7.2: Summation Notation

Math 370 Learning Objectives

• Understand and use summation notation.
• Manipulate sums using properties of summation notation.
• Compute the values of arithmetic and geometric summations.
• Use summations within applications.
• Understand series, specifically geometric series, and determine when a geometric series is convergent or divergent.

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Introduction to Summation Notation

In the previous section, we introduced sequences and now we shall present notation and theorems concerning the sum of terms of a sequence. We begin with a definition, which, while intimidating, is meant to make our lives easier.

Definition: Summation Notation

Given a sequence \(\left\{ a_{n} \right\}_{n=k}^{\infty}\) and numbers \(m\) and \(p\) satisfying \(k \leq m \leq p\), the summation from \(m\) to \(p\) of the sequence \(\left\{a_{n}\right\}\) is written

\[\sum_{n=m}^{p} a_{n}=a_{m}+a_{m+1}+\ldots+a_{p}.\nonumber\]

The variable \(n\) is called the index of summation. The number \(m\) is called the lower limit of summation while the number \(p\) is called the upper limit of summation.

In English, the definition of summation notation is simply defining a short-hand notation for adding up the terms of the sequence \(\left\{ a_{n} \right\}_{n=k}^{\infty}\) from \(a_{m}\) through \(a_{p}\). The symbol \(\Sigma\) is the capital Greek letter sigma and is shorthand for "sum." The lower and upper limits of the summation tells us which term to start with and which term to end with, respectively. For example, using the sequence \(a_{n} = 2n-1\) for \(n \geq 1\), we can write the sum \(a_{3} +a_{4} + a_{5} + a_{6}\) as

\[\begin{array}{rcl}
\displaystyle{\sum_{n=3}^{6}(2n-1) } & = & (2(3)-1) + (2(4)-1) + (2(5)-1) + (2(6)-1) \\
& = & 5 + 7 + 9 + 11 \\
& = & 32 \\
\end{array}\nonumber\]

The index variable is considered a "dummy variable" in the sense that it may be changed to any letter without affecting the value of the summation. For instance,

\[\displaystyle{\sum_{n=3}^{6}(2n-1)} = \displaystyle{\sum_{k=3}^{6}(2k-1)} = \displaystyle{\sum_{j=3}^{6}(2j-1)}\nonumber\]

One place you may encounter summation notation is in mathematical definitions. For example, summation notation allows us to define polynomials as functions of the form

\[f(x) = \displaystyle{\sum_{k=0}^{n} a_{k} x^{k}}\nonumber\]

for real numbers \(a_{k}\), \(k = 0, 1, \ldots n\). The reader is invited to compare this with what is given in the definition of a polynomial function in Section 3.1.

Our next example gives us practice with this new notation.

Example \( \PageIndex{1} \)

1. Find the following sums.
   1. \(\displaystyle{\sum_{k=1}^{4} \frac{13}{100^k} }\)
   2. \(\displaystyle{\sum_{n=0}^{4} \frac{n!}{2}}\)
   3. \(\displaystyle{\sum_{n=1}^{5} \frac{(-1)^{n+1}}{n} (x-1)^n}\)
2. Write the following sums using summation notation.
   1. \(1 + 3 + 5 + \ldots + 117\)
   2. \(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots + \frac{1}{117}\)
   3. \(0.9+0.09+0.009+\ldots 0.\underbrace{0 \cdots 0}_{n-1 \text { zeros}} 9\)

