8.6: Geometric Series
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Infinite series occur often in mathematics and physics. Two series Which occur often are the geometric series and the binomial series. We will discuss these next.
A geometric series is of the form
\[\sum_{n=0}^{\infty} a r^{n}=a+a r+a r^{2}+\ldots+a r^{n}+\ldots \nonumber \]
Here \(a\) is the first term and \(r\) is called the ratio. It is called the ratio because the ratio of two consecutive terms in the sum is \(r\).
\[1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\ldots \nonumber \]
is an example of a geometric series. We can write this using summation notation,
\[1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\ldots=\sum_{n=0}^{\infty} 1\left(\dfrac{1}{2}\right)^{n} \nonumber \]
Thus, \(a=1\) is the first term and \(r=\dfrac{1}{2}\) is the common ratio of successive terms.
Next, we seek the sum of this infinite series, if it exists.
The sum of a geometric series, when it exists, can easily be determined. We consider the \(n\)th partial sum:
\[s_{n}=a+a r+\ldots+a r^{n-2}+a r^{n-1} \nonumber \]
Now, multiply this equation by \(r\).
\[r s_{n}=a r+a r^{2}+\ldots+a r^{n-1}+a r^{n} \nonumber \]
Subtracting these two equations, while noting the many cancelations, we have
\[ \begin{aligned} (1-r) s_{n}=&\left(a+a r+\ldots+a r^{n-2}+a r^{n-1}\right) \
\[4pt] &-\left(a r+a r^{2}+\ldots+a r^{n-1}+a r^{n}\right) \
\[4pt] =& a-a r^{n} \
\[4pt] =& a\left(1-r^{n}\right) \end{aligned} \label{A.92} \]
Thus, the \(n\)th partial sums can be written in the compact form
\[s_{n}=\dfrac{a\left(1-r^{n}\right)}{1-r} . \label{A.93} \]
The sum, if it exists, is given by \(S=\lim _{n \rightarrow \infty} s_{n} .\) Letting \(n\) get large in the partial sum (Equation \ref{A.93}), we need only evaluate \(\lim _{n \rightarrow \infty} r^{n}\). From the special limits in the Appendix we know that this limit is zero for \(|r|<1\). Thus, we have
\[\begin{aligned} &\text { Geometric Series } \
\[4pt] &\sum_{n=0}^{\infty} a r^{n}=\dfrac{a}{1-r^{\prime}}, \quad|r|<1 \
\[4pt] &\text { (A.94) } \end{aligned} \nonumber \]
The sum of the geometric series exists for \(|r|<1\) and is given by
The reader should verify that the geometric series diverges for all other values of \(r\). Namely, consider what happens for the separate cases \(|r|>1\), \(r=1\) and \(r=-1\).
Next, we present a few typical examples of geometric series.
\(\sum_{n=0}^{\infty} \dfrac{1}{2^{n}}\)
Solution
In this case we have that \(a=1\) and \(r=\dfrac{1}{2}\). Therefore, this infinite series converges and the sum is
\[S=\dfrac{1}{1-\dfrac{1}{2}}=2 \nonumber \]
\(\sum_{k=2}^{\infty} \dfrac{4}{3^{k}}\)
Solution
In this example we first note that the first term occurs for \(k=2\). It sometimes helps to write out the terms of the series,
\[\sum_{k=2}^{\infty} \dfrac{4}{3^{k}}=\dfrac{4}{3^{2}}+\dfrac{4}{3^{3}}+\dfrac{4}{3^{4}}+\dfrac{4}{3^{5}}+\ldots \nonumber \]
Looking at the series, we see that \(a=\dfrac{4}{9}\) and \(r=\dfrac{1}{3} .\) Since \(|\mathrm{r}|<\mathrm{1}\), the geometric series converges. So, the sum of the series is given by
\[S=\dfrac{\dfrac{4}{9}}{1-\dfrac{1}{3}}=\dfrac{2}{3}\nonumber \]
\(\sum_{n=1}^{\infty}\left(\dfrac{3}{2^{n}}-\dfrac{2}{5^{n}}\right)\)
Solution
Finally, in this case we do not have a geometric series, but we do have the difference of two geometric series. Of course, we need to be \({ }^{\text {I }}\) A rearrangement of terms in an infinite careful whenever rearranging infinite series. In this case it is allowed series is allowed when the series is absolutely convergent. (See the Appendix.)
\[\sum_{n=1}^{\infty}\left(\dfrac{3}{2^{n}}-\dfrac{2}{5^{n}}\right)=\sum_{n=1}^{\infty} \dfrac{3}{2^{n}}-\sum_{n=1}^{\infty} \dfrac{2}{5^{n}}\nonumber \]
Now we can add both geometric series to obtain
\[\sum_{n=1}^{\infty}\left(\dfrac{3}{2^{n}}-\dfrac{2}{5^{n}}\right)=\dfrac{\dfrac{3}{2}}{1-\dfrac{1}{2}}-\dfrac{\dfrac{2}{5}}{1-\dfrac{1}{5}}=3-\dfrac{1}{2}=\dfrac{5}{2}\nonumber \]
Geometric series are important because they are easily recognized and summed. Other series which can be summed include special cases of Taylor series and telescoping series. Next, we show an example of a telescoping series.
\(\sum_{n=1}^{\infty} \dfrac{1}{n(n+1)}\)
Solution
The first few terms of this series are
\[\sum_{n=1}^{\infty} \dfrac{1}{n(n+1)}=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\ldots\nonumber \]
It does not appear that we can sum this infinite series. However, if we used the partial fraction expansion
\[\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}\nonumber \]
then we find the \(k\) th partial sum can be written as
\[ \begin{aligned} s_{k} &=\sum_{n=1}^{k} \dfrac{1}{n(n+1)} \
\[4pt] &=\sum_{n=1}^{k}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right) \
\[4pt] &=\left(\dfrac{1}{1}-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\cdots+\left(\dfrac{1}{k}-\dfrac{1}{k+1}\right) \end{aligned} \label{A.95} \]
We see that there are many cancelations of neighboring terms, leading to the series collapsing (like a retractable telescope) to something manageable:
\[s_{k}=1-\dfrac{1}{k+1}\nonumber \]
Taking the limit as \(k \rightarrow \infty\), we find \(\sum_{n=1}^{\infty} \dfrac{1}{n(n+1)}=1\).
Actually, what are now known as Taylor and Maclaurin series were known long before they were named. James Gregory (1638-1675) has been recognized for discovering Taylor series, which were later named after Brook Taylor (1685-1731). Similarly, Colin Maclaurin (1698-1746) did not actually discover Maclaurin series, but the name was adopted because of his particular use of series.