Sums and Series

Learning Outcomes

  • Explain the meaning of the sum of an infinite series

An infinite series is a sum of infinitely many terms and is written in the form

[latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}={a}_{1}+{a}_{2}+{a}_{3}+ \cdots [/latex].

 

But what does this mean? We cannot add an infinite number of terms in the same way we can add a finite number of terms. Instead, the value of an infinite series is defined in terms of the limit of partial sums. A partial sum of an infinite series is a finite sum of the form

[latex]\displaystyle\sum _{n=1}^{k}{a}_{n}={a}_{1}+{a}_{2}+{a}_{3}+ \cdots +{a}_{k}[/latex].

 

To see how we use partial sums to evaluate infinite series, consider the following example. Suppose oil is seeping into a lake such that [latex]1000[/latex] gallons enters the lake the first week. During the second week, an additional [latex]500[/latex] gallons of oil enters the lake. The third week, [latex]250[/latex] more gallons enters the lake. Assume this pattern continues such that each week half as much oil enters the lake as did the previous week. If this continues forever, what can we say about the amount of oil in the lake? Will the amount of oil continue to get arbitrarily large, or is it possible that it approaches some finite amount? To answer this question, we look at the amount of oil in the lake after [latex]k[/latex] weeks. Letting [latex]{S}_{k}[/latex] denote the amount of oil in the lake (measured in thousands of gallons) after [latex]k[/latex] weeks, we see that

[latex]\begin{array}{l}{S}_{1}=1\hfill \\ {S}_{2}=1+0.5=1+\frac{1}{2}\hfill \\ {S}_{3}=1+0.5+0.25=1+\frac{1}{2}+\frac{1}{4}\hfill \\ {S}_{4}=1+0.5+0.25+0.125=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\hfill \\ {S}_{5}=1+0.5+0.25+0.125+0.0625=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}.\hfill \end{array}[/latex]

 

Looking at this pattern, we see that the amount of oil in the lake (in thousands of gallons) after [latex]k[/latex] weeks is

[latex]{S}_{k}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+ \cdots +\frac{1}{{2}^{k - 1}}=\displaystyle\sum _{n=1}^{k}{\left(\frac{1}{2}\right)}^{n - 1}[/latex].

 

We are interested in what happens as [latex]k\to \infty [/latex]. Symbolically, the amount of oil in the lake as [latex]k\to \infty [/latex] is given by the infinite series

[latex]\displaystyle\sum _{n=1}^{\infty }{\left(\frac{1}{2}\right)}^{n - 1}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots [/latex].

 

At the same time, as [latex]k\to \infty [/latex], the amount of oil in the lake can be calculated by evaluating [latex]\underset{k\to \infty }{\text{lim}}{S}_{k}[/latex]. Therefore, the behavior of the infinite series can be determined by looking at the behavior of the sequence of partial sums [latex]\left\{{S}_{k}\right\}[/latex]. If the sequence of partial sums [latex]\left\{{S}_{k}\right\}[/latex] converges, we say that the infinite series converges, and its sum is given by [latex]\underset{k\to \infty }{\text{lim}}{S}_{k}[/latex]. If the sequence [latex]\left\{{S}_{k}\right\}[/latex] diverges, we say the infinite series diverges. We now turn our attention to determining the limit of this sequence [latex]\left\{{S}_{k}\right\}[/latex].

First, simplifying some of these partial sums, we see that

[latex]\begin{array}{l}{S}_{1}=1\hfill \\ {S}_{2}=1+\frac{1}{2}=\frac{3}{2}\hfill \\ {S}_{3}=1+\frac{1}{2}+\frac{1}{4}=\frac{7}{4}\hfill \\ {S}_{4}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}=\frac{15}{8}\hfill \\ {S}_{5}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=\frac{31}{16}.\hfill \end{array}[/latex]

 

Plotting some of these values in Figure 1, it appears that the sequence [latex]\left\{{S}_{k}\right\}[/latex] could be approaching 2.

This is a graph in quadrant 1with the x and y axes labeled n and S_n, respectively. From 1 to 5, points are plotted. They increase and seem to converge to 2 and n goes to infinity.