Solution

1. 1. We substitute \(k=1\) into the formula \(\frac{13}{100^k}\) and add successive terms until we reach \(k=4.\)\[\begin{array}{rcl}
      \displaystyle{\sum_{k=1}^{4} \dfrac{13}{100^k} } & = & \dfrac{13}{100^1} + \dfrac{13}{100^2} + \dfrac{13}{100^3} + \dfrac{13}{100^4} \\
      & = & 0.13 + 0.0013 + 0.000013 + 0.00000013 \\
      & = & 0.13131313 \\
      \end{array}\nonumber\]
   2. Proceeding as in (i), we replace every occurrence of \(n\) with the values \(0\) through \(4\). We recall the factorials, \(n!\) as defined in number Example 7.1.1, number 6 and get:\[\begin{array}{rcl}
      \displaystyle{\displaystyle{\sum_{n=0}^{4} \dfrac{n!}{2}}} & = & \dfrac{0!}{2} + \dfrac{1!}{2} + \dfrac{2!}{2} + \dfrac{3!}{2} + \dfrac{4!}{2} \\
      & = & \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{2 \cdot 1}{2} + \dfrac{3 \cdot 2 \cdot 1}{2} + \dfrac{4 \cdot 3 \cdot 2 \cdot 1 }{2} \\
      & = & \dfrac{1}{2} + \dfrac{1}{2} + 1 + 3 + 12 \\
      & = & 17 \\
      \end{array}\nonumber\]
   3. We proceed as before, replacing the index \(n\), but not the variable \(x\), with the values \(1\) through \(5\) and adding the resulting terms.\[\begin{array}{rcl}
      \displaystyle{\sum_{n=1}^{5} \dfrac{(-1)^{n+1}}{n} (x-1)^n} & = & \dfrac{(-1)^{1+1}}{1} (x-1)^1 + \dfrac{(-1)^{2+1}}{2} (x-1)^2 + \dfrac{(-1)^{3+1}}{3} (x-1)^3 \\
      & = & + \dfrac{(-1)^{1+4}}{4} (x-1)^4 + \dfrac{(-1)^{1+5}}{5} (x-1)^5 \\
      & = & (x-1) - \dfrac{(x-1)^2}{2} + \dfrac{(x-1)^3}{3} - \dfrac{(x-1)^4}{4} + \dfrac{(x-1)^5}{5} \\
      \end{array}\nonumber\]
2. The key to writing these sums with summation notation is to find the pattern of the terms. To that end, we make good use of the techniques presented in Section 7.1.
   1. The terms of the sum \(1\), \(3\), \(5\), etc., form an arithmetic sequence with first term \(a = 1\) and common difference \(d = 2\). We get the formula for the \( n^{\text{th}} \) term of the sequence \(a_{n} = 1 + (n-1)2 = 2n-1\), \(n \geq 1\). At this stage, we have the formula for the terms, namely \(2n-1\), and the lower limit of the summation, \(n=1\). To finish the problem, we need to determine the upper limit of the summation. In other words, we need to determine which value of \(n\) produces the term \(117\). Setting \(a_{n} = 117\), we get \(2n-1=117\) or \(n = 59\). Our final answer is\[\begin{array}{rcl}
      1 + 3 + 5 + \ldots + 117 & = & \displaystyle{\sum_{n=1}^{59} (2n-1)} \\
      \end{array}\nonumber\]
   2. We rewrite all of the terms as fractions, the subtraction as addition, and associate the negatives "\(-\)" with the numerators to get\[\dfrac{1}{1} + \dfrac{-1}{2} + \dfrac{1}{3} + \dfrac{-1}{4} + \ldots + \dfrac{1}{117}\nonumber\]The numerators, \(1\), \(-1\), etc. can be described by the geometric sequence1 \(c_{n} = (-1)^{n-1}\) for \(n \geq 1\), while the denominators are given by the arithmetic sequence2 \(d_{n} = n\) for \(n \geq 1\). Hence, we get the formula \(a_{n} = \frac{(-1)^{n-1}}{n}\) for our terms, and we find the lower and upper limits of summation to be \(n=1\) and \(n = 117\), respectively. Thus\[\begin{array}{rcl}
      1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + - \ldots + \dfrac{1}{117} & = & \displaystyle{\sum_{n=1}^{117} \dfrac{(-1)^{n-1}}{n}} \\
      \end{array}\nonumber\]
   3. Thanks to Example 7.1.3, we know that one formula for the \(n^{\text{th}}\) term is \(a_{n} = \frac{9}{10^{n}}\) for \(n \geq 1\). This gives us a formula for the summation as well as a lower limit of summation. To determine the upper limit of summation, we note that to produce the \(n-1\) zeros to the right of the decimal point before the \(9\), we need a denominator of \(10^{n}\). Hence, \(n\) is the upper limit of summation. Since \(n\) is used in the limits of the summation, we need to choose a different letter for the index of summation.³ We choose \(k\) and get\[0.9+0.09+0.009+\ldots 0.\underbrace{0 \cdots 0}_{n-1 \text { zeros }} 9=\sum_{k=1}^{n} \frac{9}{10^{k}}\nonumber\]

Subsection Footnotes

¹ This is indeed a geometric sequence with first term \(a = 1\) and common ratio \(r = −1\).

² It is an arithmetic sequence with first term \(a = 1\) and common difference \(d = 1\).

³ To see why, try writing the summation using "\(n\)" as the index.

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Properties of Summation Notation

The following theorem presents some general properties of summation notation. While we shall not have much need of these properties in Algebra, they do play a great role in Calculus. Moreover, there is much to be learned by thinking about why the properties hold. We invite the reader to prove these results. To get started, remember, "When in doubt, write it out!"

Theorem: Properties of Summation Notation

Suppose \(\left\{a_{n}\right\}\) and \(\left\{b_{n}\right\}\) are sequences so that the following sums are defined.

• \(\displaystyle{ \sum_{n=m}^{p} \left(a_{n} \pm b_{n} \right) = \sum_{n=m}^{p} a_{n} \pm \sum_{n=m}^{p} b_{n} }\)
• \(\displaystyle{\sum_{n=m}^{p} c \, a_{n} = c \sum_{n=m}^{p} a_{n}}\), for any real number \(c\).
• \(\displaystyle{\sum_{n=m}^{p} a_{n} = \sum_{n=m}^{j} a_{n} + \sum_{n=j+1}^{p} a_{n}}\), for any natural number \(m \leq j < j+1 \leq p\).
• \(\displaystyle{\sum_{n=m}^{p} a_{n} = \sum_{n=m+r}^{p+r} a_{n-r}}\), for any whole number \(r\).

