(S4) Geometric Series#

In this lesson we are going to see:

  • the definition of a geometric series.

  • how to determine whether such a series is convergent or divergent, and how to find its sum (if convergent).

Review Videos#

Geometric Series#

When we first introduced the idea of sequences and limits of sequences, we spent some time investigating a particular type of sequence which we called a geometric sequence. A geometric series is simply what we get when we take a geometric sequence and try to add up all the terms.

Geometric Series

A geometric series is a series of the form:

\[ \sum_{n=0}^{\infty} a r^n = a+ ar+ar^2+ar^3+ \cdots + ar^{n}+ \cdots \]

Sometimes it is helpful to shift the index values so that we start with \(n=1\). In which case our series would be written as:

\[ \sum_{n=1}^{\infty} a r^{n-1} \]

Partial Sums#

Geometric series are one of the very few series where we can actually find a nice, simple formula for the general term \(s_n\) in the sequence of partial sums. (Usually this is very difficult to do.)

Partial Sum for \(r\neq 1\)
\[ s_n = \dfrac{a(1-r^n)}{1-r} \]
Partial Sum for \(r = 1\)
\[ s_n = n\cdot a \]

Convergence#

The convergence of a geometric series depends solely on the value of \(r\) (although the value of \(a\) does influence the sum).

Convegence and Sum

The series converges provided \(|r|<1\) and diverges if \(|r|\geq 1\). Furthermore the sum is given by:

\[ \sum_{n=0}^{\infty} a r^{n} = \dfrac{a}{1-r} \qquad \text{if}\quad |r|<1 \]
Case \(r\leq -1\)

In this case we recall our previous result for geometric sequences, when \(r\leq -1\) we have:

\[ \lim_{n\to\infty} r^n = \text{DNE} \]

This means:

\[ \sum_{n=0}^{\infty} a r^{n} = \lim_{n\to\infty} s_n = \lim_{n\to\infty} \dfrac{a(1-r^n)}{1-r} = \text{DNE} \]

Because this limit does not exist, the series is divergent.

Case \(-1< r< 1\)

Here we recall again a previous limit result, when \(-1< r< 1\) we have:

\[ \lim_{n\to\infty} r^n = 0 \]

This means:

\[ \sum_{n=0}^{\infty} a r^{n} = \lim_{n\to\infty} s_n = \lim_{n\to\infty} \dfrac{a(1-r^n)}{1-r} = \dfrac{a(1-0)}{1-r}\]

Because this limit is a finite number, we say the series is convergent and its sum is equal to \(\tfrac{a}{1-r}\).

Case \(r= 1\)

Here the formula for the partial sums is slightly different:

\[ s_n=a+a+a+\cdots + a = n\cdot a \]

This means:

\[ \sum_{n=0}^{\infty} a r^{n} = \lim_{n\to\infty} s_n = \lim_{n\to\infty} n\cdot a = \pm \infty \]

(Where the sign depends on the value of \(a\).) Because this limit is infinite, we say the series is divergent.

Case \(1<r\)

One last limit result to recall:

\[ \lim_{n\to\infty} r^n = \infty \]

This means:

\[ \sum_{n=0}^{\infty} a r^{n} = \lim_{n\to\infty} s_n = \lim_{n\to\infty} \dfrac{a(1-r^n)}{1-r} = \pm \infty \]

(Where the sign depends on the value of \(a\).) Because this limit is infinite, we say the series is divergent.

Example 1#

Determine whether the following geometric series are convergent or divergent. If convergent, find the sum.

(a) \(\quad \displaystyle \sum_{n=0}^{\infty} 5\left(\dfrac{1}{2}\right)^{n}\)

If we compare the series with the standard form:

\[ \sum_{n=0}^{\infty} a r^{n} \]

We see that \(a=5\) and \(r=\tfrac{1}{2}\). Since \(|r|<1\), this means the series is convergent with sum given by:

\[ \sum_{n=0}^{\infty} 5\left(\dfrac{1}{2}\right)^{n}=\dfrac{a}{1-r}=\dfrac{5}{1-\tfrac{1}{2}}=10 \]
(b) \(\quad \displaystyle \sum_{n=0}^{\infty} \left(\dfrac{1}{2}\right)^{n+1}\)

Before we decide on the convergence and sum, we need to write this in the standard geometric form, where the exponent is just an \(n\). To do this, we use an exponent law that allows us to break up the + sign in the exponent:

\[ \sum_{n=0}^{\infty} \left(\dfrac{1}{2}\right)^{n+1}= \sum_{n=0}^{\infty} \dfrac{1}{2}\cdot \left(\dfrac{1}{2}\right)^{n} \]

Comparing this to the standard geometric form:

\[ \sum_{n=0}^{\infty} a r^{n} \]

We see that \(a=\tfrac{1}{2}\) and \(r=\tfrac{1}{2}\). Since \(|r|<1\), this means the series is convergent with sum given by:

\[ \sum_{n=0}^{\infty} \left(\dfrac{1}{2}\right)^{n+1}= \dfrac{a}{1-r}=\dfrac{\tfrac{1}{2}}{1-\tfrac{1}{2}}=1 \]
(c) \(\quad \displaystyle \sum_{n=0}^{\infty} 2^{-n}\)

Before we decide on the convergence and sum, we need to write this in the standard geometric form, where the exponent is just an \(n\). To do this, we rewrite the negative exponent as a fraction:

\[ \sum_{n=0}^{\infty} 2^{-n} = \sum_{n=0}^{\infty} \dfrac{1}{2^{n}} = \sum_{n=0}^{\infty} \left(\dfrac{1}{2}\right)^{n} \]

Comparing this to the standard geometric form:

\[ \sum_{n=0}^{\infty} a r^{n} \]

We see that \(a=1\) (there’s always a hidden 1) and \(r=\tfrac{1}{2}\). Since \(|r|<1\), this means the series is convergent with sum given by:

\[ \sum_{n=0}^{\infty} 2^{-n} = \dfrac{a}{1-r}=\dfrac{1}{1-\tfrac{1}{2}}=2 \]

Example 2#

Determine whether the geometric series is convergent or divergent. If it converges, find the sum.

\[ \sum_{n=0}^{\infty} 3^{2n+1}6^{2-n} \]
Solution (Click to see the steps.)

