Sequences and Notation

Learning goals

Define a sequence as a function on the positive integers
Separate the index from the term it names
Contrast an explicit rule with a recursive one
Supply enough seeds for a recursion that looks back $k$ steps
Read $\sum_{k=m}^{p} a_k$ with both limits included
Treat the summation index as a dummy name

A sequence is a function on the positive integers

Definition. A sequence is a function whose domain is the set of positive integers $\{1, 2, 3, \ldots\}$ . (Some sequences are more natural indexed from $0$ . When that happens we say so, and the domain is $\{0, 1, 2, \ldots\}$ instead. The idea is unchanged either way, because the inputs are still consecutive integers running upward from a stated place, and unless a lesson says otherwise that place is $1$ .) If the function is named $a$ , we write its output at the input $n$ as $a_n$ rather than $a(n)$ , and we call $a_n$ the $n$ th term of the sequence.

The only genuinely new thing in that definition is the subscript. Where you used to write $f(3)$ you now write $a_3$ , and where you used to write $f(n) = n^2 + 1$ you now write $a_n = n^2 + 1$ . The parentheses became a subscript and nothing else moved:

a_n \;=\; a(n).

The input $n$ is called the index. Because the domain is the positive integers, the inputs arrive in a natural order, $1$ then $2$ then $3$ . That natural order is exactly why a sequence can be written as a list when a general function cannot. The list is not a second object sitting beside the function; it is the function’s output table, read from left to right.

A sequence is a function. Feed it an index (1, 2, 3, and so on) and it hands back a term. The only new thing is the subscript, which is what you would otherwise write as a(n).

Two habits are worth fixing right away. First, the index and the term are different numbers: above, $a_3 = 10$ , so the index is $3$ while the term is $10$ . Second, order is part of the data. The set $\{2, 5, 10\}$ does not care which element you name first, but the sequence $2, 5, 10$ does, because every term is pinned to a particular index.

Since a sequence is a function, it has a graph: plot the point $(n, a_n)$ for each index $n$ . But only positive integers are legal inputs, so the graph is a scatter of separate dots and never a curve. There is no point above $n = 2.5$ , because $2.5$ is outside the domain, which means $a_{2.5}$ names nothing at all.

The graph of a sequence is a scatter of separate dots, one for each index. Nothing sits above 2.5, because 2.5 is not a positive integer, so joining the dots into a curve would invent points the sequence does not have.

Terms, indices, and what a list does not tell you

Some sequences stop. A finite sequence is a function on $\{1, 2, \ldots, k\}$ for some positive integer $k$ , so it has exactly $k$ terms. An infinite sequence is a function on all of the positive integers, so it never runs out. When someone writes $3, 7, 11, 15$ they usually mean a finite sequence with four terms, and the trailing dots in $3, 7, 11, 15, \ldots$ signal an infinite one.

As the definition promised, indexing does not have to begin at $1$ . Some sequences are more natural starting from $0$ (a bank balance before any interest has been added, say), and then the domain is $\{0, 1, 2, \ldots\}$ and the first term is $a_0$ . The idea survives intact, because the inputs are still consecutive integers running upward from a stated place. What you may never do is leave that place unstated. That is because $a_3$ is the fourth term when the count begins at $0$ , and the third term when it begins at $1$ . Every sequence in this lesson starts at $n = 1$ unless it says otherwise.

Now a warning that will save you a lot of grief: a few terms never determine a sequence. Shown $2, 4, 6, 8, \ldots$ you would probably answer “the even numbers”, $a_n = 2n$ . But the rule

a_n = 2n + (n-1)(n-2)(n-3)(n-4)

kills the product term at $n = 1, 2, 3, 4$ and so produces those same four terms, and then it hands you $a_5 = 10 + 24 = 34$ . Both rules are perfectly good sequences. A pattern you spot in a handful of terms is a guess, sometimes an excellent guess, but the sequence is whatever the rule says it is. That is why the rest of this lesson is about stating a rule precisely. That is also why a puzzle that shows four terms and asks “what comes next” has, strictly speaking, many right answers.

Check your understanding

For the sequence $a_n = n^2 - 3n$ , what is $a_5$ ?

4

-10

10

5

Explicit rules and recursive rules

There are two standard ways to pin a sequence down, and they trade against each other.

An explicit rule (also called a closed form) gives $a_n$ directly as an expression in $n$ :

a_n = n^2 + 1.

