Skip to main content
Logo image

APEX Calculus: for University of Lethbridge

Section 10.5 Alternating Series and Absolute Convergence

All of the series convergence tests we have used require that the underlying sequence {an} be a positive sequence. (We can relax this with Theorem 10.2.24 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 10.5.1. Alternating Series.

Let {an} be a positive sequence. An alternating series is a series of either the form
n=1(1)nan or n=1(1)n+1an.
Recall the terms of Harmonic Series come from the Harmonic Sequence {an}={1/n}. An important alternating series is the Alternating Harmonic Series:
n=1(1)n+11n=112+1314+1516+
Geometric Series can also be alternating series when r<0. For instance, if r=1/2, the geometric series is
n=0(12)n=112+1418+116132+
Theorem 10.2.7 states that geometric series converge when |r|<1 and gives the sum: n=0rn=11r. When r=1/2 as above, we find
n=0(12)n=11(1/2)=13/2=23.
A powerful convergence theorem exists for other alternating series that meet a few conditions.
Figure 10.5.3. Video presentation of Definition 10.5.1 and Theorem 10.5.2
The basic idea behind Theorem 10.5.2 is illustrated in Figure 10.5.4–10.5.5. A positive, decreasing sequence {an} is shown along with the partial sums
Sn=i=1n(1)i+1ai=a1a2+a3a4++(1)n+1an.
Because {an} is decreasing, the amount by which Sn bounces up/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 10.1.32) and the terms of {an} approach 0, one can show the odd and even terms of Sn converge to the same common limit L, the sum of the series.
Scatter plots of a positive, decreasing sequence and its alternating sequence of partial sums.
On one set of coordinate axes, two scatter plots are shown. The first is the plot of a positive, decreasing sequence an; the second is the plot of the sequence of partial sums for the corresponding alternating sequence (1)nan.
The scatter plots illustrate why an alternating series converges: as n increases, the partial sums oscillate back and forth across a horizontal line marked L (the limiting value). Since an is a decreasing sequence, the oscillations get smaller as n increases, and the points in the scatter plot for Sn get closer and closer to the line y=L.
Figure 10.5.4. Illustrating convergence with the Alternating Series Test
An illustration of the partial sums in an alternating series, using line segments to demonstrate convergence.
On a set of coordinate axes, a sequence of horizontal line segments is shown. At the top of the image is a line segment indicating in increase from 0 to a1. An arrow pointing to the right is at the end of the segment.
Below this segment is a shorter segment with an arrow pointing to the left. The right end of this segment aligns with the right end of the first segment, indicating that we obtain a1a2 by starting at a1 and then moving to the left by a distance a2.
The third segment is below the second. Its left end aligns with the left end of the second segment, and it points to the right, indicating the act of adding a3 to the partial sum. The length of this segment is shorter than that of the second, indicating the fact that the sequence an is decreasing.
As we move down, additional segments are drawn, alternating between pointing left and right, and getting shorter with each step. Below the segments is a horizontal axis, on which the values S1,S2, of the partial sums are shown, as well as the limit L. The illustration overall is intended to convey the idea that the line segments will shrink toward the limiting value L as n increases.
Figure 10.5.5. A visual representation of adding terms of an alternating series. The arrows represent the length and direction of each term of the sequence.

Example 10.5.6. 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=1(1)nln(n)n
  3. n=1(1)n+1|sin(n)|n2
Solution 1.
  1. This is the Alternating Harmonic Series as seen previously. The underlying sequence is {an}={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=ln(2).
  2. The underlying sequence is {an}={ln(n)/n}. This is positive and approaches 0 as n (use L’Hospital’s Rule). However, the sequence is not decreasing for all n. It is straightforward to compute a1=0, a20.347, a30.366, and a40.347: the sequence is increasing for at least the first 3 terms. We do not immediately conclude that we cannot apply the Alternating Series Test. Rather, consider the long-term behavior of {an}. Treating an=a(n) as a continuous function of n defined on [1,), we can take its derivative:
    a(n)=1ln(n)n2.
    The derivative is negative for all n3 (actually, for all n>e), meaning a(n)=an 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)nln(n)n converges; adding the terms with n=1 and n=2 do not change the convergence (i.e., we apply Theorem 10.2.24). 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. The underlying sequence is {an}=|sin(n)|/n. This sequence is positive and approaches 0 as n. However, it is not a decreasing sequence; the value of |sin(n)| oscillates between 0 and 1 as n. We cannot remove a finite number of terms to make {an} 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 10.5.2. We will be able to show that this series converges shortly.
Solution 2. Video solution
Key Idea 10.2.20 gives the sum of some important series. Two of these are
n=11n2=π261.64493 and n=1(1)n+1n2=π2120.82247.
These two series converge to their sums at different rates. To be accurate to two places after the decimal, we need 202 terms of the first series though only 13 of the second. To get 3 places of accuracy, we need 1069 terms of the first series though only 33 of the second. Why is it that the second series converges so much faster than the first?
While there are many factors involved when studying rates of convergence, the alternating structure of an alternating series gives us a powerful tool when approximating the sum of a convergent series.
Figure 10.5.8. Video presentation of Theorem 10.5.7
Part 1 of Theorem 10.5.7 states that the nth partial sum of a convergent alternating series will be within an+1 of its total sum. You can see this visually in Figure 10.5.5. Look at the distance between S6 and L. Clearly this distance is less than the length of the arrow corresponding to a7.
Also consider the alternating series we looked at before the statement of the theorem, n=1(1)n+1n2. Since a14=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 10.5.6 we determined the series n=1(1)n+1ln(n)n converged. With n=1001, we find ln(n)/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 10.5.9. Approximating the sum 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+1ln(n)n
Solution 1.
  1. Using Theorem 10.5.7, we want to find n where 1/n30.001. That is, we want to find the the first time a term in the sequence an is smaller than the desired level of error:
    1n30.001=11000n31000n10003n10.
    Let L be the sum of this series. By Part 1 of the theorem, |S9L|<a10=1/1000. (We found a10=an+1<0.0001, so n=9). 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)n/103=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. We want to find n where ln(n)/n<0.001. We start by solving ln(n)/n=0.001 for n. This cannot be solved algebraically, so we will use Newton’s Method to approximate a solution. (Note: we can also use a “Brute Force” technique. That is, we can guess and check numerically until we find a solution.) Let f(x)=ln(x)/x0.001; 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
    xn+1=xnf(xn)f(xn).
    We find f(x)=(1ln(x))/x2. This gives
    x2=1000ln(1000)/10000.001(1ln(1000))/10002=2000.
    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 0.159369<L<0.160369.
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.
Solution 2. Video solution
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 10.5.10. Absolute and Conditional Convergence.

