Computing The Sum of a Geometric Series Examples 1

Recall from the Computing The Sum of a Geometric Series page that a series in the form $\sum_{n=1}^{\infty} ar^{n-1}$ is a geometric series, and $\sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{1 - r}$ if $\mid r \mid < 1$ and diverges if $\mid r \mid ≥ 1$.

We will now look at some more examples involving geometric series.

Example 1

Find the sum of the series $\sum_{n=1}^{\infty} 3 \left ( \frac{1}{6} \right )^{n-1}$.

We note that $r = \frac{1}{6}$ and $\biggr \rvert \frac{1}{6} \biggr \rvert < 1$, so this series converges and:

(1)

\begin{align} \quad \sum_{n=1}^{\infty} 3 \left ( \frac{1}{6} \right )^{n-1} = \frac{3}{1 - \frac{1}{6}} = \frac{18}{5} \end{align}

Example 2

Find the sum of the series $\sum_{n=0}^{\infty} -2 \left ( \frac{1}{3} \right)^n$.

Notice that this series can be rewritten as $\sum_{n=1}^{\infty} -2 \left ( \frac{1}{3} \right)^{n-1}$, and so $r = \frac{1}{3}$ and $\biggr \rvert \frac{1}{3} \biggr \rvert < 1$ so this series converges and:

(2)

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

Example 3

Find the sum of the series $\sum_{n=-4}^{\infty} \left ( \frac{2}{3} \right )^{2n}$.

We will first rewrite this series by removing the squared power, and so:

(3)

\begin{align} \quad \sum_{n=-4}^{\infty} \left ( \frac{2}{3} \right )^{2n} = \sum_{n=-4}^{\infty} \left ( \left ( \frac{2}{3} \right )^2 \right )^{n} = \sum_{n=-4}^{\infty} \left ( \frac{4}{9} \right )^{n} \end{align}

We now want the power $n - 1$ attached to the fraction in the summation. We can achieve this by changing the start index of the sum:

(4)

\begin{align} \sum_{n=-4}^{\infty} \left ( \frac{4}{9} \right )^{n} = \sum_{n=-3}^{\infty} \left ( \frac{4}{9} \right )^{n-1} \end{align}

Now we need the start index to be at $1$, and so we will compute the first few terms of the series, that is:

(5)

\begin{align} \quad \sum_{n=-3}^{\infty} \left ( \frac{4}{9} \right )^{n-1} = \left ( \frac{4}{9} \right)^{-4} + \left ( \frac{4}{9} \right)^{-3} + \left ( \frac{4}{9} \right)^{-2} + \left ( \frac{4}{9} \right)^{-1} + \sum_{n=1}^{\infty} \left ( \frac{4}{9} \right )^{n-1} \end{align}

Now the remaining summation is a geometric series and $\sum_{n=1}^{\infty} \left ( \frac{4}{9} \right )^{n-1} = \frac{1}{1 - \frac{4}{9}} = \frac{9}{5}$, and so:

(6)

\begin{align} \quad \sum_{n=-3}^{\infty} \left ( \frac{4}{9} \right )^{n-1} = \left ( \frac{4}{9} \right)^{-4} + \left ( \frac{4}{9} \right)^{-3} + \left ( \frac{4}{9} \right)^{-2} + \left ( \frac{4}{9} \right)^{-1} + \frac{9}{5} \\ \quad \sum_{n=-3}^{\infty} \left ( \frac{4}{9} \right )^{n-1} = \frac{6561}{256} + \frac{729}{64} + \frac{81}{16} + \frac{9}{4} + \frac{9}{5} = 46 + \frac{169}{1280} \end{align}

Example 4

Find the sum of the series $\sum_{n=-2}^{\infty} \left [ \left ( \frac{1}{2} \right )^n - \left ( \frac{1}{3} \right) ^{2n} \right ]$.

Let's first split this series into two simpler series:

(7)

\begin{align} \quad \sum_{n=-2}^{\infty} \left [ \left ( \frac{1}{2} \right )^n - \left ( \frac{1}{3} \right) ^{2n} \right ] = \sum_{n=-2}^{\infty} \left ( \frac{1}{2} \right )^n - \sum_{n=-2}^{\infty} \left ( \frac{1}{3} \right )^{2n} = \sum_{n=-2}^{\infty} \left ( \frac{1}{2} \right )^n - \sum_{n=-2}^{\infty} \left ( \frac{1}{9} \right )^{n} \end{align}

We want the exponents in both of these series to be $n - 1$ so we will change the start indices:

(8)

\begin{align} \quad \sum_{n=-2}^{\infty} \left ( \frac{1}{2} \right )^n - \sum_{n=-2}^{\infty} \left ( \frac{1}{9} \right )^{n} = \sum_{n=-1}^{\infty} \left ( \frac{1}{2} \right )^{n-1} - \sum_{n=-1}^{\infty} \left ( \frac{1}{9} \right )^{n-1} \end{align}

Now we want the start indices to be $1$, so we will compute the first few terms of each series:

(9)

\begin{align} \quad \sum_{n=-1}^{\infty} \left ( \frac{1}{2} \right )^{n-1} - \sum_{n=-1}^{\infty} \left ( \frac{1}{9} \right )^{n-1} = \left [ \left ( \frac{1}{2} \right )^{-2} + \left ( \frac{1}{2} \right )^{-1} + \sum_{n=1}^{\infty} \left ( \frac{1}{2} \right )^{n-1} \right ] - \left [ \left ( \frac{1}{9} \right )^{-2} + \left ( \frac{1}{9} \right )^{-1} + \sum_{n=1}^{\infty} \left ( \frac{1}{9} \right )^{n-1} \right ] \\ \quad \sum_{n=-1}^{\infty} \left ( \frac{1}{2} \right )^{n-1} - \sum_{n=-1}^{\infty} \left ( \frac{1}{9} \right )^{n-1} = 4 +2 + \frac{1}{1 - \frac{1}{2}} - 81 - 9 - \frac{1}{1 - \frac{1}{9}} \\ \quad \sum_{n=-1}^{\infty} \left ( \frac{1}{2} \right )^{n-1} - \sum_{n=-1}^{\infty} \left ( \frac{1}{9} \right )^{n-1} = 4 +2 + 2 - 81 - 9 - \frac{9}{8} = -82 - \frac{9}{8} = -\frac{665}{8} \\ \end{align}