Want the $500$ th term? Substitute, and you are done: $a_{500} = 500^2 + 1 = 250{,}001$ . You never had to look at any other term.

A recursive rule instead gives one or more starting values, called seeds, and then a rule that builds each new term out of earlier ones:

a_1 = 2, \qquad a_n = a_{n-1} + 2n - 1 \quad \text{for } n \ge 2.

Run it: $a_2 = 2 + 3 = 5$ , then $a_3 = 5 + 5 = 10$ , then $a_4 = 10 + 7 = 17$ . Those are the same terms the explicit rule produced, which is not a coincidence, and we will prove it in a moment.

The seed is not decoration. Delete it and you no longer have a sequence at all. The bare rule $a_n = a_{n-1} + 2n - 1$ is obeyed by $2, 5, 10, 17, \ldots$ , and equally by $0, 3, 8, 15, \ldots$ , and by infinitely many other sequences, one for every choice of $a_1$ . The rule tells you how to take a step; the seed tells you where you are standing when you take the first one. Leaving it out is the most common error in this chapter.

Some recursions look back further than one step and therefore need more than one seed. To compute $a_3$ from the rule $a_n = a_{n-1} + a_{n-2}$ you need both $a_2$ and $a_1$ , so both have to be handed to you. The famous choice $a_1 = 1$ and $a_2 = 1$ produces $1, 1, 2, 3, 5, 8, 13, \ldots$ , the Fibonacci sequence. In general, a recursion that reaches back $k$ steps needs $k$ seeds, since the rule alone can never produce the first $k$ terms.

Worked example 1 List the first four terms of $a_n = \dfrac{(-1)^n (n+1)}{2n}$

Substitute $n = 1, 2, 3, 4$ in turn, and let the factor $(-1)^n$ do the sign work: it is $-1$ at odd indices and $+1$ at even ones.

At $n = 1$ and $n = 2$ ,

a_1 = \frac{(-1)^1 (2)}{2} = -1, \qquad a_2 = \frac{(-1)^2 (3)}{4} = \frac{3}{4}.

At $n = 3$ and $n = 4$ , reduce each fraction as you go:

a_3 = \frac{(-1)^3 (4)}{6} = -\frac{4}{6} = -\frac{2}{3}, \qquad a_4 = \frac{(-1)^4 (5)}{8} = \frac{5}{8}.

So the sequence opens $-1, \ \frac{3}{4}, \ -\frac{2}{3}, \ \frac{5}{8}, \ldots$ A factor of $(-1)^n$ is the standard way to make a sequence alternate in sign, and it is worth recognising on sight.

Worked example 2 Find $a_5$ for the recursion $a_1 = 2$ , $a_2 = 5$ , $a_n = a_{n-1} + 2a_{n-2}$

This rule reaches back two steps, so it comes with two seeds, and you cannot jump straight to $a_5$ . Build the terms in order.

The rule at $n = 3$ uses the two seeds:

a_3 = a_2 + 2a_1 = 5 + 2(2) = 9.

Now $a_3$ and $a_2$ give $a_4$ , and then $a_4$ and $a_3$ give $a_5$ :

a_4 = a_3 + 2a_2 = 9 + 2(5) = 19, \qquad a_5 = a_4 + 2a_3 = 19 + 2(9) = 37.

The sequence is $2, 5, 9, 19, 37, \ldots$ Notice what the answer cost: three intermediate terms for one value. Reaching $a_{100}$ this way would mean computing every one of the $98$ terms in between, which is exactly the weakness an explicit rule fixes.

The explicit rule $a_n = n^2 + 1$ and the recursion $a_1 = 2$ , $a_n = a_{n-1} + 2n - 1$ define the same sequence#

Two sequences are the same function when they have the same domain and agree at every index. So what we must show is that these two rules produce the same term at every $n$ .

Call the explicit sequence $a$ , so that $a_n = n^2 + 1$ for every positive integer $n$ , and call the recursive one $b$ , so that $b_1 = 2$ and $b_n = b_{n-1} + 2n - 1$ for every $n \ge 2$ .

Begin at the bottom of the ladder. The explicit rule gives $a_1 = 1^2 + 1 = 2$ , and the seed gives $b_1 = 2$ , so the two agree at $n = 1$ .

Next, check that $a$ obeys the very step rule that $b$ was built from. For any $n \ge 2$ ,

a_n - a_{n-1} = (n^2 + 1) - \big((n-1)^2 + 1\big) = n^2 - (n^2 - 2n + 1) = 2n - 1,

and rearranging gives $a_n = a_{n-1} + 2n - 1$ , which is exactly the rule $b$ follows.

