10 Series Basics

Highlights

Definitions of series, partial sums, and series convergence
Examples of series converging and diverging
Special properties of series with all non-negative terms.

10.1 Preparation

In this section, we build on what we learned about sequences to talk about one of their most important use cases: series. Loosely speaking, we think of a series as an “infinite sum”, and one of the primary questions we seek to address is when a series converges to a finite number and when it doesn’t. Consider for example, the four infinite sums below:

\[\begin{array}{l} 1+2+3+4+\cdots \\~\\ 1+(-1)+1+(-1)+\cdots \\~\\ 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots \\~\\ %1+\left(-\frac{1}{2}\right)+\frac{1}{4}+\left(-\frac{1}{8}\right)+\cdots 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots \end{array} \tag{10.1}\]

Some of these might strike you as obviously undefined or obviously convergent, but others may not. In what follows below, we’ll make a precise definition of what we mean by convergent so that we can prove theorems and talk coherently about examples.¹

Suppose \((a_n)\) is a sequence in \(\mathbb{R}.\) Then the series with terms \((a_n)\) is the formal expression

\[\sum_{i=1}^\infty a_i=a_1+a_2+\cdots\]

By “formal expression” I mean that we are not considering this expression to represent a number; it’s just notation meant to express the idea of an infinite sum. Sometimes we abbreviate the notation as \(\sum a_i.\)

Definition 10.1 (Partial Sums) The sequence of partial sums of the series \(\displaystyle\sum_{i=1}^\infty a_i\) is the sequence \((s_n)\) defined by \[ s_n=\sum_{i=1}^n a_i = a_1+\cdots+a_n.\]

This means that \[\begin{array}{ccl} s_1&=&a_1\\ s_2&=&a_1+a_2\\ s_3&=&a_1+a_2+a_3\\ &\vdots& \end{array}\]

We now define convergence of series in terms of the partial sums.

Definition 10.2 (Series Convergence) The series \(\displaystyle\sum_{i=1}^\infty a_i\) is said to converge or diverge when its sequence of partial sums \((s_n)\) does. If \(s_n\rightarrow S,\) we call \(S\) the sum of the series.

In the case that \(\displaystyle\sum_{i=1}^\infty a_i\) has sum \(S,\) we sometimes abuse notation slightly and use the expression \(\displaystyle\sum_{i=1}^\infty a_i\) to represent the number \(S.\)

As an example of the vocabulary in action, the series \(\displaystyle \sum_{i=1}^\infty \frac{2}{3^i}\)

has terms given by the sequence \(\left(\frac{2}{3^i}\right)\) or \(\frac{2}{3},\frac{2}{9},\frac{2}{27},\frac{2}{81},\cdots\)
has partial sums \(\frac{2}{3},\frac{2}{3}+\frac{2}{9},\frac{2}{3}+\frac{2}{9}+\frac{2}{27},\ldots.\) These simplify to \(\frac{2}{3},\frac{8}{9},\frac{26}{27},\ldots\) or \(\left(\frac{3^n-1}{3^n}\right).\)
converges with a sum of \(1.\)

Exercise 10.3 Write each of the four infinite sums in Equation 10.1 using series notation. Then write out the first 6 partial sums for each. Are there any you can tell right away converge or diverge?

10.2 Working with Partial Sums

We start with an important class of series called geometric series.

Example 10.4 (Geometric Series) For \(a\neq 0\) and any \(r\in\mathbb{R},\) a series of the form \[\sum_{i=0}^\infty ar^i\] is called a geometric series. This series converges to \(\frac{a}{1-r}\) when \(|r|<1\) and diverges otherwise.

Proof. To determine whether the series converges, we analyze its partial sums in cases.

If \(|r|\neq1,\) we can use some algebraic tricks to simplify the expression for the partial sum \(s_n:\) \[\begin{array}{rcl} s_n&=&a+ar+ar^2+\cdots+ar^n\\ &=&a(1+r+r^2+\cdots+r^n)\\ &=&\displaystyle\frac{a(1-r^{n+1})}{1-r} \end{array}\] If \(|r|<1,\) then \(r^{n+1}\rightarrow0,\) so using what we know about sequences and arithmetic, we can say \(s_n\rightarrow\frac{a}{1-r}.\)

If \(|r|>1,\) then \((r^{n+1})\) diverges. So if, by way of contradiction, we assume the partial sums \((s_n)\) did converge to some value \(S,\) then we could say that the sequence \(\left(1-s_n\frac{1-r}{a}\right)\) converges to \(1-S\frac{1-r}{a}.\) However \(1-s_n\frac{1-r}{a}=r^{n+1},\) which we already said diverges.²

We leave the cases where \(r=\pm1\) to the reader in Exercise 10.5 below.

Exercise 10.5 Finish the proof of Example 10.4 by proving that the geometric series diverges when \(r=1\) and when \(r=-1.\) (Hint: what are the partial sums in each of these cases?)

Note

In the section on sequences, we insisted that the first term in our sequence corresponded to an index of 1. For series, it is often convenient to start with an index 0 (as in Example 10.4) or something else. And now that we are more comfortable with the concept of a sequence, there is less risk of confusion.

Exercise 10.6 (A “Telescoping” Series) Show that the series \(\displaystyle\sum_{i=1}^\infty \frac{1}{i(i+1)}\) converges. It will be helfpul to note that \[\frac{1}{i(i+1)}=\frac{1}{i}-\frac{1}{i+1}\] and to write out the first several partial sums.

Both Example 10.4 and Exercise 10.6 are somewhat unusual in that in both cases we can write down an explicit formula for the partial sum sequence. More often an explicit formula is unavailable and we need to rely on other means for determining and proving the con/divergence of a series.

The so-called “convergence tests” for series that we learn in calculus give us tools for determining the con/divergence of a series by analyzing their terms rather than their partial sums.

Theorem 10.7 (Divergence Test) If a series converges, then its terms converge to \(0.\)

Exercise 10.8

State the contrapositive of the Divergence Test.
What (if anything) does the Divergence Test say about the con/divergence of the series \(\displaystyle \sum_{i=1}^\infty \frac{i}{i+1}?\)
What (if anything) does the Divergence Test say about the con/divergence of the series \(\displaystyle \sum_{i=1}^\infty \frac{1}{i^2}?\)
Prove the Divergence Test.

Hint for part 4 of Exercise 10.8

Note that \(a_{n+1}=s_{n+1}-s_n.\)

Theorem 10.9 Suppose \(\sum a_n=S\) and \(\sum b_n=T\) are convergent series and that \(k\in\mathbb{R}.\) Then

\(\sum (a_n+b_n) = S+T\)
\(\sum (ka_n)=kS\)
If \(a_n\leq b_n\) for all \(n,\) then \(S\leq T.\)

Note that part of what parts 1 and 2 in Theorem 10.9 are claiming is that these two series do in fact converge.

Note

When an equation or proposition does not depend on the starting index of a series, we sometimes just write \(\sum a_n\) as shorthand for \(\sum_{n=k}^\infty a_n\) for some \(k.\)

Exercise 10.10 Prove Theorem 10.9.

Hint

Leverage what we already know about sequence convergence to do these proofs!

10.3 Series With Non-Negative Terms

Series with all non-negative terms have some special properties that we can quickly prove using what we know about sequences.

Exercise 10.11 Suppose \(\displaystyle \sum_{i=1}^\infty a_i\) is a series all of whose terms are non-negative. Prove the following.

The sequence of partial sums is increasing.
The series converges if and only if the sequence of partial sums is bounded.
If the series diverges, then the sequence of partial sums limits to \(+\infty.\)

The upshot of Exercise 10.11 is that we can determine the con/divergence of a series just by knowing whether its sequence of partial sums is bounded or not.

Example 10.12 The series \(\displaystyle\sum_{i=1}^\infty \frac{1}{i^2}\) converges.

Proof. Let \((s_n)\) be the sequence of partial sums for \(\displaystyle\sum_{i=1}^\infty \frac{1}{i^2}.\)

In Exercise 10.6, we worked with the series \(\displaystyle\sum_{i=1}^\infty \frac{1}{i(i+1)}.\) Let \((t_n)\) be its sequence of partial sums.

As we saw, the series converges to some number \(T.\) Therefore by Theorem 10.9, the series \(\displaystyle\sum_{i=1}^\infty \frac{2}{i(i+1)}\) converges to \(2T;\) that is \(2t_n\rightarrow 2T.\) Comparing the terms of the two series, we see that for all \(i,\) \[\frac{1}{i^2}\leq \frac{2}{i(i+1)}.\] Adding up this expression for \(i=1,\ldots,n,\) we have that \(s_n\leq 2t_n\) for all \(n.\) Since \(\displaystyle\sum_{i=1}^\infty \frac{2}{i(i+1)}\) converges, \((2t_n)\) is bounded above by Exercise 10.11. Therefore \((s_n)\) is also bounded above and thus converges, again by Exercise 10.11.

Note that the above proof does not tell us what the series \(\displaystyle\sum_{i=1}^\infty \frac{1}{i^2}\) converges to. This is often the case for such proofs.

The following theorem generalizes the logic of the proof of Example 10.12.

Theorem 10.13 (Comparison Test) Suppose \(\sum a_n\) and \(\sum b_n\) are series with all non-negative terms such that \(a_n\leq b_n\) for all \(n.\)

If \(\sum b_n\) converges, then \(\sum a_n\) converges.
If \(\sum a_n\) diverges, then \(\sum b_n\) diverges.

Example 10.14 (Harmonic Series) The harmonic series \(\displaystyle \sum_{i=1}^\infty \frac{1}{i}\) diverges.

Proof. Call the terms of this series \(c_i=\frac{1}{i}\) and the partial sums \(s_n.\) Note that \(c_i\rightarrow0,\) so the Divergence Test cannot be used to prove divergence.

Consider another series \(\displaystyle \sum_{i=1}^\infty b_i\) where the terms \(b_i\) are defined by \[\begin{array}{rcl} b_1&=&1\\ b_2&=&\frac{1}{2}\\ b_3=b_4&=&\frac{1}{4}\\ b_5=\cdots=b_8&=&\frac{1}{8}\\ &\vdots&\\ b_{2^{i-1}+1}=\cdots=b_{2^i} & = & \frac{1}{2^i}\\ &\vdots& \end{array}\]

Calling the partial sums \(t_n,\) we can consider the subsequence \(t_{2^n}:\) \[\begin{array}{rcll} t_{2^1} &=& 1+\frac{1}{2}&=\frac{3}{2}\\ t_{2^2} &=& 1+\frac{1}{2}+2\cdot\frac{1}{4}&=\frac{4}{2}\\ t_{2^3} &=& 1+\frac{1}{2}+2\cdot\frac{1}{4}+4\cdot\frac{1}{8}&=\frac{5}{2}\\ &\vdots& \end{array}\] Thus \(t_{2^n}=\frac{n+2}{2},\) so \((t_n)\) has a divergent subsequence and hence \(\displaystyle \sum_{i=1}^\infty b_i\) diverges. Therefore since \(c_i\geq b_i\) for all \(i\) (and both series have all non-negative terms) \(\displaystyle \sum_{i=1}^\infty \frac{1}{i}\) diverges as well by the Comparison Test.

Exercise 10.15 Prove Theorem 10.13.

Review Questions

How is series convergence defined?
In basically all of the series convergence proofs from this chapter, the first step is rewriting the statement as a statement about the convergence of what sequence?
What are some special properties of series with all non-negative terms?

There are actually several theories of convergent series which are useful in different contexts. We’ll present the most standard version here.↩︎
I am bending over backwards here to make precise the intuition that “Since \((r^{n+1})\) diverges, so must a sequence involving it”. Unfortunately, that reasoning doesn’t always work—think about \((\sqrt{n+1} - \sqrt{n})\) which converges despite being the sum of two divergent sequences—so a proof of the style shown here is appropriate.↩︎