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8.2: Geometric Series

  • Page ID
    107847
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    Motivating Questions
    • What is a geometric series?
    • What is a partial sum of a geometric series? What is a simplified form of the \(n\)th partial sum of a geometric series?
    • Under what conditions does a geometric series converge? What is the sum of a convergent geometric series?

    Many important sequences are generated by addition. In Preview Activity \(\PageIndex{1}\), we see an example of a sequence that is connected to a sum.

    Preview Activity \(\PageIndex{1}\)

    Warfarin is an anticoagulant that prevents blood clotting; often it is prescribed to stroke victims in order to help ensure blood flow. The level of warfarin has to reach a certain concentration in the blood in order to be effective.

    Suppose warfarin is taken by a particular patient in a 5 mg dose each day. The drug is absorbed by the body and some is excreted from the system between doses. Assume that at the end of a 24 hour period, 8% of the drug remains in the body. Let \(Q(n)\) be the amount (in mg) of warfarin in the body before the \((n+1)\)st dose of the drug is administered.

    1. Explain why \(Q(1) = 5 \times 0.08\) mg.
    2. Explain why \(Q(2) = (5+Q(1)) \times 0.08\) mg. Then show that

      \[ Q(2) = (5 \times 0.08)\left(1+0.08\right) \text{mg}\text{.} \nonumber \]

    3. Explain why \(Q(3) = (5+Q(2)) \times 0.08\) mg. Then show that
      \[ Q(3) = (5 \times 0.08)\left(1+0.08+0.08^2\right) \text{mg}\text{.} \nonumber \]
    4. Explain why \(Q(4) = (5+Q(3)) \times 0.08\) mg. Then show that
      \[ Q(4) = (5 \times 0.08)\left(1+0.08+0.08^2+0.08^3\right) \text{mg}\text{.} \nonumber \]
    5. There is a pattern that you should see emerging. Use this pattern to find a formula for \(Q(n)\text{,}\) where \(n\) is an arbitrary positive integer.
    6. Complete Table \(\PageIndex{1}\) with values of \(Q(n)\) for the provided \(n\)-values (reporting \(Q(n)\) to 10 decimal places). What appears to be happening to the sequence \(Q(n)\) as \(n\) increases?
      Table \(\PageIndex{1}\). Values of \(Q(n)\) for selected values of \(n\)
      \(n\) \(1\) \(2\) \(3\) \(4\) \(5\) \(6\) \(7\) \(8\) \(9\) \(10\)
      \(Q(n)\) \(0.40\)                  

    Geometric Series

    In Preview Activity \(\PageIndex{1}\) we encountered the sum

    \[ (5 \times 0.08)\left(1+0.08+0.08^2+0.08^3+ \cdots + 0.08^{n-1}\right) \nonumber \]

    for the long-term level of Warfarin in the patient's system. This sum has the form

    \[ a+ar+ar^2+ \cdots + ar^{n-1}\label{AuQ}\tag{\(\PageIndex{1}\)} \]

    where \(a=5 \times 0.08\) and \(r=0.08\text{.}\) Such a sum is called a finite geometric series with ratio \(r\text{.}\)

    Activity \(\PageIndex{2}\)

    Let \(a\) and \(r\) be real numbers (with \(r \ne 1\)) and let

    \[ S_n = a+ar+ar^2 + \cdots + ar^{n-1}\text{.} \nonumber \]

    In this activity we will find a shortcut formula for \(S_n\) that does not involve a sum of \(n\) terms.

    1. Multiply \(S_n\) by \(r\text{.}\) What does the resulting sum look like?
    2. Subtract \(rS_n\) from \(S_n\) and explain why
      \[ S_n - rS_n = a - ar^n\text{.}\label{RRw}\tag{\(\PageIndex{2}\)} \]
    3. Solve equation (\(\PageIndex{2}\)) for \(S_n\) to find a simple formula for \(S_n\) that does not involve adding \(n\) terms.

    The sum of the terms of a sequence is called a series. We summarize the result of Activity \(\PageIndex{2}\) in the following way.

