Active Calculus · 8: Sequences and Series

8.4: Alternating Series

Motivating Questions
  • What is an alternating series?
  • Under what conditions does an alternating series converge? Why?
  • How well does the nnth partial sum of a convergent alternating series approximate the actual sum of the series? Why?

So far, we've considered series with exclusively nonnegative terms. Next, we consider series that have some negative terms. For instance, the geometric series

2−43+89−⋯+2(−23)n+⋯, 2 - \frac{4}{3} + \frac{8}{9} - \cdots + 2 \left(-\frac{2}{3} \right)^n + \cdots\text{,}

has a=2a = 2 and r=−23,r = -\frac{2}{3}\text{,} so that every other term alternates in sign. This series converges to

S=a1−r=21−(−23)=65. S = \frac{a}{1-r} = \frac{2}{1- \left(-\frac{2}{3}\right)} = \frac{6}{5}\text{.}

In Preview Activity 8.4.18.4.1 and our following discussion, we investigate the behavior of similar series where consecutive terms have opposite signs.

Preview Activity 8.4.18.4.1

Preview Activity 8.3.1 showed how we can approximate the number ee with linear, quadratic, and other polynomial approximations. We use a similar approach in this activity to obtain linear and quadratic approximations to ln⁡(2).\ln(2)\text{.} Along the way, we encounter a type of series that is different than most of the ones we have seen so far. Throughout this activity, let f(x)=ln⁡(1+x).f(x) = \ln(1+x)\text{.}

  1. Find the tangent line to ff at x=0x=0 and use this linearization to approximate ln⁡(2).\ln(2)\text{.} That is, find L(x),L(x)\text{,} the tangent line approximation to f(x),f(x)\text{,} and use the fact that L(1)≈f(1)L(1) \approx f(1) to estimate ln⁡(2).\ln(2)\text{.}
  2. The linearization of ln⁡(1+x)\ln(1+x) does not provide a very good approximation to ln⁡(2)\ln(2) since 11 is not that close to 0.0\text{.} To obtain a better approximation, we alter our approach; instead of using a straight line to approximate ln⁡(2),\ln(2)\text{,} we use a quadratic function to account for the concavity of ln⁡(1+x)\ln(1+x) for xx close to 0.0\text{.} With the linearization, both the function's value and slope agree with the linearization's value and slope at x=0.x=0\text{.} We will now make a quadratic approximation P2(x)P_2(x) to f(x)=ln⁡(1+x)f(x) = \ln(1+x) centered at x=0x=0 with the property that P2(0)=f(0),P_2(0) = f(0)\text{,} P2′(0)=f′(0),P'_2(0) = f'(0)\text{,} and P2′′(0)=f′′(0).P''_2(0) = f''(0)\text{.}
    1. Let P2(x)=x−x22.P_2(x) = x - \frac{x^2}{2}\text{.} Show that P2(0)=f(0),P_2(0) = f(0)\text{,} P2′(0)=f′(0),P'_2(0) = f'(0)\text{,} and P2′′(0)=f′′(0).P''_2(0) = f''(0)\text{.} Use P2(x)P_2(x) to approximate ln⁡(2)\ln(2) by using the fact that P2(1)≈f(1).P_2(1) \approx f(1)\text{.}
    2. We can continue approximating ln⁡(2)\ln(2) with polynomials of larger degree whose derivatives agree with those of ff at 0.0\text{.} This makes the polynomials fit the graph of ff better for more values of xx around 0.0\text{.} For example, let P3(x)=x−x22+x33.P_3(x) = x - \frac{x^2}{2}+\frac{x^3}{3}\text{.} Show that P3(0)=f(0),P_3(0) = f(0)\text{,} P3′(0)=f′(0),P'_3(0) = f'(0)\text{,} P3′′(0)=f′′(0),P''_3(0) = f''(0)\text{,} and P3′′′(0)=f′′′(0).P'''_3(0) = f'''(0)\text{.} Taking a similar approach to preceding questions, use P3(x)P_3(x) to approximate ln⁡(2).\ln(2)\text{.}
    3. If we used a degree 44 or degree 55 polynomial to approximate ln⁡(1+x),\ln(1+x)\text{,} what approximations of ln⁡(2)\ln(2) do you think would result? Use the preceding questions to conjecture a pattern that holds, and state the degree 44 and degree 55 approximation.

