In this case, the starting value is $a=3$ and the common ratio is $r=2$. Notice how each successive term in the series is multiplied by the common ratio.

An example of an infinite geometric series with the same starting value and common ratio is:

$$S=\sum^{\infty}_{i=1}3(2)^{i-1}= 3(2)^{0}+3(2)^{1}+3(2)^{2}+3(2)^{3}+\cdots$$

Theorem.

Formula for the sum of finite geometric series

The sum of a finite geometric series is given by:

$$S_n=\frac{a(1-r^{n})}{1-r}$$

Where:

$a$ is the starting value of the geometric series.
$r$ is the common ratio ($r\ne1$).
$n$ is the number of terms in the series.

Proof. Recall that the finite geometric series is defined as:

$$\begin{equation}\label{eq:LRo0GiI7V8p2kXoLDtY} S_n=\sum^{n}_{i=1}a(r)^{i-1} =a+ar+ar^2+ar^3+\cdots+ar^{n-2}+ar^{n-1} \end{equation}$$

Now, we multiply both sides by the common ratio $r$ to get:

$$\begin{equation}\label{eq:Yk1VMiOngd0BFjlTGG1} rS_n =ar+ar^2+ar^3+ar^4+\cdots+ar^{n-1}+ar^{n} \end{equation}$$

Let's write $S_n$ and $rS_n$ on top of each other:

$$\begin{align*} S_n&=a+ar+ar^2+ar^3+\cdots+ar^{n-1}\\ rS_n&=0+ar+ar^2+ar^3+\cdots+ar^{n-1}+ar^{n} \end{align*}$$

We've added a $0+$ term in the beginning of $rS_n$ to align $S_n$ and $rS_n$. Now, we subtract $rS_n$ from $S_n$ to get:

$$\begin{equation}\label{eq:ByzMoH9Wot7R2cVsndo} \begin{aligned}[b] S_n-rS_n&=a-ar^{n}\\ S_n(1-r)&=a(1-r^{n})\\ S_n&=\frac{a(1-r^{n})}{1-r} \end{aligned} \end{equation}$$

Notice how $S_n$ is undefined when $r=1$. Let's think about the case when $r=1$. When the common ratio is equal to one, the geometric series is:

$$S_n=\sum^{n}_{i=1}a(1)^{i-1} =\sum^{n}_{i=1}a =na $$

Even though our formula \eqref{eq:ByzMoH9Wot7R2cVsndo} will not allow us to compute the sum of the series when the common ratio is equal to one, the sum is still defined and is computed by $na$.

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Example.

Computing sum of finite geometric series by formula

Consider the following finite geometric series:

$$S_5=\sum^{5}_{i=1}3(2)^{i-1}= 3+3(2)^{1}+3(2)^{2}+3(2)^{3}+3(2)^{4}$$

Compute the sum of the series.

Solution. The starting value is $a=3$ and the common ratio is $r=2$. Let's use formula for the sum of finite geometric series:

$$\begin{align*} S_5&=\frac{3(1-2^{5})}{1-2}\\ &=\frac{3(-31)}{-1}\\ &=93\\ \end{align*}$$

Notice how the formula extremely handy when computing a series with a large $n$.

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Theorem.

Formula for the sum of infinite geometric series

If the common ratio $r$ is between $-1$ and $1$, that is $\vert{r}\vert\lt1$, then the sum of infinite geometric series is given by:

$$S=\frac{a}{1-r}$$

Where:

$a$ is the starting value of the geometric series.
$r$ is the common ratio between $-1$ and $1$.

Note that if $\vert{r}\vert\gt1$, then the infinite geometric series diverges to infinity.

Proof. Given $\vert{r}\vert\lt1$, we know that:

$$\begin{equation}\label{eq:mtnJgF9z0h12ZpuNlym} \lim_{n\to\infty}r^n=0 \end{equation}$$

This means that as $n$ becomes larger and larger, $r^n$ approaches zero. For instance:

$$\begin{align*} (0.5)^{1}&=0.5\\ (0.5)^{10}&\approx9.77\times10^{-4}\\ (0.5)^{100}&\approx7.89\times10^{-31}\\ \end{align*}$$

Again, this is only true when $\vert{r}\vert\lt1$.

Now, recall that the sum of a finite geometric serieslink is given by:

$$S_n=\frac{a(1-r^n)}{1-r}$$

To obtain an infinite geometric series, we let $n$ tend to infinity and use the properties of limits to simplify:

$$\begin{equation}\label{eq:otdG6obG3LCR5KjstgD} \begin{aligned}[b] \lim_{n\to\infty}S_n&= \lim_{n\to\infty}\frac{a(1-r^n)}{1-r}\\ &=\lim_{n\to\infty}\Big(\frac{a}{1-r}-\frac{ar^n}{1-r}\Big)\\ &=\lim_{n\to\infty}\Big(\frac{a}{1-r}\Big)- \lim_{n\to\infty}\Big(\frac{ar^n}{1-r}\Big)\\ &=\frac{a}{1-r}- \Big(\frac{a\cdot\lim_{n\to\infty}r^n}{1-r}\Big)\\ &=\frac{a}{1-r}- \Big(\frac{a\cdot0}{1-r}\Big)\\ &=\frac{a}{1-r} \end{aligned} \end{equation}$$

This completes the proof.

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Example.

Computing the sum of infinite geometric series by formula

Consider the following infinite geometric series:

$$S= \sum^{\infty}_{i=1}3\Big(\frac{1}{2}\Big)^{i-1}= 3+ 3\Big(\frac{1}{2}\Big)^{1}+ 3\Big(\frac{1}{2}\Big)^{2}+ 3\Big(\frac{1}{2}\Big)^{3}+ \cdots$$

Compute the sum of the series.

Solution. The starting value is $a=3$ and the common ratio is $r=1/2$, which is between $-1$ and $1$. Therefore, we can use the formula for the sum of infinite geometric series:

$$\begin{align*} S&=\frac{a}{1-r}\\ &=\frac{3}{1-(1/2)}\\ &=\frac{3}{1/2}\\ &=6 \end{align*}$$

This means that as we consider more and more terms in the series, the sum converges to $6$.

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Published by Isshin Inada

Edited by 0 others

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