    Note

    A finite geometric series \(S_n\) is a sum of the form

    \[ S_n = a + ar + ar^2 + \cdots + ar^{n-1}\text{,}\label{gBZ}\tag{\(\PageIndex{3}\)} \]

    where \(a\) and \(r\) are real numbers such that \(r \ne 1\text{.}\) The finite geometric series \(S_n\) can be written more simply as

    \[ S_n = a+ar+ar^2+ \cdots + ar^{n-1} = \frac{a(1-r^n)}{1-r}\text{.}\label{MJi}\tag{\(\PageIndex{4}\)} \]

    We now apply Equation (\(\PageIndex{4}\)) to the example involving Warfarin from Preview Activity \(\PageIndex{1}\). Recall that

    \[ Q(n)=(5 \times 0.08)\left(1+0.08+0.08^2+0.08^3+ \cdots + 0.08^{n-1}\right) \text{mg}\text{,} \nonumber \]

    so \(Q(n)\) is a geometric series with \(a=5 \times 0.08 = 0.4\) and \(r = 0.08\text{.}\) Thus,

    \[ Q(n) = 0.4\left(\frac{1-0.08^n}{1-0.08}\right) = \frac{1}{2.3} \left(1-0.08^n\right)\text{.} \nonumber \]

    Notice that as \(n\) goes to infinity, the value of \(0.08^n\) goes to 0. So,

    \[ \lim_{n \to \infty} Q(n) = \lim_{n \to \infty} \frac{1}{2.3} \left(1-0.08^n\right) = \frac{1}{2.3} \approx 0.435\text{.} \nonumber \]

    Therefore, the long-term level of Warfarin in the blood under these conditions is \(\frac{1}{2.3}\text{,}\) which is approximately 0.435 mg.

    To determine the long-term effect of Warfarin, we considered a finite geometric series of \(n\) terms, and then considered what happened as \(n\) was allowed to grow without bound. In this sense, we were actually interested in an infinite geometric series (the result of letting \(n\) go to infinity in the finite sum).

    Definition \(\PageIndex{1}\)

    An infinite geometric series is an infinite sum of the form

    \[ a + ar + ar^2 + \cdots = \sum_{n=0}^{\infty} ar^n\text{.}\label{llS}\tag{\(\PageIndex{5}\)} \]

    The value of \(r\) in the geometric series (\(\PageIndex{5}\)) is called the common ratio of the series because the ratio of the (\(n+1\))st term, \(ar^n\text{,}\) to the \(n\)th term, \(ar^{n-1}\text{,}\) is always \(r\text{:}\)

    \[ \frac{ar^n}{ar^{n-1}} = r\text{.} \nonumber \]

    Geometric series are common in mathematics and arise naturally in many different situations. As a familiar example, suppose we want to write the number with repeating decimal expansion

    \[ N=0.1212\overline{12} \nonumber \]

    as a rational number. Observe that

    \begin{align*} N & = 0.12 + 0.0012 + 0.000012 + \cdots\\[4pt] & = \left(\frac{12}{100}\right) + \left(\frac{12}{100}\right)\left(\frac{1}{100}\right) + \left(\frac{12}{100}\right)\left(\frac{1}{100}\right)^2 + \cdots\text{.} \end{align*}

    This is an infinite geometric series with \(a=\frac{12}{100}\) and \(r = \frac{1}{100}\text{.}\)

    By using the formula for the value of a finite geometric sum, we can also develop a formula for the value of an infinite geometric series. We explore this idea in the following activity.

    Activity \(\PageIndex{3}\)

    Let \(r \ne 1\) and \(a\) be real numbers and let

    \[ S = a+ar+ar^2 + \cdots ar^{n-1} + \cdots \nonumber \]

    be an infinite geometric series. For each positive integer \(n\text{,}\) let

    \[ S_n = a+ar+ar^2 + \cdots + ar^{n-1}\text{.} \nonumber \]

    Recall that

    \[ S_n = a\frac{1-r^n}{1-r}\text{.} \nonumber \]
    1. What should we allow \(n\) to approach in order to have \(S_n\) approach \(S\text{?}\)
    2. What is the value of \(\lim_{n \to \infty} r^n\) for \(|r| \gt 1\text{?}\) for \(|r| \lt 1\text{?}\) Explain.
    3. If \(|r| \lt 1\text{,}\) use the formula for \(S_n\) and your observations in (a) and (b) to explain why \(S\) is finite and find a resulting formula for \(S\text{.}\)

