10.2 Summation Notation

In Section 10.1, we showed how the formula for compound interest is a geometric sequence. In retirement planning, it is seldom the case that an investor deposits a set amount of money into an account and waits for it to grow. Usually, additional payments of principal are made at regular intervals and the value of the investment grows accordingly. This kind of investment is called an annuity and will be discussed in later in this section once we have developed more mathematical machinery that enables us to add sequences.

In the previous section, we introduced sequences. Each of the numbers in the sequence is called a `term' which implies these numbers are meant to be added. To that end, we introduce the following notation which is used to describe the sum of (some of the) terms of a sequence.

Definition

Summation Notation: Given a sequence ${a_{n}}_{n = k}^{\infty}$ and numbers $m$ and $p$ satisfying $k \leq m \leq p$ , the summation from $m$ to $p$ of the sequence ${a_{n}}$ is written

$\sum_{n = m}^{p} a_{n} = a_{m} + a_{m + 1} + \dots + a_{p}$

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, Definition 10.4 is simply defining a short-hand notation for adding up the terms of the sequence ${a_{n}}_{n = k}^{\infty}$ from $a_{m}$ through $a_{p}$ . The symbol $Σ$ 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 $a_{n} = 2 n - 1$ for $n greater than or equal to 1$ , we can write $a_{3} + a_{4} + a_{5} + a_{6}$ as

$\begin{matrix} \sum_{n = 3}^{6} (2 n - 1) & = & (2 (3) - 1) + (2 (4) - 1) + (2 (5) - 1) + (2 (6) - 1) \\ = & 5 + 7 + 9 + 11 \\ = & 32 \end{matrix}$

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,

$\sum_{n = 3}^{6} (2 n - 1) = \sum_{k = 3}^{6} (2 k - 1) = \sum_{j = 3}^{6} (2 j - 1)$

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

$f (x) = \sum_{k = 0}^{n} a_{k} x^{k}$

for real numbers $a_{k}$ , $k equals 0, 1, … ⁢ n$ . The reader is invited to compare this with what is given in Definition 2.4. Summation notation is particularly useful when talking about matrix operations. For example, we can write the product of the $i$ th row $R_{i}$ of a matrix $A = {[a_{i j}]}_{m \times n}$ and the $j^{th}$ column $C_{j}$ of a matrix $B = {[b_{i j}]}_{n \times r}$ as

$R i \cdot C j = \sum_{k = 1}^{n} a_{i k} b_{k j}$

Again, the reader is encouraged to write out the sum and compare it to Definition 9.8. Our next example gives us practice with this new notation.

Example 1

Find the following sums.
1. $\sum_{k = 1}^{4} \frac{13}{100^{k}}$
2. $\sum_{n = 0}^{4} \frac{n!}{2}$
3. $\sum_{n = 1}^{5} \frac{{(- 1)}^{n + 1}}{n} {(x - 1)}^{n}$
Write the following sums using summation notation.
1. $1 + 3 + 5 + \dots + 117$
2. $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + - \dots + \frac{1}{117}$
3. $0.9 + 0.09 + 0.009 + \dots 0 . \underset{n - 1 zeros}{\underset{⏟}{0 \dots 0}} 9$

1. We substitute $k equals 1$ into the formula $\frac{13}{100^{k}}$ and add successive terms until we reach $k equals 4.$
  $\begin{matrix} \sum_{k = 1}^{4} \frac{13}{100^{k}} & = & \frac{13}{100^{1}} + \frac{13}{100^{2}} + \frac{13}{100^{3}} + \frac{13}{100^{4}} \\ = & 0.13 + 0.0013 + 0.000013 + 0.00000013 \\ = & 0.13131313 \end{matrix}$
2. Proceeding as in (a), we replace every occurrence of $n$ with the values $0$ through $4$ . We recall the factorials, $n!$ as defined in number Example 10.1.1, number and get:
  $\begin{matrix} \sum_{n = 0}^{4} \frac{n!}{2} & = & \frac{0!}{2} + \frac{1!}{2} + \frac{2!}{2} + \frac{3!}{2} = \frac{4!}{2} \\ = & \frac{1}{2} + \frac{1}{2} + \frac{2 \cdot 1}{2} + \frac{3 \cdot 2 \cdot 1}{2} + \frac{4 \cdot 3 \cdot 2 \cdot 1}{2} \\ = & \frac{1}{2} + \frac{1}{2} + 1 + 3 + 12 \\ = & 17 \end{matrix}$
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.
  $\begin{matrix} \sum_{n = 1}^{5} \frac{{(- 1)}^{n + 1}}{n} {(x - 1)}^{n} & = & \frac{{(- 1)}^{1 + 1}}{1} {(x - 1)}^{1} + \frac{{(- 1)}^{2 + 1}}{2} {(x - 1)}^{2} + \frac{{(- 1)}^{3 + 1}}{3} {(x - 1)}^{3} \\ + \frac{{(- 1)}^{1 + 4}}{4} {(x - 1)}^{4} + \frac{{(- 1)}^{1 + 5}}{5} {(x - 1)}^{5} \\ = & (x - 1) - \frac{{(x - 1)}^{2}}{2} + \frac{{(x - 1)}^{3}}{3} - \frac{{(x - 1)}^{4}}{4} + \frac{{(x - 1)}^{5}}{5} \end{matrix}$
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 10.1.
1. The terms of the sum $1$ , $3$ , $5$ , etc., form an arithmetic sequence with first term $a equals 1$ and common difference $d equals 2$ . Using Equation 10.1, we get $a_{n} = 1 + (n - 1) 2 = 2 n - 1$ , $n greater than or equal to 1$ .
  At this stage, we have the formula for the terms, namely $2 ⁢ n minus 1$ , and the lower limit of the summation, $n equals 1$ . 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 $a sub n equals 117$ , we get $2 n - 1 = 117$ or $n equals 59$ . Our final answer is
  $\begin{matrix} 1 + 3 + 5 + \dots + 117 & = & \sum_{n = 1}^{59} (2 n - 1) \end{matrix}$
2. We rewrite all of the terms as fractions, the subtraction as addition, and associate the negatives `' with the numerators to get
  $\frac{1}{1} + \frac{- 1}{2} + \frac{1}{3} + \frac{- 1}{4} + \dots + \frac{1}{117}$
  The numerators, $1$ , $minus 1$ , etc. can be described by the geometric sequence¹ $C_{n} = {(- 1)}^{n - 1}$ for $n greater than or equal to 1$ , while the denominators are given by the arithmetic sequence² $D_{n} = n$ for $n greater than or equal to 1$ . 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 $n equals 1$ and $n equals 117$ , respectively. Thus
  $\begin{matrix} 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + - \dots + \frac{1}{117} & = & \sum_{n = 1}^{117} \frac{{(- 1)}^{n - 1}}{n} \end{matrix}$
3. Thanks to Example 10.1.3, we know that one formula for the $n^{th}$ term is $a_{n} = \frac{9}{10^{n}}$ for $n greater than or equal to 1$ . 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 $n minus 1$ zeros to the right of the decimal point before the $9$ , we need a denominator of $10 to the power n$ . 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
  $\begin{matrix} 0.9 + 0.09 + 0.009 + \dots 0 . \underset{n - 1 zeros}{\underset{⏟}{0 \dots 0}} 9 & = & \sum_{k = 1}^{n} \frac{9}{10^{k}} \end{matrix}$

