Computing The Sum of a Geometric Series

# Computing The Sum of a Geometric Series

Recall from the Geometric Series page that a series in the form $\sum_{n=1}^{\infty} ar^{n-1}$ is a geometric series, and that such a series converges/diverges under the following conditions:

• $\sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{1-r}$ if $\mid r \mid < 1$.
• $\sum_{n=1}^{\infty} ar^{n-1}$ diverges if $\mid r \mid ≥ 1$.

Therefore if $\mid r \mid < 1$ then the series converges to a finite value and we can calculate that with the formula that the sum $S$ is $S = \frac{a}{1 - r}$ provided that our geometric series starts at the index $n = 1$.

## Example 1

Determine whether the series $\sum_{n=1}^{\infty} 3 \left ( \frac{1}{2} \right )^{n-1}$ converges or diverges. If this series converges, compute its sum.

In this example we note that $a = 3$ and the common ratio $r = \frac{1}{2}$. Since $\mid r \mid = \rvert \frac{1}{2} \rvert < 1$ we have that this series converges.

(1)
\begin{align} \sum_{n=1}^{\infty} 3 \left ( \frac{1}{2} \right)^{n-1} = \frac{a}{1 - r} = \frac{3}{1 - \frac{1}{2}} = 6 \end{align}

## Example 2

Determine whether the series $\sum_{n=1}^{\infty} \frac{1}{7} ( e )^{n-1}$ converges or diverges. If this series converges, compute its sum.

In this example we have that $a = \frac{1}{7}$ and the common ratio $r = e$. But we know that $e \approx 2.718...$ and so $\mid r \mid = \mid e \mid ≥ 1$. Therefore this series diverges.

## Example 3

Determine whether the series $\sum_{n=1}^{\infty} \frac{1}{3} \left ( \frac{1}{\sqrt{2}} \right )^{n-1}$ converges or diverges. If this series converges, compute its sum.

In this example we have that $a = \frac{1}{3}$ and that $r = \frac{1}{\sqrt{2}}$. We don't know the exactly value for $r$, but we know that since $\sqrt{2} > 1$, then $\frac{1}{\sqrt{2}} < 1$ and so we have that $\mid r \mid < 1$ and so this series is convergent.

(2)
\begin{align} \quad \sum_{n=1}^{\infty} \frac{1}{3} \left ( \frac{1}{\sqrt{2}} \right )^{n-1} = \frac{a}{1-r} = \frac{\frac{1}{3}}{1 - \frac{1}{\sqrt{2}}} = \frac{1}{3 - \frac{3}{\sqrt{2}}} \end{align}

## Example 4

Determine whether the series $\sum_{n=0}^{\infty} 2 \left ( -\frac{1}{4} \right )^{n-1}$ converges or diverges. If this series converges, compute its sum.

In this example we have that $a = 2$ and $r = -\frac{1}{4}$ and so $\mid r \mid < 1$. We must be careful though. Notice that in this example our index starts at $n = 0$ while the formula to compute the sum starts at $n = 1$. This isn't a big problem though since all we need to do is split up the sum as follows:

(3)
\begin{align} \quad \sum_{n=0}^{\infty} 2 \left ( -\frac{1}{4} \right )^{n-1} = \sum_{n=0}^{0} 2 \left ( -\frac{1}{4} \right )^{n-1} + \sum_{n=1}^{\infty} 2 \left ( -\frac{1}{4} \right )^{n-1} = -8 + \frac{2}{1 + \frac{1}{4}} = -8 + \frac{8}{5} = \frac{-32}{5} \end{align}

## Example 5

Determine whether the series $\sum_{n=1}^{\infty} 3 \left ( \frac{1}{5} \right)^{n+3}$ converges or diverges. If this series converges, compute its sum.

In this example we have that $a = 3$ and $r = \frac{1}{5}$ and so $\mid r \mid < 1$ so this series converges. Once again we must be careful with applying our formula for the sum of this series because our exponent on our common ratio is $n + 3$ and not $n - 1$ like the other examples. We will need to change the indices of this sum. Let $n + 3 = m - 1$. Then $m = n + 4 = 5$ and so:

(4)
\begin{align} \sum_{n=1}^{\infty} 3 \left ( \frac{1}{5} \right)^{n+3} = \sum_{m=5}^{\infty} 3 \left ( \frac{1}{5} \right)^{m-1} \end{align}

These series are the same (as you should verify). Now we note that to apply our series formula, we needed our index to start at $m = 1$ and so we'll have to subtract some terms off as follows:

(5)
\begin{align} \quad \quad \sum_{m=5}^{\infty} 3 \left ( \frac{1}{5} \right)^{m-1} = \sum_{m=1}^{\infty} 3 \left ( \frac{1}{5} \right)^{m-1} - \sum_{m=1}^{4} 3 \left ( \frac{1}{5} \right)^{m-1} = \left ( \frac{3}{1 - \frac{1}{5}} \right ) - \left (3 + \frac{3}{5} + \frac{3}{25} + \frac{3}{125} \right ) = 3.75 - 3.744 = 0.006 \end{align}