MATH 1070: Mathematics of Change, part I

We already briefly discussed sequences. When working with sums and the "Sigma" notation for summations, you need to be able to write formulas for sequences you understand intuitively. For example, if you want to write the sequence \(7, 9, 11, 13, \ldots\) in the notation \(\{ b_n : n \geq 1 \}\text{,}\) so that \(b_1 = 7\text{,}\)\(b_2 = 9\) and so on, one choice would be to say,

Let \(\{ b_n : n \geq 1 \}\) be the sequence defined by \(b_n := 5 + 2n\text{.}\)

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The subscript \(n\) is called the index

Aside

(plural: indices). Indexing can begin at any natural number. In this case, as is most common, we began at \(n=1\text{.}\) Defining \(\{ b_n : n \geq 3 \}\) by \(b_n := 1 + 2n\) yields the same sequence: \(7, 9, 11, 13, \ldots\text{.}\)

Secondly, the informal notation \(7, 9, 11, 13, \ldots\) is not mathematically precise because it assumes we all agree exactly what the pattern is. Producing a formula for the \(n^{th}\) term removes any ambiguity. A formula is often necessary if you want to sum the sequence or to use it to define other sequences. This section considers some common types of sequences and gives you some practice writing a formula for the general term.

Our example sequence \(7, 9, 11, 13, \ldots\) is an arithmetic sequence with common difference 2. It is particularly easy to write a formula for the general term of an arithmetic sequence if you start indexing at zero. The \(n^{th}\) term is the zeroth term plus \(n\) copies of the common difference. In notation, if the common difference is \(d\) and the sequence is \(\{ a_k : k \geq 0 \}\text{,}\)this means \(a_k = a_0 + k d\text{.}\) Setting \(a_0 = 7\) and \(d = 2\) gives \(a_k = 7 + 2k\) for the sequence \(7, 9, 11, 13, \ldots\text{.}\)

For example, the sequence \(10, 20, 40, 80, 160, \ldots\) is geometric with common ratio 2.

Sequences with alternating signs appear often enough that it’s a good idea to know a way to write their general term. The key to being able to write such sequences is to notice that \((-1)^n\) bounces back and forth between \(+1\) and \(-1\text{.}\) The odd terms are negative, so starting with \(n=1\) (or 3 or 5, etc.) starts with \(-1\) whereas starting with 0 (or 2, -2, etc.) starts with \(+1\text{.}\) You can incorporate this in a sum as a multiplicative factor and it will change the sign of every second term. Thus for example, to write the sequence \(1, -2, 3, - 4 \ldots\) you can write \((-1)^{n+1} \cdot n \, .\) Note that we used \((-1)^{n+1}\) rather than \((-1)^n\) so that the term corresponding to \(n=1\) was positive rather than negative.

When the sum has a pattern that takes a couple of steps to repeat, the greatest integer function can be useful. For example, \(1, 1, 1, 2, 2, 2, 3, 3, 3, \ldots\) can be written as \(a_n := \left \lfloor \frac{n+2}{3} \right \rfloor\) for \(n \geq 1\text{.}\) Actually, it comes out a little more simply if you index starting from zero: \(a_n := \left \lfloor \frac{n}{3} \right \rfloor + 1\) for \(n \geq 0\text{.}\)

Plenty of sequences don’t fit any of these molds. Writing a formula for the general term is a matter of trying an expression, seeing if it works, then if not, tinkering to get it right.