The following theorem presents some general properties of summation notation.

Theorem

Properties of Summation Notation: Suppose ${a_{n}}$ and ${b_{n}}$ are sequences so that the following sums are defined.

Sum and Difference Property: $\sum_{n = m}^{p} (a_{n} \pm b_{n}) = \sum_{n = m}^{p} a_{n} \pm \sum_{n = m}^{p} b_{n}$
Distributive Property: $\sum_{n = m}^{p} c a_{n} = c \sum_{n = m}^{p} a_{n}$ , for any real number $c$ .
Additive Index Property: $\sum_{n = m}^{j} a_{n} + \sum_{n = j + 1}^{p} a_{n} = \sum_{n = m}^{p} a_{n}$ , for any natural number $m less than or equal to j less than j plus 1 less than or equal to p$ .
Re-indexing: $\sum_{n = m}^{p} a_{n} = \sum_{n = m + r}^{p + r} a_{n - r}$ , for any integer $r$ .

There is much to be learned by thinking about why the properties hold, so we leave the proof of these properties to the reader.⁴

Example 2

If $\sum_{n = 2}^{50} (a_{n} - 3 b_{n}) = 17$ and $\sum_{n = 2}^{50} a_{n} = 10$ , find $∑ n equals 250 b sub n$ .
If $\sum_{n = 1}^{20} a_{n} = - 3$ and $\sum_{n = 1}^{21} a_{n} = 7$ , find $a sub 21$ .
Rewrite the sum so the index starts at $0$ : $\sum_{n = 2}^{437} n (n - 1) x^{n - 2}$

Using the Sum and Difference Property along with the Distributive Property of Theorem 10.4, we get:
$\sum_{n = 2}^{50} (a_{n} - 3 b_{n}) = \sum_{n = 2}^{50} a_{n} - \sum_{n = 2}^{50} 3 b_{n} = \sum_{n = 2}^{50} a_{n} - 3 \sum_{n = 2}^{50} b_{n}$
Hence, $\sum_{n = 2}^{50} a_{n} - 3 \sum_{n = 2}^{50} b_{n} = 17$ . If $\sum_{n = 2}^{50} a_{n} = 10$ , then $10 - 3 \sum_{n = 2}^{50} b_{n} = 17$ so $\sum_{n = 2}^{50} b_{n} = - \frac{7}{3}$ .
There are at least two ways to approach this problem. By definition, $\sum_{n = 1}^{21} a_{n} = a_{1} + a_{2} + \dots + a_{21}$ . That is, we add up the first $21$ terms of the sequence $a_{n}$ . Similarly, $\sum_{n = 1}^{20} a_{n} = a_{1} + a_{2} + \dots + a_{20}$ means we add up the first $20$ terms of the sequence. Hence, $a_{21} = \sum_{n = 1}^{21} a_{n} - \sum_{n = 1}^{20} a_{n} = 7 - (- 3) = 10 .$
Alternatively, we can use the Additive Index Property:
$\sum_{n = 1}^{21} a_{n} = \sum_{n = 1}^{20} a_{n} + \sum_{n = 21}^{21} a_{n} = \sum_{n = 1}^{20} a_{n} + a_{21},$
which gives $a_{21} = \sum_{n = 1}^{21} a_{n} - \sum_{n = 1}^{20} a_{n} = 7 - (- 3) = 10$ as well.
To re-index $\sum_{n = 2}^{437} n (n - 1) x^{n - 2}$ so $n$ starts at $0$ , we follow the formula in Theorem 10.2.2 with $r = - 2$ :
$\sum_{n = 2}^{437} n (n - 1) x^{n - 2} = \sum_{n = 2 + (- 2)}^{437 + (- 2)} (n - (- 2)) (n - (- 2) - 1) x^{n - (- 2) - 2} = \sum_{n = 0}^{435} (n + 2) (n + 1) x^{n} .$
We leave it to the reader to check by writing out the first few, and last few, terms.
Alternatively, to better see why the re-indexing works in this way, we can introduce a new counter, $k$ . We want this new counter to start at $k equals 0$ whereas the current counter starts at $n equals 2$ , so we want $k = n - 2$ . When $n equals 2$ , $k equals 0$ , as required, and when $n equals 437$ , $k equals 435$ .
Moreover, $n = k + 2$ , so substituting this into the sum, we get
$\sum_{n = 2}^{437} n (n - 1) x^{n - 2} = \sum_{k = 0}^{435} (k + 2) ((k + 2) - 1) x^{(k + 2) - 2} = \sum_{k = 0}^{435} (k + 2) (k + 1) x^{k},$
which is the same sum we had before, just with a different dummy variable.