Proof of \(\displaystyle{ \sum_{n=m}^{p} \left(a_{n} + b_{n} \right) = \sum_{n=m}^{p} a_{n} + \sum_{n=m}^{p} b_{n} }\)

\[ \begin{array}{rclr}
\displaystyle \sum_{n=m}^{p} \left(a_{n} + b_{n} \right) & = & (a_m + b_m) + (a_{m + 1} + b_{m + 1}) + \cdots + (a_{p - 1} + b_{p - 1}) + (a_p + b_p) & \left( \text{Definition of Summation Notation} \right) \\
 & = & (a_m + a_{m + 1} + \cdots + a_{p - 1} + a_p) + (b_m + b_{m + 1} + \cdots + b_{p - 1} + b_p) & \left( \text{Commutative and Associative Properties of Addition} \right) \\
 & = & \displaystyle \sum_{n=m}^{p} a_{n} + \displaystyle \sum_{n=m}^{p} b_{n} & \left( \text{Definition of Summation Notation} \right) \\
\end{array} \nonumber \]

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Summations of Arithmetic and Geometric Sequences

We now turn our attention to the sums involving arithmetic and geometric sequences. Given an arithmetic sequence \(a_{k} = a + (k-1) d\) for \(k \geq 1\), we let \(S_n\) denote the sum of the first \(n\) terms. To derive a formula for \(S_n\), we write it out in two different ways

\[\begin{array}{ccccccccccc}
S_n & = & a & + & (a + d) & + & \ldots & + & (a + (n-2)d) & + & (a + (n-1)d) \\
S_n & = & (a + (n-1)d) & + & (a + (n-2)d) & + & \ldots & + & (a + d) & + & a \\
\end{array}\nonumber\]

If we add these two equations and combine the terms which are aligned vertically, we get

\[2S_n = (2a + (n-1)d) + (2a + (n-1)d) + \ldots + (2a + (n-1)d) + (2a + (n-1)d)\nonumber\]

The right hand side of this equation contains \(n\) terms, all of which are equal to \((2a + (n-1)d)\) so we get \(2S_n = n(2a + (n-1)d)\). Dividing both sides of this equation by \(2\), we obtain the formula

\[S_n = \dfrac{n}{2} (2a + (n-1)d)\nonumber\]

If we rewrite the quantity \(2a + (n-1)d\) as \(a + (a + (n-1)d) = a_{1} + a_{n}\), we get the formula

\[S_n = n \left(\dfrac{a_1 + a_{n}}{2}\right)\nonumber\]

A helpful way to remember this last formula is to recognize that we have expressed the sum as the product of the number of terms \(n\) and the average of the first and \(n^{\text{th}}\) terms.

To derive the formula for the geometric sum, we start with a geometric sequence \(a_{k} = ar^{k-1}\), \(k \geq 1\), and let \(S_n\) once again denote the sum of the first \(n\) terms. Comparing \(S_n\) and \(rS_n\), we get

\[\begin{array}{ccccccccccccccc}
S_n & = & a & + & ar & + & ar^2 & + & \ldots & + & ar^{n-2} & + & ar^{n-1} & & \\
r S_n & = & & & ar & + & ar^2 & + & \ldots & + & ar^{n-2} & + & ar^{n-1} & + & ar^{n} \\
\end{array}\nonumber\]

Subtracting the second equation from the first forces all of the terms except \(a\) and \(ar^{n}\) to cancel out and we get \(S_n - rS_n = a - ar^{n}\). Factoring, we get \(S_n(1-r) = a \left(1-r^{n}\right)\). Assuming \(r \neq 1\), we can divide both sides by the quantity \((1-r)\) to obtain

\[S_n = a \left( \dfrac{1-r^n}{1-r}\right)\nonumber\]

If we distribute \(a\) through the numerator, we get \(a - ar^{n} = a_{1} - a_{n + 1}\) which yields the formula

\[S_n=\frac{a_{1}-a_{n+1}}{1-r}\nonumber\]

In the case when \(r=1\), we get the formula

\[S_n = \underbrace{a + a + \ldots +a }_{\text{$n$ times}} = n \, a\nonumber\]

Our results are summarized below.

