9 Sequences and Series

9.5 Alternating Series and Absolute Convergence

The series convergence tests we have used require that the underlying sequence {an} be a positive sequence. (We can relax this with Theorem 9.2.5 and state that there must be an N>0 such that an>0 for all n>N; that is, {an} is positive for all but a finite number of values of n.)

In this section we explore series whose summation includes negative terms. We start with a very specific form of series, where the terms of the summation alternate between being positive and negative.

Definition 9.5.1 Alternating Series

Let {bn} be a positive sequence. An alternating series is a series of either the form

n=1(-1)nbn  or  n=1(-1)n+1bn.

We want to think that an alternating sequence {an} is related to a positive sequence {bn} by an=(-1)nbn.

Recall that the terms of Harmonic Series come from the Harmonic Sequence {bn}={1n}. An important alternating series is the Alternating Harmonic Series:


Geometric Series can also be alternating series when r<0. For instance, if r=-1/2, the geometric series is


Theorem 9.2.1 states that geometric series converge when |r|<1 and gives the sum: n=0rn=11-r. When r=-1/2 as above, we find


A powerful convergence theorem exists for other alternating series that meet a few conditions.

Theorem 9.5.1 Alternating Series Test

Let {bn} be a positive, decreasing sequence where limnbn=0. Then

n=1(-1)nbn  and  n=1(-1)n+1bn


The basic idea behind Theorem 9.5.1 is illustrated in Figure 9.5.1. A positive, decreasing sequence {bn} is shown along with the partial sums


Because {bn} is decreasing, the amount by which Sn bounces up and down decreases. Moreover, the odd terms of Sn form a decreasing, bounded sequence, while the even terms of Sn form an increasing, bounded sequence. Since bounded, monotonic sequences converge (see Theorem 9.1.6) and the terms of {bn} approach 0, we will show below that the odd and even terms of Sn converge to the same common limit L, the sum of the series.

margin: L2468100.51nybnSn Figure 9.5.1: Illustrating convergence with the Alternating Series Test. Λ
  • Proof

    Because {bn} is a decreasing sequence, we have bn-bn+10. We will consider the even and odd partial sums separately. First consider the even partial sums.

    S2 =b1-b20 since b2 b1
    S4 =b1-b2+b3-b4=S2+b3-b4S2 since b3-b4 0
    S6 =S4+b5-b6S4 since b5-b6 0
    S2n =S2n-2+b2n-1-b2nS2n-2 since b2n-1-b2n 0

    We now have


    so {S2n} is an increasing sequence. But we can also write

    S2n =b1-b2+b3-b4+b5--b2n-2+b2n-1-b2n

    Each term in parentheses is positive and b2n is positive so we have S2nb1 for all n. We now have the sequence of even partial sums, {S2n}, is increasing and bounded above so by Theorem 9.1.6 {S2n} converges. Since we know it converges, we will assume it’s limit is L or


    Next we determine the limit of the sequence of odd partial sums.

    limnS2n+1 =limn(S2n+b2n+1)

    Both the even and odd partial sums converge to L so we have limnSn=L, which means the series is convergent. ∎

Example 9.5.1 Applying the Alternating Series Test

Determine if the Alternating Series Test applies to each of the following series.

1. n=1(-1)n+11n  2. n=2(-1)nlnnn  3. n=1(-1)n+1|sinn|n2


  1. (a)

    This is the Alternating Harmonic Series as seen previously. The underlying sequence is {bn}={1/n}, which is positive, decreasing, and approaches 0 as n. Therefore we can apply the Alternating Series Test and conclude this series converges. While the test does not state what the series converges to, we will see later that n=1(-1)n+11n=ln2.

  2. (b)

    The underlying sequence is {bn}={(lnn)/n}. This is positive for n2 and limnlnnn=limn1n=0 (use L’Hôpital’s Rule). However, the sequence is not decreasing for all n. It is straightforward to compute b10.347, b20.366, and b30.347: the sequence is increasing for at least the first 2 terms. We do not immediately conclude that we cannot apply the Alternating Series Test. Rather, consider the long-term behavior of {bn}. Treating bn=b(n) as a continuous function of n defined on [2,), we can take its derivative:


    The derivative is negative for all n3 (actually, for all n>e), meaning b(n)=bn is decreasing on [3,). We can apply the Alternating Series Test to the series when we start with n=3 and conclude that n=3(-1)nlnnn converges; adding the terms with n=2 does not change the convergence (i.e., we apply Theorem 9.2.5). The important lesson here is that as before, if a series fails to meet the criteria of the Alternating Series Test on only a finite number of terms, we can still apply the test.

  3. (c)

    The underlying sequence is {bn}={|sinn|/n2}. This sequence is positive and approaches 0 as n. However, it is not a decreasing sequence; the value of |sinn| oscillates between 0 and 1 as n. We cannot remove a finite number of terms to make {bn} decreasing, therefore we cannot apply the Alternating Series Test. Keep in mind that this does not mean we conclude the series diverges; in fact, it does converge. We are just unable to conclude this based on Theorem 9.5.1.