    We can now find the value of the geometric series

    \[ N = \left(\frac{12}{100}\right) + \left(\frac{12}{100}\right)\left(\frac{1}{100}\right) + \left(\frac{12}{100}\right)\left(\frac{1}{100}\right)^2 + \cdots\text{.} \nonumber \]

    Using \(a = \frac{12}{100}\) and \(r = \frac{1}{100}\text{,}\) we see that

    \[ N = \frac{12}{100} \left(\frac{1}{1-\frac{1}{100}}\right) = \frac{12}{100} \left(\frac{100}{99}\right) = \frac{4}{33}\text{.} \nonumber \]

    The sum of a finite number of terms of an infinite geometric series is often called a partial sum of the series. Thus,

    \[ S_n = a+ar+ar^2 + \cdots + ar^{n-1} = \sum_{k=0}^{n-1} ar^k\text{.} \nonumber \]

    is called the \(n\)th partial sum of the series \(\sum_{k=0}^{\infty} ar^k\text{.}\) We summarize our recent work with geometric series as follows.

    Note
    • An infinite geometric series is an infinite sum of the form
      \[ a + ar + ar^2 + \cdots = \sum_{n=0}^{\infty} ar^n\text{,}\label{Ckd}\tag{\(\PageIndex{6}\)} \]

      where \(a\) and \(r\) are real numbers such that \(r \ne 0\text{.}\)

    • The \(n\)th partial sum \(S_n\) of an infinite geometric series is
      \[ S_n = a+ar+ar^2+ \cdots + ar^{n-1}\text{.} \nonumber \]
    • If \(|r| \lt 1\text{,}\) then using the fact that \(S_n = a\frac{1-r^n}{1-r}\text{,}\) it follows that the sum \(S\) of the infinite geometric series (\(\PageIndex{6}\)) is
      \[ S = \lim_{n \to \infty} S_n = \lim_{n \to \infty} a\frac{1-r^n}{1-r} = \frac{a}{1-r} \nonumber \]
    Activity \(\PageIndex{4}\)

    The formulas we have derived for an infinite geometric series and its partial sum have assumed we begin indexing the sums at \(n=0\text{.}\) If instead we have a sum that does not begin at \(n=0\text{,}\) we can factor out common terms and use the established formulas. This process is illustrated in the examples in this activity.

    1. Consider the sum
      \[ \sum_{k=1}^{\infty} (2)\left(\frac{1}{3}\right)^k = (2)\left(\frac{1}{3}\right) + (2)\left(\frac{1}{3}\right)^2 + (2)\left(\frac{1}{3}\right)^3 + \cdots\text{.} \nonumber \]

      Remove the common factor of \((2)\left(\frac{1}{3}\right)\) from each term and hence find the sum of the series.

    2. Next let \(a\) and \(r\) be real numbers with \(-1\lt r\lt 1\text{.}\) Consider the sum
      \[ \sum_{k=3}^{\infty} ar^k = ar^3+ar^4+ar^5 + \cdots\text{.} \nonumber \]

      Remove the common factor of \(ar^3\) from each term and find the sum of the series.

    3. Finally, we consider the most general case. Let \(a\) and \(r\) be real numbers with \(-1\lt r\lt 1\text{,}\) let \(n\) be a positive integer, and consider the sum
      \[ \sum_{k=n}^{\infty} ar^k = ar^n+ar^{n+1}+ar^{n+2} + \cdots\text{.} \nonumber \]

      Remove the common factor of \(ar^n\) from each term to find the sum of the series.

    Summary

    • An infinite geometric series is an infinite sum of the form
      \[ \sum_{k=0}^{\infty} ar^k \nonumber \]

      where \(a\) and \(r\) are real numbers and \(r \neq 0\text{.}\)

    • The \(n\)th partial sum of the geometric series \(\sum_{k=0}^{\infty} ar^k\) is
      \[ S_n = \sum_{k=0}^{n-1} ar^k\text{.} \nonumber \]

      A formula for the \(n\)th partial sum of a geometric series is

      \[ S_n = a \frac{1-r^n}{1-r}\text{.} \nonumber \]

      If \(|r| \lt 1\text{,}\) the infinite geometric series \(\sum_{k=0}^{\infty} ar^k\) has the finite sum \(\frac{a}{1-r}\text{.}\)


    This page titled 8.2: Geometric Series is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Matthew Boelkins, David Austin & Steven Schlicker (ScholarWorks @Grand Valley State University) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.