Theorem \(\PageIndex{2}\): Sums of Arithmetic and Geometric Sequences

• The sum \(S_n\) of the first \(n\) terms of an arithmetic sequence \(a_{k}= a + (k-1)d\) for \(k \geq 1\) is\[S_n = \displaystyle{\sum_{k=1}^{n} a_{k}} = n \left(\dfrac{a_1 + a_{n}}{2}\right) = \dfrac{n}{2} (2a + (n-1)d) \label{aseq} \]
• The sum \(S_n\) of the first \(n\) terms of a geometric sequence \(a_{k}= ar^{k-1}\) for \(k \geq 1\) is\[S_n = \begin{cases}
  \displaystyle \sum_{k=1}^{n} a_{k} & = & \dfrac{a_{1} - a_{n + 1}}{1-r} = a \left( \dfrac{1-r^n}{1-r}\right), & \text{ if } r \neq 1  \\ 
  \displaystyle \sum_{k=1}^{n} a_{k} & = & \sum_{k=1}^{n} a =n a, & \text{ if } r =1 \\
  \end{cases} \label{gseq} \]

While we have made an honest effort to derive the formulas in Theorem \( \PageIndex{2} \), formal proofs require the machinery in Section 7.3.

Example \(\PageIndex{2}\)

1. Find the value of the sum.\[ -7 - \frac{9}{2} - 2 + \frac{1}{2} + 3 + \cdots + 108 \nonumber \]
2. Compute \[ \displaystyle \sum_{n = 1}^{5}{7 \left(\frac{2}{3}\right)^{n - 1}} \nonumber \]

Solution

1. We can see that the first term of this summation (and, hence, the first term of the corresponding sequence) is \( a_1 = -7 \). In order to guarantee we can find the value of this summation,⁴ we must determine if the underlying sequence is an arithmetic or geometric sequence. Writing out the terms so we can clearly see what's going on, we build the following table:\[ \begin{array}{|c|c|}
   \hline k & a_k \\
   \hline 1 & -7 \\
   \hline 2 & -\frac{9}{2} \\
   \hline 3 & -2 \\
   \hline 4 & \frac{1}{2} \\
   \hline 5 & 3 \\
   \hline \vdots & \vdots \\
   \hline ? & 108 \\
   \hline \end{array} \nonumber \]
   A quick computation shows that \( a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = a_5 - a_4 = \frac{5}{2} \). Therefore, we hope that this pattern continues and that this sequence is arithmetic with common difference \( d = \frac{5}{2} \) and initial value \( a_1 = -7 \). We now turn to Theorem \( \PageIndex{2} \) to compute the value of this summation.\[ S_n = n \left( \dfrac{-7 + 108}{2} \right) = \dfrac{101}{2}n. \nonumber \]Our current issue is that we need to find \( n \). Luckily, we have a formula for the sequence. We just need to find \( n \) so that\[ 108 = -7 + \frac{5}{2}(n - 1). \nonumber \]Solving this equation for \( n \) (you should do this), we get \( n = 47 \). Therefore, the sum of these \( 47 \) terms is\[ -7 - \frac{9}{2} - 2 + \frac{1}{2} + 3 + \cdots + 108 = \frac{101}{2} \cdot 47 = 2373.5. \nonumber \]
2. The sequence that this summation is built upon is\[ a_n = 7 \left(-\frac{2}{3}\right)^{n - 1}. \nonumber \]This is definitely geometric with \( a = 7 \) and common ratio \( r = -\frac{2}{3} \). Thus, by Theorem \( \PageIndex{2} \),\[ \displaystyle \sum_{n = 1}^{5}{7 \left(\frac{2}{3}\right)^{n - 1}} = 7 \cdot \dfrac{1 - \left( -\frac{2}{3} \right)^{5}}{1 - \left(-\frac{2}{3}\right)} = 7 \cdot \dfrac{1 + \frac{2^5}{3^5}}{1 + \frac{2}{3}} = 7 \cdot \dfrac{3^5 + 2^5}{3^5 + 2 \cdot 3^4} = \dfrac{1925}{405} = \dfrac{385}{81}. \nonumber \]

Subsection Footnotes

⁴ This is a summation. It's just that we are adding negative values for the first few terms.

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Applications of Summations

Example \(\PageIndex{3}\)

One way to stack wood for the winter (if you have a wood-burning stove) is to dry stack the wood so that each layer contains one fewer log than the layer below. If I dry stacked wood so that the bottom layer had 40 logs and the top layer ended with 18 logs, how long can I keep my house warm considering that I burn one log per hour?

Solution

This is a very common application of arithmetic summations. We have the following sum:\[ 40 + 39 + 38 + \cdots + 18 \nonumber \]Therefore, the initial term of our sequence this summation is based on is \( a_1 = 40 \). The common difference (how we get to each successive term) is \( d = -1 \). Finally, there are \( n = 40 - 18 + 1 = 23 \) layers of wood (why did we add the 1?). Thus, by Theorem \( \PageIndex{2} \), we have\[ S_n = n \dfrac{a_1 + a_n}{2} = 23 \dfrac{40 + 18}{2} = (23)(29) = 667 \nonumber \]pieces of wood. That's \( 667 \) hours of heat (or almost 28 days of warmth).