One of the famous results of mathematics is that the Harmonic Series, n=11n diverges, yet the Alternating Harmonic Series, n=1(-1)n+11n, converges. The notion that alternating the signs of the terms in a series can make a series converge leads us to the following definitions.

Definition 9.5.2 Absolute and Conditional Convergence

(a) A series n=1an converges absolutely if n=1|an| converges. (b) A series n=1an converges conditionally if n=1an converges but n=1|an| diverges.

margin: Note: In Definition 9.5.2, n=1an is not necessarily an alternating series; it just may have some negative terms. Λ

Thus we say the Alternating Harmonic Series converges conditionally.

Example 9.5.2 Determining absolute and conditional convergence.

Determine if the following series converge absolutely, conditionally, or diverge.
(a) n=1(-1)nn+3n2+2n+5 (b) n=3(-1)n3n-35n-10


  1. (a)

    We can show the series


    diverges using the Limit Comparison Test, comparing with 1/n. The sequence {n+3n2+2n+5} is monotonically decreasing, so that the series n=1(-1)nn+3n2+2n+5 converges using the Alternating Series Test; we conclude it converges conditionally.

  2. (b)

    The series


    diverges using the Test for Divergence, so it does not converge absolutely. The series n=3(-1)n3n-35n-10 fails the conditions of the Alternating Series Test as (3n-3)/(5n-10) does not approach 0 as n. We can state further that this series diverges; as n, the series effectively adds and subtracts 3/5 over and over. This causes the sequence of partial sums to oscillate and not converge. Therefore the series n=1(-1)n3n-35n-10 diverges.

Knowing that a series converges absolutely allows us to make two important statements, given in the following theorem. The first is that absolute convergence is “stronger” than regular convergence. That is, just because n=1an converges, we cannot conclude that n=1|an| will converge, but knowing a series converges absolutely tells us that n=1an will converge.

Theorem 9.5.2 Absolute Convergence Theorem

Let n=1an be a series that converges absolutely.

  1. (a)

    n=1an converges.

  2. (b)

    Let {bn} be any rearrangement of the sequence {an}. Then


One reason this is important is that our convergence tests all require that the underlying sequence of terms be positive. By taking the absolute value of the terms of a series where not all terms are positive, we are often able to apply an appropriate test and determine absolute convergence. This, in turn, determines that the series we are given also converges.

The second statement relates to rearrangements of series. When dealing with a finite set of numbers, the sum of the numbers does not depend on the order which they are added. (So 1+2+3=3+1+2.) One may be surprised to find out that when dealing with an infinite set of numbers, the same statement does not always hold true: some infinite lists of numbers may be rearranged in different orders to achieve different sums. The theorem states that the terms of an absolutely convergent series can be rearranged in any way without affecting the sum.

The theorem states that rearranging the terms of an absolutely convergent series does not affect its sum. This implies that perhaps the sum of a conditionally convergent series can change based on the arrangement of terms. Indeed, it can. The Riemann Rearrangement Theorem (named after Bernhard Riemann) states that any conditionally convergent series can have its terms rearranged so that the sum is any desired value or infinity.

Before we consider an example, we state the following theorem that illustrates how the alternating structure of an alternating series is a powerful tool when approximating the sum of a convergent series.

Theorem 9.5.3 The Alternating Series Approximation Theorem

Let {bn} be a sequence that satisfies the hypotheses of the Alternating Series Test, and let Sn and L be the nth partial sum and sum, respectively, of either n=1(-1)nbn or n=1(-1)n+1bn. Then

  1. (a)

    |Sn-L|<bn+1, and

  2. (b)

    L is between Sn and Sn+1.

Part 1 of Theorem 9.5.3 states that the nth partial sum of a convergent alternating series will be within bn+1 of its total sum. Consider the alternating series we looked at before the statement of the theorem, n=1(-1)n+1n2. Since b14=1/1420.0051, we know that S13 is within 0.0051 of the total sum.

Moreover, Part 2 of the theorem states that since S130.8252 and S140.8201, we know the sum L lies between 0.8201 and 0.8252. One use of this is the knowledge that S14 is accurate to two places after the decimal.

Some alternating series converge slowly. In Example 9.5.1 we determined the series n=2(-1)n+1lnnn converged. With n=1001, we find (lnn)/n0.0069, meaning that S10000.1633 is accurate to one, maybe two, places after the decimal. Since S10010.1564, we know the sum L is 0.1564L0.1633.

Example 9.5.3 Approximating the sums of convergent alternating series

Approximate the sum of the following series, accurate to within 0.001.

1. n=1(-1)n+11n3  2. n=1(-1)n+1lnnn.


  1. (a)

    Using Theorem 9.5.3, we want to find n where 1/n30.001:

    1n3 0.001=11000
    n3 1000
    n 10003
    n 10.

    Let L be the sum of this series. By Part 1 of the theorem, |S9-L|<b10=1/1000. We can compute S9=0.902116, which our theorem states is within 0.001 of the total sum. We can use Part 2 of the theorem to obtain an even more accurate result. As we know the 10th term of the series is -1/1000, we can easily compute S10=0.901116. Part 2 of the theorem states that L is between S9 and S10, so 0.901116<L<0.902116.