The Alternating Series Test

Preview Activity 8.4.18.4.1 gives us several approximations to ln⁡(2).\ln(2)\text{.} The linear approximation is 1,1\text{,} and the quadratic approximation is 1−12=12.1 - \frac{1}{2} = \frac{1}{2}\text{.} If we continue this process, cubic, quartic (degree 44), quintic (degree 55), and higher degree polynomials give us the approximations to ln⁡(2)\ln(2) in Table 8.4.18.4.1.

Table 8.4.18.4.1.
linear 11 11
quadratic 1−121 - \frac{1}{2} 0.50.5
cubic 1−12+131 - \frac{1}{2} + \frac{1}{3} 0.83‾0.8\overline{3}
quartic 1−12+13−141 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} 0.583‾0.58\overline{3}
quintic 1−12+13−14+151 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} 0.783‾0.78\overline{3}

The pattern here shows that ln⁡(2)\ln(2) can be approximated by the partial sums of the infinite series

∑k=1∞(−1)k+11k(8.4.1) \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{k}\tag{8.4.18.4.1}

where the alternating signs are indicated by the factor (−1)k+1.(-1)^{k+1}\text{.} We call such a series an alternating series.

Using computational technology, we find that the sum of the first 100 terms in this series is 0.6881721793. As a comparison, ln⁡(2)≈0.6931471806.\ln(2) \approx 0.6931471806\text{.} This shows that even though the series (8.4.18.4.1) converges to ln⁡(2),\ln(2)\text{,} it must do so quite slowly, since the sum of the first 100 terms isn't particularly close to ln⁡(2).\ln(2)\text{.} We will investigate the issue of how quickly an alternating series converges later in this section.

Definition 8.4.18.4.1

An alternating series is a series of the form

∑k=0∞(−1)kak, \sum_{k=0}^{\infty} (-1)^k a_k\text{,}

where ak>0a_k \gt 0 for each k.k\text{.}

We have some flexibility in how we write an alternating series; for example, the series

∑k=1∞(−1)k+1ak, \sum_{k=1}^{\infty} (-1)^{k+1} a_k\text{,}

whose index starts at k=1,k = 1\text{,} is also alternating. As we will soon see, there are several very nice results that hold for alternating series, while alternating series can also demonstrate some unusual behaivior.

It is important to remember that most of the series tests we have seen in previous sections apply only to series with nonnegative terms. Alternating series require a different test.

Activity 8.4.28.4.2

Remember that, by definition, a series converges if and only if its corresponding sequence of partial sums converges.

  1. Calculate the first few partial sums (to 10 decimal places) of the alternating series
    ∑k=1∞(−1)k+11k. \sum_{k=1}^{\infty} (-1)^{k+1}\frac{1}{k}\text{.}

    Label each partial sum with the notation Sn=∑k=1n(−1)k+11kS_n = \sum_{k=1}^{n} (-1)^{k+1}\frac{1}{k} for an appropriate choice of n.n\text{.}

  2. Plot the sequence of partial sums from part (a). What do you notice about this sequence?

Activity 8.4.28.4.2 illustrates the general behavior of any convergent alternating series. We see that the partial sums of the alternating harmonic series oscillate around a fixed number that turns out to be the sum of the series.