Before we decide on the convergence and sum, we need to write this in the standard geometric form, where we have just a single number raised to just the single power \(n\). To do this, we use an exponent law that allows us to break up the + and - signs in the exponent:

\[ \sum_{n=0}^{\infty} 3^{2n+1}6^{2-n} = \sum_{n=0}^{\infty} \left(3\cdot 3^{2n}\right) \cdot \left(6^2\cdot 6^{-n}\right) \]

That helps, but we still need to do something with that \(2n\) in the exponent of the numerator. Another exponent law to the rescue! Note that

\[ 3^{2n}=\left(3^2\right)^n= 9^n \]

Putting this into our series and grouping all terms without \(n\) we get:

\[ \sum_{n=0}^{\infty} \left(3\cdot 6^2\right)\cdot {9^n}{6^{-n}} \]

We can rewrite that negative exponent as a fraction (just like we normally do):

\[ \sum_{n=0}^{\infty} \left(3\cdot 6^2\right)\cdot \dfrac{9^n}{6^{n}} \]

And then one final exponent law, essentially allow us to factor off that common exponent (which only works now, since the exponents are the same).

\[ \sum_{n=0}^{\infty} \left(3\cdot 6^2\right)\cdot \left(\dfrac{9}{6}\right)^n \]

At this point, we can see that \(r=\tfrac{9}{6}\) which is greater than 1, and so this series diverges.

Example 3#

Determine whether the series is convergent or divergent. If it converges, find the sum.

\[ 2-\dfrac{4}{3}+\dfrac{8}{9}-\dfrac{16}{27}+\cdots \]
Solution (Click to see the steps.)

We see that each term in this series increases by a factor of 2 in the numerator and 3 in the denominator. This leads us to believe that this is a geometric series. So we start by writing out the first few terms of a general geometric series:

\[ a+ ar+ar^2+ar^3+\cdots \]

From this, we notice:

\[\begin{split} a & = \text{1st Term}\\[10pt] r & = \dfrac{ar}{r} = \dfrac{\text{2nd Term}}{\text{1st Term}} \end{split}\]

By comparing terms we calculate that \(a=2\) and \(r = \dfrac{-4/3}{2}= -\dfrac{2}{3}\). Since \(|r|<1\), this means the series is convergent with sum given by:

\[ \sum_{n=0}^{\infty} 2\left(-\dfrac{2}{3}\right)^n = \dfrac{a}{1-r}=\dfrac{2}{1-\tfrac{-2}{3}}=\dfrac{6}{5} \]

It’s also a good idea to confirm that the \(a\) and \(r\) values we found also agre with the other terms that are listed in the original representation:

  • 3rd Term: \(ar^2= 2\left(\tfrac{-2}{3}\right)^2 = \tfrac{8}{9}\) which does match up correctly!

  • 4th Term: \(ar^3= 2\left(\tfrac{-2}{3}\right)^3 = -\tfrac{16}{27}\) which also matches up correctly!

Example 4#

Express the following decimal as a rational number.

\[ 3.\overline{9}=3.99999999\dots \]
Solution (Click to see the steps.)

We write this repeated decimal as a geometric series:

\[\begin{split} 3.99999\dots & = 3+0.9+0.09+0.009+0.0009+\cdots \\[10pt] & = 3+\dfrac{9}{10}+\dfrac{9}{100}+\dfrac{9}{1000}+\dfrac{9}{10000}+\cdots \\[10pt] & = 3+\dfrac{9}{10}+\dfrac{9}{10^2}+\dfrac{9}{10^3}+\dfrac{9}{10^4}+\cdots \\[10pt] \end{split}\]

From here we see that the 3 does not fit the geometric pattern so we keep this separate and instead view the fractions as the geometric series (and we see that the denominators are increasing by a factor of 10 confirming our suspicion). We calculate:

\[\begin{split} a & = \text{1st Term} = \dfrac{9}{10}\\[10pt] r & = \dfrac{\text{2nd Term}}{\text{1st Term}} = \dfrac{9/10^2}{9/10} = \dfrac{1}{10} \end{split}\]

Based on this \(r\) value, we see that the geometric series is convergent because \(|r|=\left|\tfrac{1}{10}\right|<1\) and then calculating the sum:

\[ 0.9+0.09+0.009+0.0009+\cdots = \dfrac{a}{1-r} = \dfrac{\tfrac{9}{10}}{1-\tfrac{1}{10}} = 1 \]

Therefore we combine this with the 3 that did not originally fit the geometric pattern to get:

\[ 3.99999\dots = 3+0.9+0.09+0.009+0.0009+\cdots = 3+1 =4 \]

Therefore this repeated decimal can also be written as rational number 4, or more explicitly as \(4/1\).