Figure 1. The graph shows the sequence of partial sums [latex]\left\{{S}_{k}\right\}[/latex]. It appears that the sequence is approaching the value [latex]2[/latex].

Let’s look for more convincing evidence. In the following table, we list the values of [latex]{S}_{k}[/latex] for several values of [latex]k[/latex].

[latex]k[/latex] [latex]5[/latex] [latex]10[/latex] [latex]15[/latex] [latex]20[/latex]
[latex]{S}_{k}[/latex] [latex]1.9375[/latex] [latex]1.998[/latex] [latex]1.999939[/latex] [latex]1.999998[/latex]

These data supply more evidence suggesting that the sequence [latex]\left\{{S}_{k}\right\}[/latex] converges to [latex]2[/latex]. Later we will provide an analytic argument that can be used to prove that [latex]\underset{k\to \infty }{\text{lim}}{S}_{k}=2[/latex]. For now, we rely on the numerical and graphical data to convince ourselves that the sequence of partial sums does actually converge to [latex]2[/latex]. Since this sequence of partial sums converges to [latex]2[/latex], we say the infinite series converges to [latex]2[/latex] and write

[latex]\displaystyle\sum _{n=1}^{\infty }{\left(\frac{1}{2}\right)}^{n - 1}=2[/latex].

 

Returning to the question about the oil in the lake, since this infinite series converges to [latex]2[/latex], we conclude that the amount of oil in the lake will get arbitrarily close to [latex]2000[/latex] gallons as the amount of time gets sufficiently large.

This series is an example of a geometric series. We discuss geometric series in more detail later in this section. First, we summarize what it means for an infinite series to converge.

Definition

An infinite series is an expression of the form

[latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}={a}_{1}+{a}_{2}+{a}_{3}+\cdots [/latex].

 

For each positive integer [latex]k[/latex], the sum

[latex]{S}_{k}=\displaystyle\sum _{n=1}^{k}{a}_{n}={a}_{1}+{a}_{2}+{a}_{3}+\cdots +{a}_{k}[/latex]

 

is called the [latex]k\text{th}[/latex] partial sum of the infinite series. The partial sums form a sequence [latex]\left\{{S}_{k}\right\}[/latex]. If the sequence of partial sums converges to a real number [latex]S[/latex], the infinite series converges. If we can describe the convergence of a series to [latex]S[/latex], we call [latex]S[/latex] the sum of the series, and we write

[latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}=S[/latex].

 

If the sequence of partial sums diverges, we have the divergence of a series.

Note that the index for a series need not begin with [latex]n=1[/latex] but can begin with any value. For example, the series

[latex]{\displaystyle\sum _{n=1}^{\infty }\left(\frac{1}{2}\right)}^{n - 1}[/latex]

 

can also be written as

[latex]{\displaystyle\sum _{n=0}^{\infty }\left(\frac{1}{2}\right)}^{n}\text{or}{\displaystyle\sum _{n=5}^{\infty }\left(\frac{1}{2}\right)}^{n - 5}[/latex].

 

Often it is convenient for the index to begin at [latex]1[/latex], so if for some reason it begins at a different value, we can reindex by making a change of variables. For example, consider the series

[latex]\displaystyle\sum _{n=2}^{\infty }\frac{1}{{n}^{2}}[/latex].

 

By introducing the variable [latex]m=n - 1[/latex], so that [latex]n=m+1[/latex], we can rewrite the series as

[latex]\displaystyle\sum _{m=1}^{\infty }\frac{1}{{\left(m+1\right)}^{2}}[/latex].

 

Example: Evaluating Limits of Sequences of Partial Sums

For each of the following series, use the sequence of partial sums to determine whether the series converges or diverges.

  1. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{n}{n+1}[/latex]
  2. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n}[/latex]
  3. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{n\left(n+1\right)}[/latex]

try it

Determine whether the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{\left(n+1\right)}{n}[/latex] converges or diverges.

Watch the following video to see the worked solution to the above Try IT.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “5.2.1” here (opens in new window).

Try It