Sequences and Notation

A sequence is a function on consecutive integers

Definition. A sequence is a function whose domain is a run of consecutive integers, starting from some stated place and continuing upward, either forever, which makes an infinite sequence, or until it has exactly $k$ terms, which makes a finite sequence. Unless a lesson says otherwise, that starting place is $1$, so the default domain is $\{1, 2, 3, \ldots\}$ (infinite) or $\{1, 2, \ldots, k\}$ (finite, with $k$ terms). Some sequences are more natural indexed from $0$ instead, with domain $\{0, 1, 2, \ldots\}$ or $\{0, 1, \ldots, k-1\}$ for the same $k$ terms; the idea is unchanged, because the inputs are still consecutive integers running upward from wherever the lesson says to start. 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 term of the sequence at index $n$.

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:

The input $n$ is called the index. Because the domain is always a run of consecutive integers, the inputs arrive in a natural order, one after the next with nothing skipped, whether that run is finite, infinite, or starts at $0$. That built-in order is exactly why a sequence is so naturally written as a list, in a way that most functions, whose inputs come in no particular order, are not. The list is not a second object sitting beside the function; it is the function’s output table, read from left to right.

Two habits are worth fixing right away. First, the index and the term play different roles and need not match: 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 the consecutive integers in the domain 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.

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

You can usually tell which kind of sequence is meant from how the list is written. When someone writes $3, 7, 11, 15$ they mean a finite sequence with four terms, full stop. The trailing dots in $3, 7, 11, 15, \ldots$ signal an infinite one instead: the list keeps going by the same rule.

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 how a sequence continues. This is about predicting more terms, not about a complete finite list: written all the way out, $3, 7, 11, 15$ is the whole sequence, full stop, exactly as the paragraph above says. But shown the opening $2, 4, 6, 8, \ldots$ with no promise that it stops there, you would probably guess “the even numbers”, $a_n = 2n$, so the next term is $10$. That is a reasonable guess, but it is still only a guess: the same four terms $2, 4, 6, 8$ are also the start of the sequence that simply repeats, $2, 4, 6, 8, 2, 4, 6, 8, \ldots$, and of many other sequences besides. A pattern you spot in a handful of terms is a guess, sometimes an excellent guess, but how the sequence continues 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.

Explicit rules and recursive rules

There are two standard ways to pin a sequence down: state $a_n$ directly, or build it from earlier terms. Each has a different cost, as the rest of this section shows.

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

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:

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 here is why, in a moment.

The seed is not decoration. Delete it and the rule no longer pins down one particular sequence. The bare rule $a_n = a_{n-1} + 2n - 1$ for $n \ge 2$ 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 $a_n = a_{n-1} + a_{n-2}$ for $n \ge 3$ 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.

Here is why that is no coincidence, in plain terms. Go back to the explicit rule $a_n = n^2 + 1$ and its recursion $a_1 = 2$, $a_n = a_{n-1} + 2n - 1$: both start from the same value, $2$. And at every step after that, both do the exact same thing to get the next term. The explicit rule’s value goes up by $2n - 1$ each time $n$ increases by $1$, since $(n^2+1) - \big((n-1)^2+1\big) = 2n - 1$, and that is exactly the step the recursive rule takes. Two rules that start in the same place and always take the same step land on the same value at every step after that, all the way out. That is also why the labor differs even though the answer does not: 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.

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 indices $m \le p$ in the sequence’s domain,

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.

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

The letter under the sigma is only a temporary name, a local label for the index. 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. The formal name for a letter used this way is a dummy index:

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

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 $3, 4, 5, 6, 7, 8$.

The summand need not mention the index. In

the index still runs $1, 2, 3, 4, 5$ and the summand is still evaluated five times. It simply returns $7$ every time. The limits, not the summand, control how many terms there are, even when the summand never mentions the index at all.

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.

Common mistakes