nonumber \]

We see that adding each term takes the sum closer and closer to the number \(2\). More precisely, adding a term \(a_n\) to the partial sum \(a_1+\dots+a_{n-1}\) cuts the distance between \(2\) and \(a_1+\dots+a_{n-1}\) in half. For this reason we can, in fact, get arbitrarily close to \(2\), so that it is reasonable to expect that

\[1+ \dfrac 1 2+ \dfrac 1 4+ \dfrac 1 8+\dfrac 1 {16}+ \dots =2 \nonumber \]

In the next definition and observation, this equation will be justified and made more precise. We start by providing the definition of an infinite series.

Definition: Infinite Series

An infinite series is given by the

\[\sum_{i=1}^\infty a_i = a_1+a_2+a_3+\dots\]

To be more precise, the infinite sum is defined as the limit \(\sum\limits_{i=1}^\infty a_i := \lim\limits_{k\to \infty} \bigg(\sum\limits_{i=1}^k a_i\bigg)\). Therefore, an infinite sum is defined, precisely when this limit exists.

Observation: Infinite Geometric Series

Let \(\{a_n\}\) be a geometric sequence with \(a_n=a_1\cdot r^{n-1}\). Then the infinite geometric series is defined whenever \(-1<r<1\). In this case, we have:

\[\label{EQU:inf-geo-series} \boxed{ \sum_{i=1}^\infty a_i = a_1\cdot \dfrac{1}{1-r} }\]

Proof

Informally, this follows from the formula \(\sum_{i=1}^k a_i = a_1\cdot \dfrac{1-r^{k}}{1-r}\) and the fact that \(r^k\) approaches zero when \(k\) increases without bound.

More formally, the proof uses the notion of limits, and goes as follows:

\[\sum_{i=1}^\infty a_i = \lim_{k\to \infty}\bigg( \sum_{i=1}^k a_i \bigg) = \lim_{k\to \infty}\bigg( a_1\cdot \frac{1-r^{k}}{1-r}\bigg) = a_1\cdot \frac{1-\lim\limits_{k\to\infty } (r^{k})}{1-r} = a_1\cdot \dfrac{1}{1-r} \nonumber \]

Example \(\PageIndex{2}\)

Find the value of the infinite geometric series.

1. \(\sum_{j=1}^\infty a_j\), for \(a_j=5\cdot \left(\dfrac 1 3\right)^{j-1}\)
2. \(\sum_{n=1}^\infty 3\cdot \left(0.71\right)^n\)
3. \(500-100+20-4+\dots\)
4. \(3+6+12+24+48+\dots\)

Solution

1. We use formula \(\ref{EQU:inf-geo-series}\) for the geometric series \(a_n=5\cdot \left(\dfrac 1 3\right)^{n-1}\), that is \(a_1=5\cdot \left(\dfrac 1 3\right)^{1-1}=5\cdot \left(\dfrac 1 3\right)^{0}=5\cdot 1 =5\) and \(r=\dfrac 1 3\). Therefore,

\[\sum_{j=1}^\infty a_j=a_1\cdot \dfrac {1}{1-r}=5\cdot \dfrac {1}{1-\frac 1 3}=5\cdot \dfrac 1 {\frac{3-1}{3}}=5\cdot \dfrac 1 {\frac 2 3}=5\cdot \dfrac 3 2 =\dfrac {15} 2 \nonumber \]

2. In this case, \(a_n=3\cdot (0.71)^n\), so that \(a_1=3\cdot 0.71^1=3\cdot 0.71=2.13\) and \(r=0.71\). Using again formula \(\ref{EQU:inf-geo-series}\), we can find the infinite geometric series as

\[\sum_{n=1}^\infty 3\cdot \left(0.71\right)^n=a_1\cdot \dfrac{1}{1-r}=2.13\cdot \dfrac 1 {1-0.71}=2.13\cdot \dfrac 1 {0.29}=\dfrac {2.13}{0.29}=\dfrac{213}{29} \nonumber \]

In the last step we simplified the fraction by multiplying both numerator and denominator by \(100\), which had the effect of eliminating the decimals.

3. Our first task is to identify the given sequence as an infinite geometric sequence:

\[\{a_n\} \text{ is given by } 500, -100, 20, -4, \dots \nonumber \]

Notice that the first term is \(500\), and each consecutive term is given by dividing by \(-5\), or in other words, by multiplying by the common ratio \(r=-\dfrac 1 5\). Therefore, this is an infinite geometric series, which can be evaluated as

\[\begin{aligned} 500-100+20-4+\dots &= \sum_{n=1}^\infty a_n\\&=a_1\cdot \dfrac 1 {1-r}\\&=500\cdot \dfrac 1 {1-\left(-\frac 1 5\right)} \\ &= 500 \cdot \dfrac{1}{1+\frac 1 5} \\&=\dfrac {500}{\frac {1+5}{5}}\\&=\dfrac{500}{\frac 6 5}\\&=500\cdot \dfrac 5 6 \\ &= \dfrac {2500}{6} \\&= \dfrac{1250}{3}\end{aligned} \nonumber \]

4. We want to evaluate the infinite series \(3+6+12+24+48+\dots\). The sequence \(3, 6, 12, 24, 48, \dots\) is a geometric sequence, with \(a_1=3\) and common ratio \(r=2\). Since \(r\geq 1\), we see that formula \(\ref{EQU:inf-geo-series}\) cannot be applied, as \(\ref{EQU:inf-geo-series}\) only applies to \(-1<r<1\). However, since we add larger and larger terms, the series gets larger than any possible bound, so that the whole sum becomes infinite.

\[3+6+12+24+48+\dots=\infty \nonumber \]

Example \(\PageIndex{3}\)

The fraction \(0.55555\dots\) may be written as:

\[0.55555\dots = 0.5+0.05+0.005+0.0005+0.00005+\dots \nonumber \]

Noting that the sequence

\[0.5,\underset{\times 0.1}{\hookrightarrow } 0.05, \underset{\times 0.1}{\hookrightarrow } 0.005, \underset{\times 0.1}{\hookrightarrow } 0.0005, \underset{\times 0.1}{\hookrightarrow } 0.00005, \dots \nonumber \]

is a geometric sequence with \(a_1=0.5\) and \(r=0.1\), we can calculate the infinite sum as:

\[0.55555\dots = \sum_{i=1}^\infty 0.5 \cdot \left(0.1\right)^{i-1}= 0.5\cdot \dfrac{1}{1-0.1} = 0.5 \cdot\dfrac{1}{0.9}=\dfrac{0.5}{0.9}=\dfrac 5 9 \nonumber \]

Here we multiplied numerator and denominator by \(10\) in the last step in order to eliminate the decimals.