  1. A series n=1an converges absolutely if n=1|an| converges.
  2. A series n=1an converges conditionally if n=1an converges but n=1|an| diverges.
Thus we say the Alternating Harmonic Series converges conditionally.

Example 10.5.11. Determining absolute and conditional convergence.

Determine if the following series converge absolutely, conditionally, or diverge.
  1. n=1(1)nn+3n2+2n+5
  2. n=1(1)nn2+2n+52n
  3. n=3(1)n3n35n10
Solution 1.
  1. We can show the series
    n=1|(1)nn+3n2+2n+5|=n=1n+3n2+2n+5
    diverges using the Limit Comparison Test, comparing with 1/n. The series n=1(1)nn+3n2+2n+5 converges using the Alternating Series Test; we conclude it converges conditionally.
  2. We can show the series
    n=1|(1)nn2+2n+52n|=n=1n2+2n+52n
    converges using the Ratio Test. Therefore we conclude n=1(1)nn2+2n+52n converges absolutely.
  3. The series
    n=3|(1)n3n35n10|=n=33n35n10
    diverges using the nth Term Test, so it does not converge absolutely. The series n=3(1)n3n35n10 fails the conditions of the Alternating Series Test as (3n3)/(5n10) 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)n3n35n10 diverges.
Solution 2. Video solution
Knowing that a series converges absolutely allows us to make two important statements, given in Theorem 10.5.13 below. 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.
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.
Figure 10.5.12. Video presentation of Definition 10.5.10 and Theorem 10.5.13

Proof.

We will provide a proof for Part 1 of Absolute Convergence Theorem. Suppose that n=1|an| converges. We start by noting that for any sequence an, we have
|an|an|an|
If we add |an| to all three sides:
0an+|an|2|an|.
We are now in a position to apply the Direct Comparison Test to the series n=1(an+|an|). Since n=1|an| converges by our supposition, so does n=12|an| (the scalar multiple of a convergent series also converges by Theorem 10.2.19). Therefore n=1(an+|an|) converges by the Direct Comparison Test.
Now we turn our attention to n=1an. We can say
n=1an=n=1(an+|an||an|)=n=1(an+|an|)n=1|an|.
The last line is the difference between two convergent series, which is also convergent by Theorem 10.2.19. Therefore n=1an converges.
In Example 10.5.11, we determined the series in Part 2 converges absolutely. Theorem 10.5.13 tells us the series converges (which we could also determine using the Alternating Series Test).
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, including !
As an example, consider the Alternating Harmonic Series once more. We have stated that
n=1(1)n+11n=112+1314+1516+17=ln(2),
Consider the rearrangement where every positive term is followed by two negative terms:
11214+131618+15110112
(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:
(112)14+(1316)18+(15110)112+=1214+1618+110112+=12(112+1314+1516+)=12ln(2).
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 ln(2), 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 end here our study of tests to determine convergence. The end of this text contains a table summarizing the tests that one may find useful.
While series are worthy of study in and of themselves, our ultimate goal within calculus is the study of Power Series, which we will consider in the next section. We will use power series to create functions where the output is the result of an infinite summation.

Exercises Exercises

Terms and Concepts

1.
Why is n=1sin(n) not an alternating series?
2.
A series n=1(1)nan converges when {an} is ,  and limnan =.
3.
Give an example of a series where n=0an converges but n=0|an| does not.
4.
The sum of a  convergent series can be changed by rearranging the order of its terms.

Problems

Exercise Group.
In the following exercises, an alternating series n=ian is given.
  1. Determine if the series converges or diverges.
  2. Determine if n=0|an| converges or diverges.
  3. If n=0an converges, determine if the convergence is conditional or absolute.
9.
n=0(1)n+13n+5n23n+1
12.
n=1(1)n+11+3+5++(2n1)
14.
n=2sin((n+1/2)π)nln(n)
Exercise Group.
Let Sn be the nth partial sum of a series. In the following exercises 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.
21.
n=1(1)nln(n+1), n=5
22.
n=1(1)n+1n4, n=4
Exercise Group.
In the following exercises, a convergent alternating series is given along with its sum and a value of ε. Use Theorem 10.5.7 to find n such that the nth partial sum of the series is within ε of the sum of the series.
25.
n=1(1)n+1n4=7π4720, ε=0.001
26.
n=0(1)nn!=1e, ε=0.0001
27.
n=0(1)n2n+1=π4, ε=0.001
28.
n=0(1)n(2n)!=cos(1), ε=108