An application of the arithmetic sum formula which proves useful in Calculus results in formula for the sum of the first \(n\) natural numbers. The natural numbers themselves are a sequence⁵ \(1, 2, 3, \ldots \) which is arithmetic with \(a = d = 1\). Applying Theorem \( \PageIndex{2} \),

\[\begin{array}{rcl}
1 + 2 + 3 + \ldots + n & = & \dfrac{n(n+1)}{2}
\end{array}\nonumber\]

So, for example, the sum of the first \(100\) natural numbers⁶ is \(\frac{100(101)}{2} = 5050\).

This formula, and some related formulas, are so important that we need to make a theorem out of it.

Theorem: Common Summations

\[\begin{array}{rcl}
\displaystyle \sum_{k = 1}^{n} k & = & \dfrac{n(n+1)}{2} \\
\displaystyle \sum_{k = 1}^{n} k^2 & = & \dfrac{n(n+1)(2n + 1)}{6} \\
\displaystyle \sum_{k = 1}^{n} k^3 & = & \dfrac{n^2(n+1)^2}{4} \\
\end{array}\nonumber\]

The proofs of these will have to wait until we cover Mathematical Induction.

An important application of the geometric sum formula is the investment plan called an annuity. Annuities differ from the kind of investments we studied in Section 6.5 in that payments are deposited into the account on an on-going basis, and this complicates the mathematics a little.⁷ Suppose you have an account with annual interest rate \(r\) which is compounded \(n\) times per year. We let \(i = \frac{r}{n}\) denote the interest rate per period. Suppose we wish to make ongoing deposits of \(P\) dollars at the end of each compounding period. Let \(A_{k}\) denote the amount in the account after \(k\) compounding periods. Then \(A_{1} = P\), because we have made our first deposit at the end of the first compounding period and no interest has been earned. During the second compounding period, we earn interest on \(A_{1}\) so that our initial investment has grown to \(A_{1}(1+i) = P(1+i)\) in accordance with the Simple Interest formula. When we add our second payment at the end of the second period, we get

\[A_2 = A_1(1+i) + P = P(1+i) + P = P(1+i)\left(1 + \dfrac{1}{1+i}\right)\nonumber\]

The reason for factoring out the \(P(1+i)\) will become apparent in short order. During the third compounding period, we earn interest on \(A_{2}\) which then grows to \(A_{2}(1+i)\). We add our third payment at the end of the third compounding period to obtain

\[A_3 = A_2(1+i) + P = P(1+i)\left(1 + \dfrac{1}{1+i}\right)(1+i) + P = P(1+i)^2\left(1 + \dfrac{1}{1+i} + \dfrac{1}{(1+i)^2}\right)\nonumber\]

During the fourth compounding period, \(A_{3}\) grows to \(A_{3}(1+i)\), and when we add the fourth payment, we factor out \(P(1+i)^3\) to get

\[A_4 = P(1+i)^3 \left(1 + \dfrac{1}{1+i} + \dfrac{1}{(1+i)^2} + \dfrac{1}{(1+i)^3}\right)\nonumber\]

This pattern continues so that at the end of the \(k^{\text{th}}\) compounding, we get

\[A_{k} = P(1+i)^{k-1} \left(1 + \dfrac{1}{1+i} + \dfrac{1}{(1+i)^2} + \ldots + \dfrac{1}{(1+i)^{k-1}}\right)\nonumber\]

The sum in the parentheses above is the sum of the first \(k\) terms of a geometric sequence with \(a = 1\) and \(r = \frac{1}{1+i}\). Using Theorem \( \PageIndex{2} \), we get

\[1 + \dfrac{1}{1+i} + \dfrac{1}{(1+i)^2} + \ldots + \dfrac{1}{(1+i)^{k-1}} = 1 \left(\dfrac{1 - \dfrac{1}{(1+i)^k}}{1 - \dfrac{1}{1+i}}\right) = \ \dfrac{(1+i)\left(1 - (1+i)^{-k}\right)}{i}\nonumber\]

Hence, we get

\[A_{k} = P(1+i)^{k-1} \left(\dfrac{(1+i)\left(1 - (1+i)^{-k}\right)}{i}\right) = \dfrac{P\left((1+i)^k - 1\right)}{i}\nonumber\]

If we let \(t\) be the number of years this investment strategy is followed, then \(k = nt\), and we get the formula for the future value of an ordinary annuity.

Theorem \(\PageIndex{3}\): Future Value of an Ordinary Annuity

Suppose an annuity offers an annual interest rate \(r\) compounded \(n\) times per year. Let \(i = \frac{r}{n}\) be the interest rate per compounding period. If a deposit \(P\) is made at the end of each compounding period, the amount \(A\) in the account after \(t\) years is given by

\[A = \dfrac{P\left((1+i)^{nt} - 1\right)}{i}\nonumber\]

The reader is encouraged to substitute \(i = \frac{r}{n}\) into the formula for the future value of an ordinary annuity and simplify. Some familiar equations arise which are cause for pause and meditation.