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

$\begin{matrix} S & = & a & + & (a + d) & + & \dots & + & (a + (n - 2) d) & + & (a + (n - 1) d) \\ S & = & (a + (n - 1) d) & + & (a + (n - 2) d) & + & \dots & + & (a + d) & + & a \end{matrix}$

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

$2 S = (2 a + (n - 1) d) + (2 a + (n - 1) d) + \dots + (2 a + (n - 1) d) + (2 a + (n - 1) d)$

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

$S = \frac{n}{2} (2 a + (n - 1) d)$

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

$S = n (\frac{a_{1} + a_{n}}{2})$

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 $n^{th}$ terms.

To derive the formula for the geometric sum, we start with a geometric sequence $a_{k} = a r^{k - 1}$ , $k greater than or equal to 1$ , and let $S$ once again denote the sum of the first $n$ terms. Comparing $S$ and $r S$ , we get

$\begin{matrix} S & = & a & + & a r & + & a r^{2} & + & \dots & + & a r^{n - 2} & + & a r^{n - 1} \\ r S & = & a r & + & a r^{2} & + & \dots & + & a r^{n - 2} & + & a r^{n - 1} & + & a r^{n} \end{matrix}$

Subtracting the second equation from the first forces all of the terms except $a$ and $a r^{n}$ to cancel out and we get $S - r S = a - a r^{n}$ . Factoring, we get $S (1 - r) = a (1 - r^{n})$ . Assuming $r not equal to 1$ , we can divide both sides by the quantity $open parenthesis 1 minus r close parenthesis$ to obtain

$S = a (\frac{1 - r^{n}}{1 - r})$

If we distribute $a$ through the numerator, we get $a - a r^{n} = a_{1} - a_{n + 1}$ which yields the formula

$S = \frac{a_{1} - a_{n + 1}}{1 - r}$

In the case when $r equals 1$ , we get the formula

$S = \underset{n times}{\underset{⏟}{a + a + \dots + a}} = n a$

Our results are summarized below.⁵

Equation

Sums of Arithmetic and Geometric Sequences:

The sum $S$ of the first $n$ terms of an arithmetic sequence $a_{k} = a + (k - 1) d$ for $k greater than or equal to 1$ is
$S = \sum_{k = 1}^{n} a_{k} = n (\frac{a_{1} + a_{n}}{2}) = \frac{n}{2} (2 a + (n - 1) d)$
The sum $S$ of the first $n$ terms of a geometric sequence $a_{k} = a r^{k - 1}$ for $k greater than or equal to 1$ is
1. $S = \sum_{k = 1}^{n} a_{k} = \frac{a_{1} - a_{n + 1}}{1 - r} = a (\frac{1 - r^{n}}{1 - r})$ , if $r not equal to 1$ .
2. $S = \sum_{k = 1}^{n} a_{k} = \sum_{k = 1}^{n} a = n a$ , if $r equals 1$ .

While we have made an honest effort to derive the formulas in Equation 10.2, formal proofs require the machinery in Section 10.3.

Example 3

1. Find the sum: $1 + 3 + 5 + \dots + 117$
2. Find a formula for the sum $∑ k equals 1 n k$ .
The classic wheat and chessboard problem asks the following question. Given a chessboard with its squares numbered $1$ to $64$ , suppose on the first square was placed one grain of wheat, the second square, two grains, the third square, four grains, and so on, each square receiving twice the number of grains as its predecessor. How many total grains of wheat would end up on the chessboard?

1. Recognizing the terms of $1 + 3 + 5 + \dots + 117$ as $1$ , $3$ , $5$ , and so on, we see we have an arithmetic sequence with $a equals 1$ and $d equals 2$ . Using Equation 10.1, we get a formula for the terms $a_{n} = 1 + 2 (n - 1) = 2 n - 1$ for $n greater than or equal to 1$ . In order to use the formula in Equation 10.2, we need to determine the number of terms being added, $n$ . Setting $2 n - 1 = 117$ , we find $n equals 59$ . Feeding in all of our data into Equation 10.2, we get $1 + 3 + 5 + \dots + 117 = 59 (\frac{1 + 117}{2}) = 3481$ .
2. Applying the adage `when in doubt, write it out,' we have $\sum_{k = 1}^{n} k = 1 + 2 + 3 + 4 + \dots + n$ . We see the terms here form an arithmetic sequence with $a = d = 1$ . Moreover, we are adding exactly $n$ terms, so Equation 10.2 gives $\sum_{k = 1}^{n} k = 1 + 2 + 3 + 4 + \dots + n = \frac{n (n + 1)}{2}$ .
  As a side note, the special case: $1 + 2 + 3 + \dots + 100$ was allegedly given to Carl Friedrich Gauss while he was in elementary school. Instead of computing the sum in a brute force method, he arrived at the answer by grouping $1 + 99 = 100$ , $2 + 98 = 100$ , etc. so that he had $50$ groups of $100$ with $50$ left over for a total of $5050$ . This is the exact same methodology we used to prove the sum of the arithmetic sequence formula in Equation 10.2.
Since we are doubling the number of grains of wheat as we move from one square to the next, a geometric sequence with $r equals 2$ describes the number of grains on each individual square.
Since we start with one grain on the first square, the number of grains on the $k$ th square is $a_{k} = (1) {(2)}^{k - 1} = 2^{k - 1}$ for $k greater than or equal to 1$ .
Adding up the number of grains on each square gives:
$1 + 2 + \dots + 2^{64 - 1} = 1 + 2 + \dots 2^{63} = \frac{1 - 2^{64}}{1 - 2} = 2^{64} - 1 \approx 1.8 \times 10^{19},$
in accordance with Equation 10.2. (The weight of these grains would total approximately $2.6 \times 10^{15}$ pounds which is approximately $15$ times the entire biomass of the planet.)

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 7.6 in that payments are deposited into the account on an on-going basis, and this complicates the mathematics a little.⁶

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 $A sub 1 equals P$ , 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 $A sub 1$ so that our initial investment has grown to $A_{1} (1 + i) = P (1 + i)$ in accordance with Equation 7.1. Adding our second payment at the end of the second period, we get

$A_{2} = A_{1} (1 + i) + P = P (1 + i) + P = P (1 + i) (1 + \frac{1}{1 + i})$

The reason for factoring out the $P ⁢ open parenthesis 1 plus i close parenthesis$ will become apparent in short order. During the third compounding period, we earn interest on $A sub 2$ which then grows to $A sub 2 ⁢ open parenthesis 1 plus i close parenthesis$ . We add our third payment at the end of the third compounding period to obtain

$A_{3} = A_{2} (1 + i) + P = P (1 + i) (1 + \frac{1}{1 + i}) (1 + i) + P = P {(1 + i)}^{2} (1 + \frac{1}{1 + i} + \frac{1}{{(1 + i)}^{2}})$

During the fourth compounding period, $A sub 3$ grows to $A sub 3 ⁢ open parenthesis 1 plus i close parenthesis$ , and when we add the fourth payment, we factor out $P ⁢ open parenthesis 1 plus i close parenthesis cubed$ to get

$A_{4} = P {(1 + i)}^{3} (1 + \frac{1}{1 + i} + \frac{1}{{(1 + i)}^{2}} + \frac{1}{{(1 + i)}^{3}})$

This pattern continues so that at the end of the $k$ th compounding, we get

$A_{k} = P {(1 + i)}^{k - 1} (1 + \frac{1}{1 + i} + \frac{1}{{(1 + i)}^{2}} + \dots + \frac{1}{{(1 + i)}^{k - 1}})$

The sum in the parentheses above is the sum of the first $k$ terms of a geometric sequence with $a equals 1$ and $r = \frac{1}{1 + i}$ . Using Equation 10.2, we get

$1 + \frac{1}{1 + i} + \frac{1}{{(1 + i)}^{2}} + \dots + \frac{1}{{(1 + i)}^{k - 1}} = 1 (\frac{1 - \frac{1}{{(1 + i)}^{k}}}{1 - \frac{1}{1 + i}}) = \frac{(1 + i) (1 - {(1 + i)}^{- k})}{i}$

Hence, we get

$A_{k} = P {(1 + i)}^{k - 1} (\frac{(1 + i) (1 - {(1 + i)}^{- k})}{i}) = \frac{P ({(1 + i)}^{k} - 1)}{i}$

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

Equation

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

$A = \frac{P ({(1 + i)}^{n t} - 1)}{i}$

The reader is encouraged to substitute $i = \frac{r}{n}$ into Equation 10.3 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 4

An ordinary annuity offers a $6 %$ annual interest rate, compounded monthly.

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

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

Geometric Series

As defined in Section 10.1, sequences are an infinite list of numbers. So far in this section, we have concerned ourselves with adding only finitely many terms. In Calculus, infinite sums, called series are studied at great length. While we do not have the mathematical machinery to embark upon an exhaustive study here, we can nevertheless focus our attention on what is arguably one of the most prevalent and useful types of series, geometric series.

As a motivating example, consider the number $0.9 bar$ . We can write this number as

$0 . \bar{9} = 0.9999 \dots = 0.9 + 0.09 + 0.009 + 0.0009 + \dots$

From Example 10.2.1, we know we can write the sum of the first $n$ of these terms as

$0 . \underset{n nines}{\underset{⏟}{9 \dots 9}} = .9 + 0.09 + 0.009 + \dots 0 . \underset{n - 1 zeros}{\underset{⏟}{0 \dots 0}} 9 = \sum_{k = 1}^{n} \frac{9}{10^{k}}$

Using Equation 10.2, we have

$\sum_{k = 1}^{n} \frac{9}{10^{k}} = \sum_{k = 1}^{n} \frac{9}{10} (\frac{1}{10^{k - 1}}) = \sum_{k = 1}^{n} \frac{9}{10} {(\frac{1}{10})}^{k - 1} = \frac{9}{10} (\frac{1 - \frac{1}{10^{n}}}{1 - \frac{1}{10}}) = 1 - \frac{1}{10^{n}}$

It stands to reason that we should define $0.9 bar equals lim n → ∞$ $(1 - \frac{1}{10^{n}})$ . Passing to a continuous variable along with our knowledge of exponential functions gives $lim n → ∞$ $(1 - \frac{1}{10^{n}}) =$ $lim x → ∞$ $(1 - \frac{1}{10^{x}}) = 1 - 0 = 1$ .

We have just argued that $0.9 bar equals 1$ , which may shock some readers.⁷

Note that in this manner, 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 have the following theorem.

Theorem

Geometric Series: Given the sequence $a_{k} = a r^{k - 1}$ for $k greater than or equal to 1$ , where $vertical bar r vertical bar less than 1$ ,

$a + a r + a r^{2} + \dots = \sum_{k = 1}^{\infty} a r^{k - 1} = lim_{n \to \infty} \sum_{k = 1}^{n} a r^{k - 1} = \frac{a}{1 - r}$

If $vertical bar r vertical bar greater than or equal to 1$ , the sum $a + a r + a r^{2} + \dots$ does not exist.

The justification of the result in Theorem 10.5 comes from taking the formula in Equation 10.2 for the sum of the first $n$ terms of a geometric sequence and taking the limit as $n → ∞$ .

Assuming $vertical bar r vertical bar less than 1$ means $minus 1 less than r less than 1$ , so per Theorem 10.3, $lim n → ∞$ $r^{n} = 0$ . Using this fact along with the Limit Properties listed in Theorem 6.2:

$lim_{n \to \infty} \sum_{k = 1}^{n} a r^{k - 1} = lim_{n \to \infty} a (\frac{1 - r^{n}}{1 - r}) = \frac{a}{1 - r}$

We'll explore what goes wrong when $vertical bar r vertical bar greater than or equal to 1$ in some of the Exercises. For now, we put this theorem to good use in the following example.

Example 5

Find the sum: $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$ .
Represent $4.2 ⁢ 17 bar$ as a fraction in lowest terms.

We recognize $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$ as a geometric series with $a = r = \frac{1}{2}$ . Using Theorem 10.5, we get
$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1 .$
The interested reader is invited to research this sum as it relates to Zeno's Dichotomy Paradox .
To use Theorem 10.5 as it applies to the repeating decimal $4.2 ⁢ 17 bar$ , we first need to rewrite this decimal in terms of a geometric series. Expanding $4.2 \bar{17} = 4.2 + 0.017 + 0.00017 + 0.0000017 + \dots,$ we see the series $0.017 + 0.00017 + 0.0000017 + \dots$ is geometric with $a equals 0.017$ and $r equals 0.01$ . Hence, we can apply Theorem 10.5 to that part of the decimal to get:
$0.017 + 0.00017 + 0.0000017 + \dots = \frac{0.017}{1 - 0.01} = \frac{\frac{17}{1000}}{\frac{99}{100}} = \frac{17}{990}$
Hence, $4.2 \bar{17} = 4.2 + \frac{17}{990} = \frac{42}{10} + \frac{17}{990} = \frac{835}{198}$ .

We note that another popular method for converting repeating decimals to fractions goes something like this: let $x equals 4.2 ⁢ 17 bar$ . Then, $100 ⁢ x equals 421.7 ⁢ 17 bar$ . Hence, $99 x = 100 x - x = 421.7 \bar{17} - 4.2 \bar{17} = 417.5$ . Hence, $x = \frac{417.5}{99} = \frac{835}{198}$ . While this procedure results in the same (correct!) answer, the manipulations involved (such as the multiplication and subtraction) are actually using some of the properties listed in Theorem 10.4 extended to infinite sums via Theorem 6.2.

Area

One of the (two) major geometric problems studied in Calculus is finding the area under a curve⁸ (more specifically, the area between the graph of a function and the $x$ -axis.)⁹ In this section, we explore how summation notation is used to help better formulate this problem, and, as with our study of Geometric Series, sneak a peak into Calculus itself.

Suppose we wish to determine the area between the graph of a continuous function $y = f (x)$ over the interval $[a, b]$ and the $x$ -axis as shown below on the left. Since we don't know any area formulas for arbitrary regions, we stick to what we know - rectangles.

To keep things simple, we divide $[a, b]$ into $n$ equal pieces (subintervals), and use the right-endpoints of each piece to determine the height of the rectangles.¹⁰ We let $x_{k}$ represent the right endpoint of the $k$ th subinterval, so the height of the $k$ th rectangle is $f (x_{k})$ .

The width of the $k$ th rectangle is the length of the $k$ th subinterval. Since the interval itself is $b - a$ units long and we are dividing the interval into $n$ equal pieces, each piece is $\frac{b - a}{n}$ units long. For brevity, we'll call this length ` $Δ x$ .' Below on the right is a depiction of $R ⁢ S sub 7$ , a `right endpoint sum' using $7$ (equally spaced) subintervals.¹¹

Figure: Area under the graph of — **Figure 10.7** Area under the graph of $y = f (x)$

Figure: Visualizing , a `right endpoint sum.' — **Figure 10.8** Visualizing $R ⁢ S sub 7$ , a `right endpoint sum.'

The idea here is to approximate the area of the shaded region by the sum of the areas of the rectangles. In symbols:

$Area \approx f (x_{1}) Δ x + f (x_{2}) Δ x + f (x_{3}) Δ x + \dots + f (x_{7}) Δ x = \sum_{k = 1}^{7} f (x_{k}) Δ x$

Our ultimate goal is to find a formula for the area approximation as described above as a function of the number of rectangles $n$ and look to see what happens as $n → ∞$ .

We first note that the right endpoints $x_{k}$ , are terms in an arithmetic sequence: the first right endpoint, $x sub 1$ is $Δ x$ to the right of $a equals x sub 0$ , so $x_{1} = x_{0} + Δ x$ ; the second right endpoint, $x sub 2$ is $Δ x$ units to the right of $x sub 1$ , so $x_{2} = x_{1} + Δ x$ ; the third right endpoint $x_{3} = x_{2} + Δ x$ and so on. In general, $x_{k} = x_{k - 1} + Δ x$ , proving the $x_{k}$ are terms of an arithmetic sequence with common difference $d = Δ x$ . It follows that $x_{k}$ , the $k$ th right endpoint is $k Δ x$ units to the right of $x sub 0 equals a$ , so that $x_{k} = a + k Δ x$ . We summarize the notation and formulas for right endpoint sums below.

Summary of Formulas for Right Endpoint Sums, $R S_{n}$

Number of rectangles: $n$
Width of each rectangle: $Δ x = \frac{b - a}{n}$
Right endpoint: $x_{k} = a + k Δ x$
Height of $k$ th rectangle: $f (x_{k})$
$Area \approx R S_{n} = the sum of the area of the rectangles = \sum_{k = 1}^{n} f (x_{k}) Δ x_{k}$

Below we summarize some common summation formulas we'll need when actually computing these sums. Formal proofs of these require the machinery of Section 10.3 and are found there.

Summation Formulas

$\sum_{k = 1}^{n} c = c n$
$\sum_{k = 1}^{n} k = \frac{n (n + 1)}{2}$

$\sum_{k = 1}^{n} k^{2} = \frac{n (n + 1) (2 n + 1)}{6}$
$\sum_{k = 1}^{n} k^{3} = \frac{n^{2} {(n + 1)}^{2}}{4}$

It is high time for an example.

Example 6

Consider $f (x) = 4 x - x^{2}$ over the interval $open bracket 0, 4 close bracket$ .

Graph $f$ over this interval and shade the area between the graph of $f$ and the $x$ -axis.
Compute $R S_{n}$ for $n equals 4$ and $n equals 8$ . Interpret your results graphically.
Find a formula for $R S_{n}$ in terms of $n$ and determine $lim n → ∞ R ⁢ S sub n$ .

