Reading Indexed Terms and Shifting the Index, foundations for calculus lesson

A list like $1, -\frac{1}{2}, \frac{1}{3}, -\frac{1}{4}, \ldots$ goes on forever, so nobody writes it out. Instead you write one formula with a counter in it and say where the counter starts. Series, power series and the ratio test in calculus all rest on this habit, and on one small move that causes many errors, writing the next term by putting $n + 1$ in place of $n$ .

Key terms

Index and indexed term

Definition Let $m$ be an integer. An indexed list assigns to each integer $n \ge m$ (or to each integer $n$ with $m \le n \le M$ for a finite list) exactly one number, written $a_n$ . The integer $n$ is the index, and $a_n$ is the indexed term, or the term with index $n$ . The index takes only integer values, one after another, never values like $\frac{1}{2}$ or $\sqrt{2}$ .

In plain words The index is a counter that says which term you are looking at. The term is the number sitting at that position.

Example Take $a_n = n^2 - 1$ for integers $n \ge 1$ .

Index $n$	$1$	$2$	$3$	$4$	$5$
Term $a_n$	$0$	$3$	$8$	$15$	$24$

The term with index $4$ is $a_4 = 15$ . There is no term with index $2.5$ , because the index is always an integer.

Subscript

Definition The subscript in $a_n$ is a label for the position, not a number that takes part in arithmetic with the term. Whatever expression sits in the subscript is the index of the term being named. So $a_{n+1}$ is the term at index $n + 1$ , while $a_n + 1$ is the term at index $n$ with $1$ added to its value. In the same way $a_{2n}$ is the term at index $2n$ , which is usually different from $2a_n$ .

In plain words What is inside the subscript tells you where to look. What is outside it tells you what to do with the number you found.

Example With $a_n = n^2 - 1$ and $n = 3$ , the two readings give different numbers.

\begin{gathered} a_{n+1} = a_4 = 4^2 - 1 = 15, \\[0.4em] a_n + 1 = a_3 + 1 = 8 + 1 = 9. \end{gathered}

Similarly $a_{2 \cdot 3} = a_6 = 35$ , while $2a_3 = 16$ .

Starting index

Definition The starting index is the smallest index $m$ in the stated set of indices, and $a_m$ is the first term. It is part of the description of the list, so it must be stated, and the first term is the term formula evaluated at $n = m$ . A starting index of $0$ only means the first term is labelled $a_0$ , and $a_0$ need not be $0$ . A finite list with indices $m, m + 1, \ldots, M$ has $M - m + 1$ terms.

In plain words The starting index tells you where counting begins. Starting at zero changes the labels, not the values.

Example Let $b_n = 3 \cdot 2^n$ for integers $n \ge 0$ . The first term is $b_0 = 3 \cdot 2^0 = 3$ , not $0$ . The first four terms are

\begin{gathered} b_0 = 3, \quad b_1 = 6, \\[0.4em] b_2 = 12, \quad b_3 = 24, \end{gathered}

so the fourth term carries the label $b_3$ . A list running from $b_0$ to $b_{10}$ has $10 - 0 + 1 = 11$ terms.

Term formula

Definition A term formula is an expression in the index $n$ , together with a stated set of integer indices, such that substituting each allowed $n$ gives the term $a_n$ . The formula must make sense, as a real number, at every allowed index. Any index where a denominator would be zero or an even root would get a negative radicand has to be left out of the set.

In plain words The formula is a machine that takes a position and returns the term there, and the stated starting index tells you which positions the machine is allowed to take.

Example The formula

c_n = \frac{n + 2}{n - 1}

has a zero denominator at $n = 1$ , so $1$ cannot be an index. The formula does give a number at $n = 0$ , but a list that runs through every integer from its start onward without reaching $1$ has to start at $2$ or later. With $n \ge 2$ the first terms are $c_2 = 4$ , $c_3 = \frac{5}{2}$ , $c_4 = 2$ and $c_5 = \frac{7}{4}$ .

Shifted index

Definition Suppose $a_n$ is given by a term formula for integers $n \ge m$ . The shifted term $a_{n+1}$ is found by replacing every occurrence of $n$ in the formula by $(n + 1)$ , keeping the parentheses until you simplify. It is defined for the integers $n$ with $n + 1 \ge m$ , that is $n \ge m - 1$ . In the same way $a_{n-1}$ comes from replacing every $n$ by $(n - 1)$ and is defined for $n \ge m + 1$ .

In plain words Every $n$ in the formula moves together. If even one copy is left behind, you have written a different formula, not the next term.

Example Let $a_n = \dfrac{n}{3n - 1}$ for $n \ge 1$ . Replacing both copies of $n$ by $(n + 1)$ gives

\begin{aligned} a_{n+1} &= \frac{(n + 1)}{3(n + 1) - 1} \\[0.4em] &= \frac{n + 1}{3n + 2}, \qquad n \ge 0. \end{aligned}

At $n = 1$ this gives $\frac{2}{5}$ , which matches $a_2 = \frac{2}{3 \cdot 2 - 1} = \frac{2}{5}$ . Two tempting mistakes are $\frac{n + 1}{3n - 1}$ , where the denominator was never shifted, and $\frac{n + 1}{3n}$ , where $1$ was added to $3n$ instead of to $n$ inside it. Powers need the same care. Shifting $n^2$ gives $(n + 1)^2$ , not $n^2 + 1$ , and shifting $2^n$ gives $2^{n+1}$ , not $2^n + 1$ . Expanded, these are

\begin{gathered} (n + 1)^2 = n^2 + 2n + 1 \quad \text{and} \\[0.4em] 2^{n+1} = 2 \cdot 2^n. \end{gathered}

Shifting down works the same way. Replacing $n$ by $(n - 1)$ gives $a_{n-1} = \frac{n - 1}{3n - 4}$ for $n \ge 2$ .