One last note: if the deposit \(P\) is made a the beginning of the compounding period instead of at the end, the annuity is called an annuity-due. We leave the derivation of the formula for the future value of an annuity-due as an exercise for the reader.

Example \( \PageIndex{4} \)

An ordinary annuity offers a \(6 \%\) annual interest rate, compounded monthly.

1. If monthly payments of \(\$50\) are made, find the value of the annuity in \(30\) years.
2. How many years will it take for the annuity to grow to \(\$100,\! 000\)?

Solution

1. We have \(r = 0.06\) and \(n = 12\) so that \(i = \frac{r}{n} = \frac{0.06}{12} = 0.005\). With \(P=50\) and \(t=30\),\[A = \dfrac{50\left((1+0.005)^{(12)(30)} - 1\right)}{0.005} \approx 50225.75\nonumber\]Our final answer is \(\$50,\!225.75\).
2. To find how long it will take for the annuity to grow to \(\$100,\!000\), we set \(A = 100000\) and solve for \(t\). We isolate the exponential and take natural logs of both sides of the equation.\[\begin{array}{rcl}
   100000 & = & \dfrac{50\left((1+0.005)^{12t} - 1\right)}{0.005} \\
   10 & = & (1.005)^{12t} - 1 \\
   (1.005)^{12t} & = & 11 \\
   \ln\left((1.005)^{12t}\right) & = & \ln(11) \\
   12t \ln(1.005) & = & \ln(11) \\
   t & = & \dfrac{\ln(11)}{12 \ln(1.005)} \approx 40.06 \\
   \end{array}\nonumber\]This means that it takes just over \(40\) years for the investment to grow to \(\$100,\!000\). Comparing this with our answer to part a, we see that in just \(10\) additional years, the value of the annuity nearly doubles. This is a lesson worth remembering.

Subsection Footnotes

⁵ This is the identity function on the natural numbers!

⁶ There is an interesting anecdote which says that the famous mathematician Carl Friedrich Gauss was given this problem in primary school and devised a very clever solution.

⁷ The reader may wish to re-read the discussion on compound interest in Section 6.5 before proceeding

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Series

We close this section with a peek into Calculus by considering infinite sums, called series. Consider the number \(0.\overline{9}\). We can write this number as

\[0.\overline{9} = 0.9999... = 0.9 + 0.09 + 0.009 + 0.0009 + \ldots\nonumber\]

From Example \( \PageIndex{1} \), we know we can write the sum of the first \(n\) of these terms as

\[0 . \underbrace{9 \cdots 9}_{n \text { nines }}=.9+0.09+0.009+\ldots 0.\underbrace{0 \cdots 0}_{n-1 \text { zeros }} 9=\sum_{k=1}^{n} \frac{9}{10^{k}}\nonumber\]

Using Theorem \( \PageIndex{2} \), we have

\[\displaystyle{\sum_{k=1}^{n} \dfrac{9}{10^{k}}} = \dfrac{9}{10} \left( \dfrac{1 - \dfrac{1}{10^{n+1}}}{1 - \dfrac{1}{10}} \right) = 1 - \dfrac{1}{10^{n+1}}\nonumber\]

It stands to reason that \(0.\overline{9}\) is the same value of \(1 - \frac{1}{10^{n+1}}\) as \(n \to \infty\). Our knowledge of exponential expressions from Section 6.1 tells us that \(\frac{1}{10^{n+1}} \to 0\) as \(n \to \infty\), so \(1 - \frac{1}{10^{n+1}} \to 1\). We have just argued that \(0.\overline{9} = 1\), which may cause some distress for some readers.⁸ Any non-terminating decimal can be thought of as an infinite sum whose denominators are the powers of \(10\), so the phenomenon of adding up infinitely many terms and arriving at a finite number is not as foreign of a concept as it may appear. We end this section with a theorem concerning geometric series.

Theorem \(\PageIndex{4}\): Sums of Geometric Series

Given the sequence \(a_{k} = ar^{k-1}\) for \(k \geq 1\), where \(|r| < 1\),

\[a + ar + ar^2 + \ldots = \displaystyle{\sum_{k=1}^{\infty} ar^{k-1}} = \dfrac{a}{1-r}\nonumber\]

If \(|r| \geq 1\), the sum \(a + ar + ar^2 + \ldots\) is not defined.

The justification of the result in Theorem \( \PageIndex{4} \) comes from taking the formula in Theorem \( \PageIndex{2} \) for the sum of the first \(n\) terms of a geometric sequence and examining the formula as \(n \to \infty\). Assuming \(|r|<1\) means \(-1 < r < 1\), so \(r^{n} \to 0\) as \(n \to \infty\). Hence as \(n \to \infty\),

\[\displaystyle{\sum_{k=1}^{n} a r^{k-1}} = a \left( \dfrac{1-r^n}{1-r}\right) \to \dfrac{a}{1-r}\nonumber\]

As to what goes wrong when \(|r| \geq 1\), we leave that to Calculus as well, but will explore some cases in the exercises. Before we jump into exercises, it's in our best interest to introduce a little bit of mathematical language that is used frequently in Calculus.