The graph of $f (x) = 4 x - x^{2}$ is a parabola with intercepts $open parenthesis 0, 0 close parenthesis$ and $open parenthesis 4, 0 close parenthesis$ with a vertex at $open parenthesis 2, 4 close parenthesis$ .
Figure 10.9 Area under the graph of $y = f (x)$
To find $R ⁢ S sub 4$ , we begin by chopping up the interval $open bracket 0, 4 close bracket$ into $4$ equal pieces so each subinterval has length $Δ x = \frac{4}{4} = 1$ unit. Our right endpoints are: $x sub 1 equals 1$ , $x sub 2 equals 2$ , $x sub 3 equals 3$ , and $x sub 4 equals 4$ . We find $f ⁢ open parenthesis 1 close parenthesis equals 3$ , $f ⁢ open parenthesis 2 close parenthesis equals 4$ , $f ⁢ open parenthesis 3 close parenthesis equals 3$ , and $f ⁢ open parenthesis 4 close parenthesis equals 0$ . Hence,
$R S_{4} = \sum_{k = 1}^{4} f (x_{k}) Δ x = f (x_{1}) Δ x + f (x_{2}) Δ x + f (x_{3}) Δ x + f (x_{4}) Δ x = (3) (1) + (4) (1) + 3 (1) + (0) (1) = 10 .$
Geometrically we have approximated the area under the graph of $f$ to be $10$ square units by the adding the areas of the shaded rectangles shaded below on the left. (Note that since $f (x_{4}) = f (4) = 0$ , the fourth `rectangle' has $0$ height.)
Figure 10.10 Visualizing $R ⁢ S sub 4$
Figure 10.11 Visualizing $R ⁢ S sub 8$
To find $R ⁢ S sub 8$ , we divide the interval $open bracket 0, 4 close bracket$ into $8$ equal pieces, so each has length $Δ x = \frac{4}{8} = 0.5$ units. This produces the right endpoints: $x sub 1 equals 0.5$ , $x sub 2 equals 1$ , $x sub 3 equals 1.5$ , $x sub 4 equals 2$ , $x sub 5 equals 2.5$ , $x sub 6 equals 3$ , $x sub 7 equals 3.5$ , $x sub 8 equals 4$ . In addition to the function values we used to compute $R ⁢ S sub 4$ , we need $f ⁢ open parenthesis 0.5 close parenthesis equals 1.75$ , $f ⁢ open parenthesis 1.5 close parenthesis equals 3.75$ , $f ⁢ open parenthesis 2.5 close parenthesis equals 3.75$ , and $f ⁢ open parenthesis 3.5 close parenthesis equals 1.75$ . Hence,
$\begin{matrix} R S_{8} & = & \sum_{k = 1}^{8} f (x_{k}) Δ x \\ = & f (x_{1}) Δ x + f (x_{2}) Δ x + f (x_{3}) Δ x + f (x_{4}) Δ x \\ + f (x_{5}) Δ x + f (x_{6}) Δ x + f (x_{7}) Δ x + f (x_{8}) Δ x \\ = & (1.75) (0.5) + (3) (0.5) + (3.75) (0.5) + (4) (0.5) \\ + (3.75) (0.5) + (3) (0.5) + (1.75) (0.5) + (0) (0.5) \\ = & 10.5 \end{matrix}$
Hence, the area under the graph $f$ is approximately $10.5$ square units as approximated by the sum of the rectangles above on the right. (Again, since $f (x_{8}) = f (4) = 0$ , the eighth `rectangle' has $0$ height.)
To find a formula for $R S_{n}$ , we imagine dividing the interval $open bracket 0, 4 close bracket$ into $n$ equal pieces each of length $Δ x = \frac{4}{n}$ . We have $n$ right endpoints, $x sub 1$ , $x sub 2$ , … $x_{n}$ where $x_{k} = 0 + k Δ x = \frac{4 k}{n}$ . Since $f (x) = 4 x - x^{2}$ ,
$f (x_{k}) = 4 x_{k} - x_{k}^{2} = 4 (\frac{4 k}{n}) - {(\frac{4 k}{n})}^{2} = \frac{16 k}{n} - \frac{16 k^{2}}{n^{2}} .$
Hence,

Note: the counter here is ` $k$ ,' not ` $n$ ,' so as far as $k$ is concerned, ` $n$ ' is a constant so we can factor it out of the summation.
Note we can partially check our answer at thus point by substituting $n equals 4$ and $n equals 8$ to our formula to $R S_{n}$ to see if we recover our answers from above. We get $R S_{4} = \frac{32 {(4)}^{2} - 32}{3 (4^{2})} = \frac{480}{48} = 10$ and $R S_{8} = \frac{32 {(8)}^{2} - 32}{3 {(8)}^{2}} = \frac{2016}{192} = 10.5$ , as required.
To find $lim n → ∞ R ⁢ S sub n equals lim n → ∞$ $\frac{32 n^{2} - 32}{3 n^{2}}$ , we compare the leading term of the numerator and denominator. As $n → ∞$ , $\frac{32 n^{2} - 32}{3 n^{2}} \approx \frac{32 n^{2}}{3 n^{2}} = \frac{32}{3}$ . Hence, $lim n → ∞ R ⁢ S sub n equals$ $\frac{32}{3}$ . Hence, as we use more and more rectangles,¹² the sum total of the area of those rectangles approaches $\frac{32}{3}$ square units. In Calculus, we more or less define the area under $f$ to be $\frac{32}{3}$ square units.

It is worth noting that, as with other examples in the text, Example 10.2.6 is more or less lifted straight out of a Calculus lecture. That being said, the vast majority of the mechanics here involve precalculus notions.¹³ In general, the machinations in Calculus amount to applying the limit concept to the mechanics of precalculus.

Exercises

In Exercises -, find the value of each sum using Definition 10.4.

$\sum_{g = 4}^{9} (5 g + 3)$
$\sum_{k = 3}^{8} \frac{1}{k}$
$\sum_{j = 0}^{5} 2^{j}$
$\sum_{k = 0}^{2} (3 k - 5) x^{k}$
$\sum_{i = 1}^{4} \frac{1}{4} (i^{2} + 1)$
$\sum_{n = 1}^{100} {(- 1)}^{n}$
$\sum_{n = 1}^{5} \frac{(n + 1)!}{n!}$
$\sum_{j = 1}^{3} \frac{5!}{j! (5 - j)!}$

In Exercises -, rewrite the sum using summation notation.