  2. (b)

    We want to find n where (lnn)/n0.001. We start by solving (lnn)/n=0.001 for n. This cannot be solved algebraically, so we will use Newton’s Method to approximate a solution. Let f(x)=ln(x)/x-0.001; we want to know where f(x)=0. We make a guess that x must be “large,” so our initial guess will be x1=1000. Recall how Newton’s Method works: given an approximate solution xn, our next approximation xn+1 is given by


    We find f(x)=(1-ln(x))/x2. This gives

    x2 =1000-ln(1000)/1000-0.001(1-ln(1000))/10002

    Using a computer, we find that Newton’s Method seems to converge to a solution x=9118.01 after 8 iterations. Taking the next integer higher, we have n=9119, where ln(9119)/9119=0.000999903<0.001. Again using a computer, we find S9118=-0.160369. Part 1 of the theorem states that this is within 0.001 of the actual sum L. Already knowing the 9,119th term, we can compute S9119=-0.159369, meaning


Notice how the first series converged quite quickly, where we needed only 10 terms to reach the desired accuracy, whereas the second series took over 9,000 terms.

We now consider the Alternating Harmonic Series once more. We have stated that


(see Example 9.5.1).

Consider the rearrangement where every positive term is followed by two negative terms:


(Convince yourself that these are exactly the same numbers as appear in the Alternating Harmonic Series, just in a different order.) Now group some terms and simplify:

(1-12)-14+(13-16)-18+(15-110)-112+ =
12-14+16-18+110-112+ =
12(1-12+13-14+15-16+) =12ln2.

By rearranging the terms of the series, we have arrived at a different sum. (One could try to argue that the Alternating Harmonic Series does not actually converge to ln2, because rearranging the terms of the series shouldn’t change the sum. However, the Alternating Series Test proves this series converges to L, for some number L, and if the rearrangement does not change the sum, then L=L/2, implying L=0. But the Alternating Series Approximation Theorem quickly shows that L>0. The only conclusion is that the rearrangement did change the sum.) This is an incredible result.

We mentioned earlier that the Integral Test did not work well with series containing factorial terms. The next section introduces the Ratio Test, which does handle such series well. We also introduce the Root Test, which is good for series where each term is raised to a power.

Exercises 9.5


Terms and Concepts

  1. 1.

    Why is n=1sinn not an alternating series?

  2. 2.

    A series n=1(-1)nan converges when {an} is xxxxxx, xxxxxx and limnan=xxxxxx.

  3. 3.

    Give an example of a series where n=0an converges but n=0|an| does not.

  4. 4.

    The sum of a xxxxxx convergent series can be changed by rearranging the order of its terms.


In Exercises 5–18., an alternating series n=ian is given.

  1. (a)

    Determine if the series converges or diverges.

  2. (b)

    Determine if n=0|an| converges or diverges.

  3. (c)

    If n=0an converges, determine if the convergence is conditional or absolute.

  1. 5.


  2. 6.


  3. 7.


  4. 8.


  5. 9.


  6. 10.


  7. 11.


  8. 12.


  9. 13.


  10. 14.


  11. 15.


  12. 16.


  13. 17.


  14. 18.


Let Sn be the nth partial sum of a series. In Exercises 19–22., a convergent alternating series is given and a value of n. Compute Sn and Sn+1 and use these values to find bounds on the sum of the series.

  1. 19.

    n=1(-1)nln(n+1),  n=5

  2. 20.

    n=1(-1)n+1n4,  n=4

  3. 21.

    n=0(-1)nn!,  n=6

  4. 22.

    n=0(-12)n,  n=9

In Exercises 23–26., a convergent alternating series is given along with its sum and a value of ϵ. Use Theorem 9.5.3 to find n such that the nth partial sum of the series is within ϵ of the sum of the series.

  1. 23.

    n=1(-1)n+1n4=7π4720,  ϵ=0.001

  2. 24.

    n=0(-1)nn!=1e,  ϵ=0.0001

  3. 25.

    n=0(-1)n2n+1=π4,  ϵ=0.001

  4. 26.

    n=0(-1)n(2n)!=cos1,  ϵ=10-8

  1. 27.

    The partial sums in problems 23. and 25. can be used to approximate π. Using the values of n from these problems, compute the respective partial sums and then use them to approximate π. Which gives a better estimate of π?

  2. 28.
    The book shows a rearrangement of the Alternating Harmonic Series, (1-12)-14+(13-16)-18+(15-110)-112+ which gives the sum 12ln2. Note that the terms in parentheses are positive, so if we simplified those terms we would have an alternating series. Without actually simplifying, show that the same scheme of rearranging terms so that each positive term is followed by two successive negative terms yields (1-12)-14-18+(13-16)-112-116+(15-110)-120-124+ and then show that this new rearrangement of the Alternating Harmonic Series has the sum 14ln2. Hint: move each right parenthesis one term to the right and then simplify inside the parentheses.
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