Introduction to Series: Learn It 1

  • Understand what we mean by the sum of an infinite series
  • Find the sum of a geometric series
  • Calculate the sum of a telescoping series

Sums and Series

You’ve already worked with sequences—ordered sets of terms like [latex]a_1, a_2, a_3, \ldots[/latex]. When you add the terms of a sequence together, you get a series.

An infinite series is the sum of infinitely many terms, written as:

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

But what does it mean to add infinitely many terms? We can’t simply add them the way we add a finite number of terms. Instead, we define the value of an infinite series using partial sums.

A partial sum is a finite sum of the first [latex]k[/latex] terms:

[latex]\displaystyle\sum _{n=1}^{k}{a}_{n}={a}_{1}+{a}_{2}+{a}_{3}+ \cdots +{a}_{k}[/latex].
The value of the infinite series is then defined as the limit of these partial sums as [latex]k \to \infty[/latex].

Let’s see how this works with a real scenario.

Suppose oil seeps into a lake following this pattern:

  • Week 1: [latex]1000[/latex] gallons enter the lake
  • Week 2: [latex]500[/latex] gallons enter (half of previous week)
  • Week 3: [latex]250[/latex] gallons enter (half of previous week)
  • And so on…

Each week, half as much oil enters compared to the previous week. The question is: if this continues forever, will the total amount of oil grow without bound, or will it approach some finite limit?

Let [latex]S_k[/latex] represent the total oil (in thousands of gallons) after [latex]k[/latex] weeks:

[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, the total oil 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].

As [latex]k \to \infty[/latex], the total amount of oil is represented 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].

The behavior of an infinite series depends entirely on what happens to its sequence of partial sums [latex]{S_k}[/latex] as [latex]k \to \infty[/latex].

To find [latex]\lim_{k \to \infty} S_k[/latex], let’s calculate some partial sums:

[latex]S_1 = 1[/latex] [latex]S_2 = 1 + \frac{1}{2} = \frac{3}{2} = 1.5[/latex] [latex]S_3 = 1 + \frac{1}{2} + \frac{1}{4} = \frac{7}{4} = 1.75[/latex] [latex]S_4 = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{15}{8} = 1.875[/latex] [latex]S_5 = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} = \frac{31}{16} = 1.9375[/latex]

Plotting some of these values in Figure 1, it appears the sequence [latex]{S_k}[/latex] might be approaching [latex]2[/latex].

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].

More 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].

This numerical evidence strongly suggests that [latex]\lim_{k \to \infty} S_k = 2[/latex].

Since the sequence of partial sums converges to [latex]2[/latex], we say the infinite series converges to [latex]2[/latex] and write:

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

Returning to our question about the oil in the lake, since this series converges to [latex]2[/latex], the total amount of oil in the lake will approach [latex]2000[/latex] gallons (remember our units were thousands of gallons), no matter how long the spill continues.

infinite series

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 [latex]\lim_{k \to \infty} S_k = S[/latex] (a finite number), the series converges to [latex]S[/latex]
  • If [latex]\lim_{k \to \infty} S_k[/latex] does not exist or is infinite, the series diverges

The starting index of a series can be adjusted for convenience. For example:

[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].

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]