Now climb. Suppose the two sequences agree at some index $n - 1$ , so $a_{n-1} = b_{n-1}$ . Each one then adds the same quantity $2n - 1$ to that same value to produce its next term, so $a_n = b_n$ . Agreement at any index therefore forces agreement at the next index. We already have agreement at $n = 1$ , so it propagates to $n = 2$ , and from there to $n = 3$ , and on through every positive integer. No index can be the first place the two disagree, because the value there is computed from the index below it, where by assumption they already agree.

The two rules describe the same function, so they describe the same sequence. What they do not share is the labour. The explicit rule hands you $a_{100} = 100^2 + 1 = 10{,}001$ in a single substitution, while the recursion reaches the same number only after $99$ additions.

That is the trade in general. A recursive rule is usually the easier one to write down, because it often states plainly what is happening. Such a rule often says: add the next odd number, double the previous term, add the last two. An explicit rule is the one you want when you need a distant term, because it does not care how far out you go. Neither kind is more correct than the other. They are two descriptions of one function, and a good deal of the skill in this chapter is moving between them.

Check your understanding

A sequence is given by $a_1 = 5$ and $a_n = 2a_{n-1} - 3$ for $n \ge 2$ . What is $a_3$ ?

7

11

17

14

Sigma notation

Sequences get added. To say “add the first fifty terms of $a$ ” you could write $a_1 + a_2 + \cdots + a_{50}$ . But those dots are an instruction to the reader rather than a definition, and they get worse as soon as the pattern of the terms is not obvious. Mathematics has a compact and unambiguous symbol for the job.

Definition. For integers $m \le p$ ,

\sum_{k=m}^{p} a_k \;=\; a_m + a_{m+1} + a_{m+2} + \cdots + a_p .

The symbol $\sum$ is a capital Greek sigma, the Greek $S$ , standing for sum. Read the notation as an instruction: let $k$ start at the lower limit $m$ and walk up through the integers one at a time. Keep going until $k$ reaches the upper limit $p$ . Evaluate the expression to the right of the sigma at each value of $k$ , and add every result. The letter $k$ is the index of summation, and the expression being evaluated is the summand. Both limits are included, at the bottom and at the top.

The anatomy of a sigma sum. The index runs through every integer from the lower limit up to the upper limit, the summand is evaluated at each one, and the results are added. The capital Greek sigma stands for sum.

Three features of that notation deserve to be said out loud.

The index is a dummy. Its name is a local label. It appears only under the sigma and inside the summand, it vanishes the moment the sum is worked out, and the answer contains no trace of it:

\sum_{k=1}^{4} k^2 \;=\; \sum_{i=1}^{4} i^2 \;=\; \sum_{j=1}^{4} j^2 \;=\; 1 + 4 + 9 + 16 \;=\; 30 .

Rename the index to any fresh letter and the value cannot change, because the terms being added are untouched. The one restriction is that the new name must not already be doing another job in the expression. For instance, renaming $k$ to $n$ inside $\sum_{k=1}^{n} k$ would be a disaster, since $n$ is the upper limit there.

Counting the terms is an off-by-one trap. Running the index from $m$ up to $p$ produces the indices $m, m+1, \ldots, p$ , and when $m \le p$ there are

p - m + 1

of them. Subtracting the limits counts the steps between them, and a list always has one more entry than it has steps. So from $k = 3$ to $k = 8$ there are five steps but six terms, namely $3, 4, 5, 6, 7, 8$ . If you want the argument rather than the picture, shift every index down by $m - 1$ . That turns the list $m, m+1, \ldots, p$ into $1, 2, \ldots, p - m + 1$ , which plainly has $p - m + 1$ entries, and shifting does not change how many there are.

The summand need not mention the index. In

\sum_{k=1}^{5} 7 \;=\; 7 + 7 + 7 + 7 + 7 \;=\; 35 ,

the index still runs $1, 2, 3, 4, 5$ and the summand is still evaluated five times. It simply returns $7$ every time. The index controls how many terms there are even when it never appears in one.

A sum of terms of a sequence is called a series. Adding finitely many terms always produces an ordinary number, which is all this lesson needs. What an infinite sum could possibly mean is a real question, and a later lesson answers it.

