Investigating Alternating Sequences

Sometimes sequences have terms that are alternate. In fact, the terms may actually alternate in sign. The steps to finding terms of the sequence are the same as if the signs did not alternate. However, the resulting terms will not show increase or decrease as [latex]n[/latex] increases. Let’s take a look at the following sequence.

[latex]\left\{2,-4,6,-8\right\}[/latex]

Notice the first term is greater than the second term, the second term is less than the third term, and the third term is greater than the fourth term. This trend continues forever. Do not rearrange the terms in numerical order to interpret the sequence.

How To: Given an explicit formula with alternating terms, write the first [latex]n[/latex] terms of a sequence.

  1. Substitute each value of [latex]n[/latex] into the formula. Begin with [latex]n=1[/latex] to find the first term, [latex]{a}_{1}[/latex]. The sign of the term is given by the [latex]{\left(-1\right)}^{n}[/latex] in the explicit formula.
  2. To find the second term, [latex]{a}_{2}[/latex], use [latex]n=2[/latex].
  3. Continue in the same manner until you have identified all [latex]n[/latex] terms.

Example 2: Writing the Terms of an Alternating Sequence Defined by an Explicit Formula

Write the first five terms of the sequence.

[latex]{a}_{n}=\frac{{\left(-1\right)}^{n}{n}^{2}}{n+1}[/latex]

Solution

Substitute [latex]n=1[/latex], [latex]n=2[/latex], and so on in the formula.

[latex]\begin{array}{lll}n=1\hfill & \begin{array}{cc}& \end{array}\hfill & {a}_{1}=\frac{{\left(-1\right)}^{1}{2}^{2}}{1+1}=-\frac{1}{2}\hfill \\ n=2\hfill & \begin{array}{cc}& \end{array}\hfill & {a}_{2}=\frac{{\left(-1\right)}^{2}{2}^{2}}{2+1}=\frac{4}{3}\hfill \\ n=3\hfill & \begin{array}{cc}& \end{array}\hfill & {a}_{3}=\frac{{\left(-1\right)}^{3}{3}^{2}}{3+1}=-\frac{9}{4}\hfill \\ n=4\hfill & \begin{array}{cc}& \end{array}\hfill & {a}_{4}=\frac{{\left(-1\right)}^{4}{4}^{2}}{4+1}=\frac{16}{5}\hfill \\ n=5\hfill & \hfill & {a}_{5}=\frac{{\left(-1\right)}^{5}{5}^{2}}{5+1}=-\frac{25}{6}\hfill \end{array}[/latex]

The first five terms are [latex]\left\{-\frac{1}{2},\frac{4}{3},-\frac{9}{4},\frac{16}{5},-\frac{25}{6}\right\}[/latex].

Analysis of the Solution

The graph of this function, shown in Figure 4, looks different from the ones we have seen previously in this section because the terms of the sequence alternate between positive and negative values.

Graph of a scattered plot with labeled points: (1, -1/2), (2, 4/3), (3, -9/4), (4, 16/5), and (5, -25/6). The x-axis is labeled n and the y-axis is labeled a_n.

Figure 4

Q & A

In Example 2, does the (–1) to the power of [latex]n[/latex] account for the oscillations of signs?

Yes, the power might be [latex]n,n+1,n - 1[/latex], and so on, but any odd powers will result in a negative term, and any even power will result in a positive term.

Try It 2

Write the first five terms of the sequence:

[latex]{a}_{n}=\frac{4n}{{\left(-2\right)}^{n}}[/latex]