Definition: Convergent and Divergent Geometric Series

The geometric series

\[\displaystyle{\sum_{k=1}^{\infty} ar^{k-1}} \nonumber\]

is called convergent if \( |r| \lt 1 \) and, in such a case, it is said to converge to \( \frac{a}{1 - r} \). If \( |r| \ge 1 \), the series is said to be divergent.

Caution

The formula for a convergent geometric series

\[\displaystyle \sum_{k=1}^{\infty} ar^{k-1} = \dfrac{a}{1 - r}, \nonumber\]

only works if \( |r| \lt 1 \). Moreover, due to the fact that series will often be given to you with a lower limit not equal to \( 1 \), or with exponents not equal to \( k - 1 \), it is very important to rephrase the required "structure" of the geometric series form. Essentially, you want to manipulate your given series (through factoring) to the form

\[ \displaystyle b \sum r^{p}, \nonumber \]

where the power, \( p \), is \( 0 \) when evaluated at the lower limit of the summation.

That "Cautionary" statement needs some clarification, so let's see it in an example. 

Example \(\PageIndex{5}\)

Determine if the series converges or diverges. If it is convergent, find its value.

1. \( \displaystyle \sum_{n = 0}^{\infty} \frac{3 (2)^{n + 1}}{(-5)^{n+3}} \)
2. \( \frac{3}{2} - \frac{15}{4} + \frac{75}{8} - \frac{375}{16} + \cdots \)

Solution

1. Note that the sequence being summed, \( \frac{3 (2)^{n + 1}}{(-5)^{n+3}} \), is fairly complex. Moreover, the lower limit of the summation is not \( 1 \) as seems to be required by Theorem \(\PageIndex{4}\). However, fear not! The "Cautionary" statement asserts that, as long as the first power of our summation is \( 0 \), we should be fine.
   
   Unfortunately, when we get the first term of the expansion, \[ \displaystyle \sum_{n = 0}^{\infty} \dfrac{3 (2)^{n + 1}}{(-5)^{n+3}} = \dfrac{3 \cdot 2^1}{(-5)^3} + \cdots, \nonumber \]we can clearly see that those exponents are definitely not \( 0 \) and they don't even match each other! What do we do?
   
   A little algebra goes a long way here.\[ \begin{array}{rcl}
   \displaystyle \sum_{n = 0}^{\infty} \dfrac{3 (2)^{n + 1}}{(-5)^{n+3}} & = & 3 \displaystyle \sum_{n = 0}^{\infty} \dfrac{(2)^{n + 1}}{(-5)^{n+3}} \\
   & = & 3 \left(\dfrac{2^1}{(-5)^3}\right) \displaystyle \sum_{n = 0}^{\infty} \dfrac{(2)^{n}}{(-5)^{n}} \\
   & = & -\dfrac{6}{125} \displaystyle \sum_{n = 0}^{\infty} \left( - \dfrac{2}{5} \right)^n \\
   \end{array} \nonumber \]We now have our series in a form where the exponent of the first term of the expansion will be \( 0 \) (try it out and see!). Moreover, we can see that \( |r| = \left| -\frac{2}{5} \right| = \frac{2}{5} \lt 1 \). Therefore, this series converges. Specifically,\[ \begin{array}{rcl}
   \displaystyle \sum_{n = 0}^{\infty} \dfrac{3 (2)^{n + 1}}{(-5)^{n+3}} & = & -\dfrac{6}{125} \displaystyle \sum_{n = 0}^{\infty} \left( - \dfrac{2}{5} \right)^n \\
   & = & -\dfrac{6}{125} \cdot \dfrac{1}{1 - \left(- \frac{2}{5} \right)} \\
   & = & -\dfrac{6}{125} \cdot \dfrac{1}{1 + \frac{2}{5}} \\
   & = & -\dfrac{6}{125} \cdot \dfrac{5}{5 + 2} \\
   & = & -\dfrac{6}{125} \cdot \dfrac{5}{7} \\
   & = & -\dfrac{6}{25} \cdot \dfrac{1}{7} \\
   & = & -\dfrac{6}{175} \\
   \end{array} \nonumber \]
2. Rewriting this summation in series form, we get\[ \displaystyle \sum_{k = 0}^{\infty} \frac{3}{2} \left( -\frac{5}{2} \right)^n. \nonumber \]Note that I used the lower index of \( k = 0 \) here to reinforce the point that you don't need your lower index to always start at \( 1 \) as long as your first power in your expansion has an exponent of \( 0 \) on \( r = -\frac{5}{2} \). Speaking of which, since \( \left| r \right| = \left| -\frac{5}{2} \right| = \frac{5}{2} \ge 1 \), this geometric series is divergent.