Worked example 3 Evaluate $\displaystyle\sum_{k=2}^{5} (k^2 - k)$

Read the limits first. The index runs from $2$ up to $5$ , so the sum has $5 - 2 + 1 = 4$ terms, and the summand $k^2 - k$ has to be evaluated at each of $k = 2, 3, 4, 5$ .

Working left to right, $k = 2$ gives $4 - 2 = 2$ , then $k = 3$ gives $9 - 3 = 6$ , then $k = 4$ gives $16 - 4 = 12$ , and $k = 5$ gives $25 - 5 = 20$ . Adding those four results:

\sum_{k=2}^{5} (k^2 - k) = 2 + 6 + 12 + 20 = 40 .

The sum does not begin at $k = 1$ , and nothing requires it to. The lower limit is part of the notation, so reading it as a $1$ out of habit would have thrown in the extra term $1^2 - 1 = 0$ . That term is harmless here, but it will not be next time.

Worked example 4 Write $\dfrac{1}{2} + \dfrac{2}{3} + \dfrac{3}{4} + \dfrac{4}{5} + \dfrac{5}{6}$ in sigma notation

Describe a single term before you describe the sum. Each numerator counts up from $1$ , and each denominator is one more than its numerator, so the term whose numerator is $k$ is $\frac{k}{k+1}$ . The numerators run from $1$ to $5$ , which fixes the limits:

\frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \frac{4}{5} + \frac{5}{6} = \sum_{k=1}^{5} \frac{k}{k+1} .

Check the ends before you trust it. At $k = 1$ the summand is $\frac{1}{2}$ , and at $k = 5$ it is $\frac{5}{6}$ , which are the first and last terms of the original sum.

That answer is correct, but it is not the only correct one. Starting the index at $0$ and adjusting the summand to compensate,

\sum_{j=0}^{4} \frac{j+1}{j+2} ,

produces exactly the same five terms. Both are right. Sigma notation is a way of writing a sum, not a fingerprint of it, so a sum can always be written in many equivalent ways. Prefer the version with the simplest summand, which is almost always the one whose index starts at $1$ .

Check your understanding

Evaluate $\displaystyle\sum_{k=3}^{6} (k^2 + 1)$ .

90

86

97

53

Common mistakes

Writing a recursion with no seed. The rule $a_n = 3a_{n-1}$ describes infinitely many sequences, one for every starting value. Without $a_1$ it defines nothing at all. A rule that reaches back two steps needs two seeds.
Confusing the index with the term. In $5, 9, 13, 17$ the third term is $a_3 = 13$ , not $3$ . The index says which term; the term is the value.
Off by one when counting terms. A sum from $k = 4$ to $k = 17$ has $17 - 4 + 1 = 14$ terms, not $13$ . Both limits are included, so subtracting them counts steps, not terms.
Believing the index name matters. Every one of $\sum_{k=1}^{4} k^2$ , $\sum_{i=1}^{4} i^2$ , and $\sum_{j=1}^{4} j^2$ is $30$ . If your answer still contains the index letter, you have not finished.
Ignoring the index when the summand hides it. In $\sum_{k=1}^{6} 5$ the value is $30$ , not $5$ : six terms are added, and each one happens to be $5$ .
Joining the dots on the graph. A sequence has no term between $a_2$ and $a_3$ , so its graph is a set of isolated points. Drawing a curve through them invents values that do not exist.
Trusting a pattern. Four terms never determine a sequence. Until a rule is stated, “what comes next” has many correct answers.

Key takeaways

A sequence is a function whose domain is the positive integers (or the integers upward from another stated starting point, such as $0$ ). The notation $a_n$ is just $a(n)$ with the parentheses replaced by a subscript, so everything you know about functions still applies.
The index $n$ and the term $a_n$ are different numbers, and the graph of a sequence is a set of separate dots, one above each positive integer.
An explicit rule gives $a_n$ as an expression in $n$ and reaches any term in one substitution. A recursive rule gives seed values plus a rule for building each term from earlier ones, and it has to be climbed one rung at a time.
A recursion without its seed defines nothing. A rule that looks back $k$ steps needs $k$ seeds.
The notation $\sum_{k=m}^{p} a_k$ means $a_m + a_{m+1} + \cdots + a_p$ . Both limits are included, so the sum has $p - m + 1$ terms.
The index of summation is a dummy: renaming it changes nothing, and it never appears in the answer.
A handful of terms never determines a sequence. Only a rule does.

Sequences and Notation

Learning goals

A sequence is a function on the positive integers

Terms, indices, and what a list does not tell you

Explicit rules and recursive rules

Sigma notation

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)