$8 + 11 + 14 + 17 + 20$
$1 - 2 + 3 - 4 + 5 - 6 + 7 - 8$
$x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7}$
$1 + 2 + 4 + \dots + 2^{29}$
$2 + \frac{3}{2} + \frac{4}{3} + \frac{5}{4} + \frac{6}{5}$
$- \ln (3) + \ln (4) - \ln (5) + \dots + \ln (20)$
$1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16} + \frac{1}{25} - \frac{1}{36}$
$\frac{1}{2} (x - 5) + \frac{1}{4} {(x - 5)}^{2} + \frac{1}{6} {(x - 5)}^{3} + \frac{1}{8} {(x - 5)}^{4}$

In Exercises -, use the formulas in Equation 10.2 to find the sum.

$\sum_{n = 1}^{10} 5 n + 3$
$\sum_{n = 1}^{20} 2 n - 1$
$\sum_{k = 0}^{15} 3 - k$
$\sum_{n = 1}^{10} {(\frac{1}{2})}^{n}$
$\sum_{n = 1}^{5} {(\frac{3}{2})}^{n}$
$\sum_{k = 0}^{5} 2 {(\frac{1}{4})}^{k}$
$1 + 4 + 7 + \dots + 295$
$4 + 2 + 0 - 2 - \dots - 146$
$1 + 3 + 9 + \dots + 2187$
$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{256}$
$3 - \frac{3}{2} + \frac{3}{4} - \frac{3}{8} + - \dots + \frac{3}{256}$
$\sum_{n = 1}^{10} - 2 n + {(\frac{5}{3})}^{n}$

In Exercises -, use Theorem 10.5 to find the sum of the given geometric series.¹⁴

$\sum_{n = 1}^{\infty} {(\frac{1}{2})}^{n - 1}$
$\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} 3^{n - 1}}{4^{n}}$
$\sum_{m = 2}^{\infty} \frac{3}{2^{m - 1}}$
$\sum_{k = 0}^{\infty} x^{k}$ , $vertical bar x vertical bar less than 1$ .

In Exercises -, use Theorem 10.5 to express each repeating decimal as a fraction of integers.

$0.7 bar$
$0.13 bar$
$10.159 bar$
$minus 5.8 ⁢ 67 bar$

In Exercises -, use Equation 10.3 to compute the future value of the annuity with the given terms. In all cases, assume the payment is made monthly, the interest rate given is the annual rate, and interest is compounded monthly.

payments are $300, interest rate is 2.5%, term is 17 years.
payments are $50, interest rate is 1.0%, term is 30 years.
payments are $100, interest rate is 2.0%, term is 20 years
payments are $100, interest rate is 2.0%, term is 25 years
payments are $100, interest rate is 2.0%, term is 30 years
payments are $100, interest rate is 2.0%, term is 35 years
Suppose an ordinary annuity offers an annual interest rate of $2 %$ , compounded monthly, for 30 years. What should the monthly payment be to have $$ 100, 000$ at the end of the term?
In this exercise, we Theorem 10.5 to represent $f (x) = \frac{1}{x^{2} + 4}$ as a series.
1. Show that $f (x) = \frac{\frac{1}{4}}{1 - (- \frac{x^{2}}{4})}$ .
2. Use the formula in Theorem 10.5: $\frac{a}{1 - r} = \sum_{k = 1}^{\infty} a r^{k - 1}$ to write $f (x)$ as an infinite series.
3. Graph $y = f (x)$ along with some partial sums of the series. What do you notice?
Using Example 10.2.6 as a guide, find and simplify formula for the right endpoint sum, $R S_{n}$ , for each of the functions below on the specified interval. Find $lim n → ∞ R ⁢ S sub n$ to find the area between the graph of $f$ and the $x$ -axis.
1. $f (x) = 4 - x$ over the interval $open bracket 0, 4 close bracket$ .
2. $f ⁢ open parenthesis x close parenthesis equals 3 ⁢ x squared$ over the interval $open bracket 1, 3 close bracket$ .
3. $f (x) = 12 - x - x^{2}$ over the interval $open bracket 0, 3 close bracket$ .
Prove the properties listed in Theorem 10.4.
Show that the formula for the future value of an annuity due is
$A = P (1 + i) [\frac{{(1 + i)}^{n t} - 1}{i}]$
Discuss with your classmates what goes wrong when trying to find the following sums.¹⁵
1. $\sum_{k = 1}^{\infty} 2^{k - 1}$
2. $\sum_{k = 1}^{\infty} {(1.0001)}^{k - 1}$
3. $\sum_{k = 1}^{\infty} {(- 1)}^{k - 1}$
In this exercise, we walk through the proof of Cauchy's Bound, Theorem 2.10 in Section 2.3.
Let $f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}$ be a polynomial of degree $n$ and let $Z$ be the largest zero of $f$ in absolute value and let $M$ be the largest of the numbers: $\frac{| a_{0} |}{| a_{n} |}$ , $\frac{| a_{1} |}{| a_{n} |}$ , …, $\frac{| a_{n - 1} |}{| a_{n} |}$ .
1. Since $P ⁢ open parenthesis Z close parenthesis equals 0$ , solve for $Z^{n}$ : $Z^{n} = \frac{a_{n - 1}}{a_{n}} Z^{n - 1} + \dots + \frac{a_{1}}{a_{n}} Z + \frac{a_{0}}{a_{n}}$ .
2. If $minus 1 less than or equal to Z less than or equal to 1$ , then Cauchy's Bound is immediately satisfied since $Z$ would automatically lie in the interval $[- (M + 1), M + 1]$ . So we assume $vertical bar Z vertical bar greater than 1$ .
  Under the assumption $vertical bar Z vertical bar greater than 1$ . explain why¹⁶
  $\begin{matrix} {| Z |}^{n} & = & | \frac{a_{n - 1}}{a_{n}} Z^{n - 1} + \dots + \frac{a_{1}}{a_{n}} Z + \frac{a_{0}}{a_{n}} | \\ \leq & \frac{| a_{n - 1} |}{| a_{n} |} {| Z |}^{n - 1} + \dots + \frac{| a_{1} |}{| a_{n} |} | Z | + \frac{| a_{0} |}{| a_{n} |} \end{matrix}$
3. Use the definition of $M$ along with the Geometric Sum Formula, Equation 10.2 to show:
  ${| Z |}^{n} \leq M ({| Z |}^{n - 1} + \dots + | Z | + 1) = M \frac{1 - {| Z |}^{n}}{1 - | Z |} = M \frac{{| Z |}^{n} - 1}{| Z | - 1}$
4. Now use the fact that $vertical bar Z vertical bar greater than 1$ to rearrange the above inequality to get:
  $| Z | - 1 \leq M \frac{{| Z |}^{n} - 1}{{| Z |}^{n}} = M (1 - \frac{1}{{| Z |}^{n}})$
5. Use the fact that $1 - \frac{1}{{| Z |}^{n}} < 1$ to get:
  $| Z | - 1 \leq M (1 - \frac{1}{{| Z |}^{n}}) < M (1) = M$
6. From $vertical bar Z vertical bar minus 1 less than M$ , we get $vertical bar Z vertical bar less than M plus 1$ . Hence, $Z$ lies in the interval $[- (M + 1), M + 1]$ .