Alternating sign

Definition For an integer $k$ ,

(-1)^k = \begin{cases} 1 & \text{if } k \text{ is even,} \\ -1 & \text{if } k \text{ is odd.} \end{cases}

A factor $(-1)^n$ or $(-1)^{n+1}$ in a term formula therefore flips the sign from each term to the next. Since $(-1)^{n+1} = -(-1)^n$ , the two choices give opposite signs at every index. The parentheses matter, because $-1^n$ means $-(1^n) = -1$ for every $n$ .

In plain words The power of $-1$ is a sign switch that flips at every step, and the starting index decides whether the list opens with a plus or a minus.

Example Let $d_n = \dfrac{(-1)^n}{n + 1}$ for $n \ge 0$ .

Index $n$	$0$	$1$	$2$	$3$
Sign $(-1)^n$	$1$	$-1$	$1$	$-1$
Term $d_n$	$1$	$-\frac{1}{2}$	$\frac{1}{3}$	$-\frac{1}{4}$

The sign and the denominator change separately, but the term is their product. To write the same list with indices starting at $1$ , shift the whole formula, which gives $\frac{(-1)^{n-1}}{n}$ for $n \ge 1$ , and $(-1)^{n-1} = (-1)^{n+1}$ .

Consecutive-term ratio

Definition The consecutive-term ratio of an indexed list is

\frac{a_{n+1}}{a_n},

defined for each index $n$ such that $n$ and $n + 1$ are both allowed indices and $a_n \ne 0$ . To simplify it, write out $a_{n+1}$ by the shift above, then divide by multiplying by the reciprocal of $a_n$ and cancel common nonzero factors.

In plain words The ratio says by what factor you multiply one term to get the next. For a geometric list it is the same number every time, and for most other lists it depends on $n$ .

Example Let $a_n = \dfrac{(-1)^n n}{3^n}$ for $n \ge 1$ . All these terms are nonzero, and

\begin{aligned} \frac{a_{n+1}}{a_n} &= \frac{(-1)^{n+1}(n + 1)}{3^{n+1}} \cdot \frac{3^n}{(-1)^n n} \\[0.4em] &= -\frac{n + 1}{3n}, \qquad n \ge 1. \end{aligned}

Here $\frac{(-1)^{n+1}}{(-1)^n} = -1$ and $\frac{3^n}{3^{n+1}} = \frac{1}{3}$ . As a check, $a_1 = -\frac{1}{3}$ and $a_2 = \frac{2}{9}$ , and their ratio is $-\frac{2}{3}$ , which matches the formula at $n = 1$ . The minus sign records the alternation. The ratio test in calculus looks at the absolute value $\frac{n + 1}{3n}$ for large $n$ . For the geometric list $b_n = 3 \cdot 2^n$ the same work gives the constant ratio $2$ .

Check yourself

Let $a_n = \dfrac{2n - 1}{n + 2}$ for $n \ge 0$ . Find the first term, write $a_{n+1}$ and $a_n + 1$ as single fractions, and evaluate both at $n = 2$ .

Show the answer

The first term is $a_0 = -\frac{1}{2}$ . Replacing every $n$ by $(n + 1)$ gives the first expression, and adding $1$ to the current term gives the second.

\begin{aligned} a_{n+1} &= \frac{2(n + 1) - 1}{(n + 1) + 2} = \frac{2n + 1}{n + 3}, \\[0.4em] a_n + 1 &= \frac{3n + 1}{n + 2}. \end{aligned}

At $n = 2$ the first is $a_3 = 1$ and the second is $\frac{3}{4} + 1 = \frac{7}{4}$ .

Write a term formula, with its starting index, for the list $1, -\frac{1}{4}, \frac{1}{9}, -\frac{1}{16}, \ldots$ that starts at index $1$ .

Show the answer

One formula is $a_n = \dfrac{(-1)^{n+1}}{n^2}$ for integers $n \ge 1$ . At $n = 1$ the sign factor is $(-1)^2 = 1$ , so the list opens with $+1$ , and each later term flips sign while the denominators run through the squares.

Simplify the consecutive-term ratio of $a_n = \dfrac{(-1)^n 2^n}{n^2 + 1}$ for $n \ge 0$ .

Show the answer

Every term is nonzero, and for $n \ge 0$

\begin{aligned} \frac{a_{n+1}}{a_n} &= \frac{(-1)^{n+1} 2^{n+1}}{(n + 1)^2 + 1} \cdot \frac{n^2 + 1}{(-1)^n 2^n} \\[0.4em] &= -\frac{2(n^2 + 1)}{n^2 + 2n + 2}. \end{aligned}

At $n = 0$ this gives $-1$ , and indeed $a_0 = 1$ and $a_1 = -1$ .

Reading Indexed Terms and Shifting the Index

Key terms

Index and indexed term

Subscript

Starting index

Term formula

Shifted index

Alternating sign

Consecutive-term ratio

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus