9 Sequences and Series

9.2 Infinite Series

Given the sequence {an}={1/2n}=1/2, 1/4, 1/8,, consider the following sums:


Later, we will be able to show that


Let Sn be the sum of the first n terms of the sequence {1/2n}. From the above, we see that S1=1/2, S2=3/4, and that Sn=1-1/2n.

Now consider the following limit: limnSn=limn(1-1/2n)=1. This limit can be interpreted as saying something amazing: the sum of all the terms of the sequence {1/2n} is 1.

This example illustrates some interesting concepts that we explore in this section. We begin this exploration with some definitions.

Definition 9.2.1 Infinite Series, n𝐭𝐡 Partial Sums, Convergence, Divergence

Let {an} be a sequence.

  1. (a)

    The sum n=1an is an infinite series (or, simply series).

  2. (b)

    Let Sn=i=1nai ; the sequence {Sn} is the sequence of n𝐭𝐡 partial sums of {an}.

  3. (c)

    If the sequence {Sn} converges to L, we say the series n=1an converges to L, and we write n=1an=L.

  4. (d)

    If the sequence {Sn} diverges, the series n=1an diverges.

Using our new terminology, we can state that the series n=11/2n converges, and n=11/2n=1.

We will explore a variety of series in this section. We start with two series that diverge, showing how we might discern divergence.

Example 9.2.1 Showing series diverge

(a) Let {an}={n2}. Show n=1an diverges. (b) Let {bn}={(-1)n+1}. Show n=1bn diverges.


  1. (a)

    Consider Sn, the nth partial sum. margin: 510100200300nyanSn Figure 9.2.1: Scatter plots relating to the series of Example 9.2.1 part 1. Λ

    Sn =a1+a2+a3++an
    =n(n+1)(2n+1)6.  by Theorem 5.3.1

    Since limnSn=, we conclude that the series n=1n2 diverges. It is instructive to write n=1n2= for this tells us how the series diverges: it grows without bound.

    A scatter plot of the sequences {an} and {Sn} is given in Figure 9.2.1. The terms of {an} are growing, so the terms of the partial sums {Sn} are growing even faster, illustrating that the series diverges.

  2. (b)

    The sequence {bn} starts with 1,-1,1,-1,. Consider some of the partial sums Sn of {bn}:

    S1 =1
    S2 =0
    S3 =1
    S4 =0

    This pattern repeats; we find that Sn={1n is odd0n is even. As {Sn} oscillates, repeating 1,0,1,0,, we conclude that limnSn does not exist, hence n=1(-1)n+1 diverges.

    margin: 510-1-0.50.51nybnSn Figure 9.2.2: Scatter plots relating to the series of Example 9.2.1 part 2. Λ

    A scatter plot of the sequence {bn} and the partial sums {Sn} is given in Figure 9.2.2. When n is odd, bn=Sn so the marks for bn are drawn oversized to show they coincide.

While it is important to recognize when a series diverges, we are generally more interested in the series that converge. In this section we will demonstrate a few general techniques for determining convergence; later sections will delve deeper into this topic.

Geometric Series

One important type of series is a geometric series.

Definition 9.2.2 Geometric Series

A geometric series is a series of the form


Note that the index starts at n=0. If the index starts at n=1 we have n=1arn-1.

We started this section with a geometric series, although we dropped the first term of 1. One reason geometric series are important is that they have nice convergence properties.

Theorem 9.2.1 Convergence of Geometric Series

Consider the geometric series n=0arn.

  1. (a)

    If r1, the nth partial sum is: Sn=k=0n-1ark=a(1-rn)1-r.

  2. (b)

    The series converges if, and only if, |r|<1. When |r|<1,

  • Proof

    If r=1, then Sn=a+a+a++a=na. Since limnSn=±, the geometric series diverges.
    If r1, we have


    Multiply each term by r and we have


    Subtract these two equations and solve for Sn.

    Sn-rSn =a-arn
    Sn =a(1-rn)1-r

    From Theorem 9.1.4, we know that if -1<r<1, then limnrn=0 so


    So when |r|<1 the geometric series converges and its sum is a1-r.

    If either r-1 or r>1, the sequence {rn} is divergent by Theorem 9.1.4. Thus limnSn does not exist, so the geometric series diverges if r-1 or r>1. ∎

According to Theorem 9.2.1, the series


converges as r=1/2, and n=012n=11-1/2=2. This concurs with our introductory example; while there we got a sum of 1, we skipped the first term of 1.

margin: 24681012nyanSn (a) 246810-1-0.50.51nyanSn (b) 2465001,000nyanSn (c) Figure 9.2.3: Scatter plots relating to the series in Example 9.2.2. Λ
Example 9.2.2 Exploring geometric series

Check the convergence of the following series. If the series converges, find its sum.

1. n=2(34)n  2. n=0(-12)n  3. n=03n


  1. (a)

    Since r=3/4<1, this series converges. By Theorem 9.2.1, we have that


    However, note the subscript of the summation in the given series: we are to start with n=2. Therefore we subtract off the first two terms, giving:


    This is illustrated in Figure 9.2.3(a).

  2. (b)

    Since |r|=1/2<1, this series converges, and by Theorem 9.2.1,


    The partial sums of this series are plotted in Figure 9.2.3(b). Note how the partial sums are not purely increasing as some of the terms of the sequence {(-1/2)n} are negative.

  3. (c)

    Since r>1, the series diverges. (This makes “common sense”; we expect the sum


    to diverge.) This is illustrated in Figure 9.2.3(c).

Later sections will provide tests by which we can determine whether or not a given series converges. This, in general, is much easier than determining what a given series converges to. There are many cases, though, where the sum can be determined.

Example 9.2.3 Telescoping series

Evaluate the sum  n=1(1n-1n+1).

SolutionIt will help to write down some of the first few partial sums of this series.

S1 =11-12 =1-12
S2 =(11-12)+(12-13) =1-13
S3 =(11-12)+(12-13)+(13-14) =1-14
S4 =(11-12)+(12-13)+(13-14)+(14-15) =1-15
margin: 2468100.51nyanSn Figure 9.2.4: Scatter plots relating to the series of Example 9.2.3. Λ

Note how most of the terms in each partial sum subtract out. In general, we see that Sn=1-1n+1. This means that the sequence {Sn} converges, as


and so we conclude that n=1(1n-1n+1)=1. Partial sums of the series are plotted in Figure 9.2.4.

The series in Example 9.2.3 is an example of a telescoping series. Informally, a telescoping series is one in which the partial sums reduce to just a fixed number of terms. The partial sum Sn did not contain n terms, but rather just two: 1 and 1/(n+1).

When possible, seek a way to write an explicit formula for the nth partial sum Sn. This makes evaluating the limit limnSn much more approachable. We do so in the next example.

Example 9.2.4 Evaluating series

Evaluate each of the following infinite series.

1.n=12n2+2n    2.n=1ln(n+1n)


  1. (a)

    We can decompose the fraction 2/(n2+2n) as


    (See Section 8.4, Partial Fraction Decomposition, to recall how this is done, if necessary.)

    Expressing the terms of {Sn} is now more instructive:

    S1 =1-13 =1-13
    S2 =(1-13)+(12-14) =1+12-13-14
    S3 =(1-13)+(12-14)+(13-15) =1+12-14-15
    S4 =(1-13)+(12-14)+(13-15)+(14-16) =1+12-15-16
    S5 =(1-13)+(12-14)+(13-15)+(14-16)+(15-17) =1+12-16-17
    margin: 2468100.511.5nyanSn Figure 9.2.5: Scatter plots relating to the series of Example 9.2.4 part 1. Λ

    We again have a telescoping series. In each partial sum, most of the terms pair up to add to zero and we obtain the formula Sn=1+12-1n+1-1n+2. Taking limits allows us to determine the convergence of the series:


    so n=11n2+2n=32. This is illustrated in Figure 9.2.5.

  2. (b)

    We begin by writing the first few partial sums of the series:

    S1 =ln(2)
    S2 =ln(2)+ln(32)
    S3 =ln(2)+ln(32)+ln(43)
    S4 =ln(2)+ln(32)+ln(43)+ln(54)

    At first, this does not seem helpful, but recall the logarithmic identity: lnx+lny=ln(xy). Applying this to S4 gives:


    We must generalize this for Sn.

    margin: 5010024nyanSn Figure 9.2.6: Scatter plots relating to the series of Example 9.2.4 part 2. Λ

    We can conclude that {Sn}={ln(n+1)}. This sequence does not converge, as limnSn=. Therefore n=1ln(n+1n)=; the series diverges. Note in Figure 9.2.6 how the sequence of partial sums grows slowly; after 100 terms, it is not yet over 5. Graphically we may be fooled into thinking the series converges, but our analysis above shows that it does not.

We are learning about a new mathematical object, the series. As done before, we apply “old” mathematics to this new topic.

Theorem 9.2.2 Properties of Infinite Series

Suppose that  n=1an and  n=1bn are convergent series, and that  n=1an=L, n=1bn=K, and c is a constant.

  1. (a)

    Constant Multiple Rule: n=1can=cn=1an=cL.

  2. (b)

    Sum/Difference Rule: n=1(an±bn)=n=1an±n=1bn=L±K.

Before using this theorem, we will consider the harmonic series n=11n.

Example 9.2.5 Divergence of the Harmonic Series

Show that the harmonic series n=11n diverges.

SolutionWe will use a proof by contradiction here. Suppose the harmonic series converges to S. That is


We then have

S 1+12+14+14+16+16+18+18+

This gives us S12+S which can never be true, thus our assumption that the harmonic series converges must be false. Therefore, the harmonic series diverges.

It may take a while before one is comfortable with this statement, whose truth lies at the heart of the study of infinite series: it is possible that the sum of an infinite list of nonzero numbers is finite. We have seen this repeatedly in this section, yet it still may “take some getting used to.”

As one contemplates the behavior of series, a few facts become clear.

  1. (a)

    In order to add an infinite list of nonzero numbers and get a finite result, “most” of those numbers must be “very near” 0.

  2. (b)

    If a series diverges, it means that the sum of an infinite list of numbers is not finite (it may approach ± or it may oscillate), and:

    1. (a)

      The series will still diverge if the first term is removed.

    2. (b)

      The series will still diverge if the first 10 terms are removed.

    3. (c)

      The series will still diverge if the first 1,000,000 terms are removed.

    4. (d)

      The series will still diverge if any finite number of terms from anywhere in the series are removed.

These concepts are very important and lie at the heart of the next two theorems.

Theorem 9.2.3 Convergence of Sequence

If the series n=1an converges, then limnan=0.

  • Proof

    Let Sn=a1+a2++an. We have

    Sn =a1+a2++an-1+an
    Sn =Sn-1+an
    an =Sn-Sn-1

    Since nan converges, the sequence {Sn} converges. Let limnSn=S. As n, n-1 also goes to , so limnSn-1=S. We now have

    limnan =limn(Sn-Sn-1)
Theorem 9.2.4 Test for Divergence

If limnan does not exist or limnan0, then the series n=1an diverges.

The Test for Divergence follows from Theorem 9.2.3. If the series does not diverge, it must converge and therefore limnan=0.

Note that the two statements in Theorems 9.2.3 and 9.2.4 are really the same. In order to converge, the terms of the sequence must approach 0; if they do not, the series will not converge.

Looking back, we can apply this theorem to the series in Example 9.2.1. In that example, we had {an}={n2} and {bn}={(-1)n+1}.



limnbn=limn(-1)n+1 which does not exist.

Thus by the Test for Divergence, both series will diverge.

Important! This theorem does not state that if limnan=0 then n=1an converges. The standard example of this is the Harmonic Series, as given in Example 9.2.5. The Harmonic Sequence, {1/n}, converges to 0; the Harmonic Series, n=11/n, diverges.

Theorem 9.2.5 Infinite Nature of Series

The convergence or divergence of a series remains unchanged by the insertion or deletion of any finite number of terms. That is:

  1. (a)

    A divergent series will remain divergent with the insertion or deletion of any finite number of terms.

  2. (b)

    A convergent series will remain convergent with the insertion or deletion of any finite number of terms. (Of course, the sum will likely change.)

In other words, when we are only interested in the convergence or divergence of a series, it is safe to ignore the first few billion terms.

Example 9.2.6 Removing Terms from the Harmonic Series

Consider once more the Harmonic Series n=11n which diverges; that is, the partial sums SN=n=1N1n grow (very, very slowly) without bound. One might think that by removing the “large” terms of the sequence that perhaps the series will converge. This is simply not the case. For instance, the sum of the first 10 million terms of the Harmonic Series is about 16.7. Removing the first 10 million terms from the Harmonic Series changes the partial sums, effectively subtracting 16.7 from the sum. However, a sequence that is growing without bound will still grow without bound when 16.7 is subtracted from it.

The equation below illustrates this. Even though we have subtracted off the first 10 million terms, this only subtracts a constant off of an expression that is still growing to infinity. Therefore, the modified series is still growing to infinity.


This section introduced us to series and defined a few special types of series whose convergence properties are well known. We know when a geometric series converges or diverges. Most series that we encounter are not one of these types, but we are still interested in knowing whether or not they converge. The next three sections introduce tests that help us determine whether or not a given series converges.

Exercises 9.2


Terms and Concepts

  1. 1.

    Use your own words to describe how sequences and series are related.

  2. 2.

    Use your own words to define a partial sum.

  3. 3.

    Given a series n=1an, describe the two sequences related to the series that are important.

  4. 4.

    Use your own words to explain what a geometric series is.

  5. 5.

    T/F: If {an} is convergent, then n=1an is also convergent.

  6. 6.

    T/F: If {an} converges to 0, then n=0an converges.


In Exercises 7–14., a series n=1an is given.

  1. (a)

    Give the first 5 partial sums of the series.

  2. (b)

    Give a graph of the first 5 terms of an and Sn on the same axes.

  1. 7.


  2. 8.


  3. 9.


  4. 10.


  5. 11.


  6. 12.


  7. 13.


  8. 14.


In Exercises 15–32., state whether the given series converges or diverges and provide justification for your conclusion.

  1. 15.


  2. 16.


  3. 17.


  4. 18.


  5. 19.


  6. 20.


  7. 21.


  8. 22.


  9. 23.


  10. 24.


  11. 25.


  12. 26.


  13. 27.


  14. 28.


  15. 29.


  16. 30.


  17. 31.


  18. 32.


In Exercises 33–48., a series is given.

  1. (a)

    Find a formula for Sn, the nth partial sum of the series.

  2. (b)

    Determine whether the series converges or diverges. If it converges, state what it converges to.

  1. 33.


  2. 34.


  3. 35.


  4. 36.


  5. 37.


  6. 38.


  7. 39.


  8. 40.


  9. 41.


  10. 42.


  11. 43.


  12. 44.


  13. 45.


  14. 46.


  15. 47.


  16. 48.


In Exercises 49–52., find the values of x for which the series converges.

  1. 49.


  2. 50.


  3. 51.


  4. 52.


In Exercises 53–58., use Theorem 9.2.4 to show the given series diverges.

  1. 53.


  2. 54.


  3. 55.


  4. 56.


  5. 57.


  6. 58.


  1. 59.

    Show the series n=1n(2n-1)(2n+1) diverges.

  2. 60.

    Rewrite 0.121212¯ as an infinite series and then express the sum as the quotient of two integers (thus showing that it is a rational number — this method can be generalized to show that every repeating decimal is rational).

  3. 61.

    A ball falling from a height of h meters is known to rebound to a height of rh meters, where the proportionality constant r satisfies 0<r<1. Find the total distance traveled (vertically) by the ball if it is dropped initially from a height of 2 meters.

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