Answers

$213$
$\frac{341}{280}$
$63$
$- 5 - 2 x + x^{2}$
$\frac{17}{2}$
$0$
$20$
$25$
$\sum_{k = 1}^{5} (3 k + 5)$
$\sum_{k = 1}^{8} {(- 1)}^{k - 1} k$
$\sum_{k = 1}^{4} {(- 1)}^{k - 1} \frac{x^{2 k - 1}}{2 k - 1}$
$\sum_{k = 1}^{30} 2^{k - 1}$
$\sum_{k = 1}^{5} \frac{k + 1}{k}$
$\sum_{k = 3}^{20} {(- 1)}^{k} \ln (k)$
$\sum_{k = 1}^{6} \frac{{(- 1)}^{k - 1}}{k^{2}}$
$\sum_{k = 1}^{4} \frac{1}{2 k} {(x - 5)}^{k}$
$305$
$400$
$minus 72$
$\frac{1023}{1024}$
$\frac{633}{32}$
$\frac{1365}{512}$
$14652$
$minus 5396$
$3280$
$\frac{255}{256}$
$\frac{513}{256}$
$\frac{17771050}{59049}$
$\sum_{n = 1}^{\infty} {(\frac{1}{2})}^{n - 1} = 2$
$\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} 3^{n - 1}}{4^{n}} = \frac{4}{21}$
$\sum_{m = 2}^{\infty} \frac{3}{2^{m - 1}} = 3$
$\sum_{k = 0}^{\infty} x^{k} = \frac{1}{1 - x}$
$\frac{7}{9}$
$\frac{13}{99}$
$\frac{3383}{333}$
$- \frac{5809}{990}$
$76,163.67
$$ 20, 981.40$
$$ 29, 479.69$
$$ 38, 882.12$
$49, 272.55$
$60, 754.80$
For $$ 100, 000$ , the monthly payment is $approximately equals $ 202.95$ .
1. $f (x) = \frac{\frac{1}{4}}{1 - (- \frac{x^{2}}{4})} = \sum_{k = 1}^{\infty} \frac{1}{4} {(- \frac{x^{2}}{4})}^{k - 1} = \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k - 1} x^{2 k - 2}}{4^{k}}$
2. No matter how many terms are added, the graph of the series seems to only account for a portion of the graph of $y = f (x)$ . This is due to the fact that geometric series converge only when the ratio $vertical bar r vertical bar less than 1$ . In this case, $r = - \frac{x^{2}}{4}$ so $vertical bar r vertical bar less than 1$ corresponds to the interval $open parenthesis minus 2, 2 close parenthesis$ .
Using Example 10.2.6 as a guide, find the area between the graph of each function below and the $x$ -axis by evaluating the limit of a right endpoint sum.
1. $R S_{n} = 8 - \frac{8}{n}$ ; Area is 8 ${units}^{2}$
2. $R S_{n} = 26 + \frac{24}{n} + \frac{4}{n^{2}}$ ; Area is 26 ${units}^{2}$
3. $R S_{n} = \frac{45}{2} - \frac{18}{n} - \frac{9}{2 n^{2}}$ ; Area is $\frac{45}{2}$ ${units}^{2}$

This is indeed a geometric sequence with first term $a equals 1$ and common ratio $r = - 1$ . ↩
It is an arithmetic sequence with first term $a equals 1$ and common difference $d equals 1$ . ↩
To see why, try writing the summation using ` $n$ ' as the index. ↩
To get started, remember the mantra “When in doubt, write it out!” ↩
Alternatively, we can use Exercise in Section 2.2. ↩
The reader may wish to re-read the discussion on compound interest in Section 7.6 before proceeding. ↩
To make this more palatable, it is usually accepted that $0 . \bar{3} = \frac{1}{3}$ so that $0 . \bar{9} = 3 (0 . \bar{3}) = 3 (\frac{1}{3}) = 1$ . ↩
The other is the concept of tangent lines which we touch on (awesome pun!) in Section 6.2. ↩
The area can actually represent a wide variety of things such as displacement, probability, or, as odd as it sounds, volume. ↩
In Calculus, you'll also use left endpoints and midpoints …. ↩
On intervals over which the function is increasing, we find the area of rectangles overestimates the area we want; on intervals over which the function is decreasing, we find the area of the rectangles underestimates the area we want. ↩
even though they become skinnier and skinner and hence, individually have smaller and smaller areas … ↩
The only Calculus bit is the limit concept which I guess is precalculus now … ↩
Remember, when in doubt … ↩
When in doubt … ↩
Feel free to use the Triangle Inequality , as needed. See Exercise in Section 1.3. ↩

Adapted from Precalculus, Preliminary 4th Edition (integrated calculus), by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.