Example \(\PageIndex{6}\)

A common starting discussion in Calculus about the topic of limits concerns walking "half-distances." Suppose you are playing a game and the goal is to move your game piece so it eventually touches the edge of the board. The rules of the game are that you can always move in any direction, but you must move at most a distance that is half the current distance between your game piece and the edge. How can you win?

Solution

You can't.

Okay, we need more discussion than that. Suppose you take the most direct route to the edge. On your first turn, you must half the distance from the current position to the edge. On your second turn, you move half the remaining distance. You repeat this process ad infinitum.

You can compute the total distance traveled using a series. Let \( D \) be the initial distance the piece is from the edge of the board, and let \( i \) be the number of the turn.

So, on turn \( i = 1 \), you move your piece a total of \( \frac{1}{2} D \) directly toward the edge of the board.

On your second turn, you move half of the remaining distance. Since the remaining distance is \( \frac{1}{2} D \), you move \( \frac{1}{2} \left( \frac{1}{2} D \right) = \frac{1}{2^2} D \).¹⁰ 

On the third turn, you move half of the remaining distance. Let's see, you have moved \( \frac{1}{2} D + \frac{1}{4} D = \frac{3}{4} D \). Therefore, the remaining distance is \( \frac{1}{4} D \). So you move your piece \( \frac{1}{2} \left( \frac{1}{4} D \right) = \frac{1}{2^3} D \).

The fourth turn comes and you now see a pattern. To check, let's do one last computation. You have already moved your piece \( \frac{1}{2} D + \frac{1}{4} D + \frac{1}{8} D = \frac{7}{8} D \). So you only have \( \frac{1}{8} D \) to go. You move your piece half this distance, \( \frac{1}{2} \left( \frac{1}{8} D \right) = \frac{1}{2^4} D \), toward the edge.

This pattern continues on and on.

Building a table of our "step sizes," we get

\[ \begin{array}{|c|c|}
\hline i & S_i \\
\hline 1 & \frac{1}{2}D \\
\hline 2 & \frac{1}{2^2} D \\
\hline 3 & \frac{1}{2^3} D \\
\hline 4 & \frac{1}{2^4} D \\
\hline \vdots & \vdots \\ \hline
\end{array} \nonumber \]

Summing the sequence \( S_i \) in the table, we get\[ \text{Total distance moved } = \displaystyle \sum_{i = 1}^{\infty}{\frac{1}{2^i} D} = \displaystyle \sum_{i = 1}^{\infty}{D \left(\frac{1}{2}\right)^i}. \nonumber \]At this point, we have the common ratio, \( r = \frac{1}{2} \), and what looks to be the value of \( a \) (it looks like \( a = D \)); however, there is a problem. The form of this series does not match the form in Theorem \( \PageIndex{4} \). We need the exponent on the common ratio, \( r = \frac{1}{2} \), to be \( i - 1 \). The easiest way to reduce the power on an expression is to factor off one power. That is, \( \frac{1}{2^i} = \frac{1}{2} \left( \frac{1}{2^{i - 1}} \right) \). Hence, our series becomes\[ \begin{array}{rcl}
\text{Total distance moved } & = & \displaystyle \sum_{i = 1}^{\infty}{\frac{1}{2^{i}} D} \\
& = & \displaystyle \sum_{i = 1}^{\infty}{\frac{D}{2} \cdot \frac{1}{2^{i - 1}}} \\
& = & \displaystyle \sum_{i = 1}^{\infty}{\frac{D}{2} \left( \frac{1}{2} \right)^{i - 1}} \\
& = & \dfrac{D/2}{1 - \frac{1}{2}} \\
& = & \dfrac{D}{2 - 1} \\
& = & D. \\
\end{array} \nonumber \]Thus, given an infinite amount of turns, we will have moved the game piece a total distance of \( D \) (which is the full length of the initial distance the piece was from the edge of the board); however, there is no possibility for us to have an infinite number of moves! Moreover, pretending that we had \( 10^{100} \) turns,¹¹ we would still have half of a remaining distance before the piece would actually reach the true edge.

Hence, despite theoretically being able to move the piece the full distance \( D \) to the edge of the board, the moment we believe we have actually reached the edge of the board would be on some finite turn, say the \( n^{\text{th}} \) turn, and at that moment, we would still have a sliver of space left between the game piece and the edge - and we can only move half that remaining distance.

Subsection Footnotes

⁸ To make this more palatable, it is usually accepted that \(0 . \overline{3}=\frac{1}{3}\) so that \(0 . \overline{9}=3(0 . \overline{3})=3\left(\frac{1}{3}\right)=1\). Feel better?

⁹ When in doubt, write them out!

¹⁰ It is a great idea when dealing with sequences, summations, and series to not simplify expressions. This allows you to see the patterns within sums.

¹¹ You should Google this number. The namesake is very familiar to you!