Recall that if lim⁡k→∞ak≠0,\lim_{k \to \infty} a_k \neq 0\text{,} then the series ∑ak\sum a_k diverges by the Divergence Test. From this point forward, we will thus only consider alternating series

∑k=1∞(−1)k+1ak \sum_{k=1}^{\infty} (-1)^{k+1} a_k

in which the sequence aka_k consists of positive numbers that decrease to 0.0\text{.} The nnth partial sum SnS_n is

Sn=∑k=1n(−1)k+1ak. S_n = \sum_{k=1}^n (-1)^{k+1} a_k\text{.}

Notice that

  • S2=a1−a2,S_2 = a_1 - a_2\text{,} and since a1>a2a_1 \gt a_2 we have 0<S2<S1.0 \lt S_2 \lt S_1 \text{.}
  • S3=S2+a3S_3 = S_2+a_3 and so S2<S3.S_2 \lt S_3\text{.} But a3<a2,a_3 \lt a_2\text{,} so S3<S1.S_3 \lt S_1\text{.} Thus, 0<S2<S3<S1.0 \lt S_2 \lt S_3 \lt S_1 \text{.}
  • S4=S3−a4S_4 = S_3-a_4 and so S4<S3.S_4 \lt S_3\text{.} But a4<a3,a_4 \lt a_3\text{,} so S2<S4.S_2 \lt S_4\text{.} Thus, 0<S2<S4<S3<S1.0 \lt S_2 \lt S_4 \lt S_3 \lt S_1 \text{.}
  • S5=S4+a5S_5 = S_4+a_5 and so S4<S5.S_4 \lt S_5\text{.} But a5<a4,a_5 \lt a_4\text{,} so S5<S3.S_5 \lt S_3\text{.} Thus, 0<S2<S4<S5<S3<S1.0 \lt S_2 \lt S_4 \lt S_5 \lt S_3 \lt S_1 \text{.}

This pattern continues as illustrated in Figure 8.4.18.4.1 (with nn odd) so that each partial sum lies between the previous two partial sums.

8_4_Alternating_Series_Test.svg

Figure 8.4.18.4.1. Partial sums of an alternating series

Note further that the absolute value of the difference between the (n−1)(n-1)st partial sum Sn−1S_{n-1} and the nnth partial sum SnS_n is

∣Sn−Sn−1∣=an. \left\lvert S_n - S_{n-1} \right\rvert = a_n\text{.}

Because the sequence {an}\{a_n\} converges to 0,0\text{,} the distance between successive partial sums becomes as close to zero as we'd like, and thus the sequence of partial sums converges (even though we don't know the exact value to which it converges).

The preceding discussion has demonstrated the truth of the Alternating Series Test.

The Alternating Series Test

Given an alternating series ∑(−1)kak,\sum (-1)^k a_k \text{,} if the sequence {ak}\{a_k\} of positive terms decreases to 0 as k→∞,k \to \infty\text{,} then the alternating series converges.

Note that if the limit of the sequence {ak}\{a_k\} is not 0, then the alternating series diverges.

Activity 8.4.38.4.3

Which series converge and which diverge? Justify your answers.

  1. ∑k=1∞(−1)kk2+2\displaystyle \displaystyle\sum_{k=1}^{\infty} \frac{(-1)^k}{k^2+2}
  2. ∑k=1∞(−1)k+12kk+5\displaystyle \displaystyle\sum_{k=1}^{\infty} \frac{(-1)^{k+1}2k}{k+5}
  3. ∑k=2∞(−1)kln⁡(k)\displaystyle \displaystyle\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\ln(k)}

Estimating Alternating Sums

If the series converges, the argument for the Alternating Series Test also provides us with a method to determine how close the nnth partial sum SnS_n is to the actual sum of the series. To see how this works, let SS be the sum of a convergent alternating series, so

S=∑k=1∞(−1)kak. S = \sum_{k=1}^{\infty} (-1)^k a_k\text{.}

Recall that the sequence of partial sums oscillates around the sum SS so that

∣S−Sn∣<∣Sn+1−Sn∣=an+1. \left|S - S_n \right| \lt \left| S_{n+1} - S_n \right| = a_{n+1}\text{.}

Therefore, the value of the term an+1a_{n+1} provides an error estimate for how well the partial sum SnS_n approximates the actual sum S.S\text{.} We summarize this fact in the statement of the Alternating Series Estimation Theorem.

Alternating Series Estimation Theorem

If the alternating series ∑k=1∞(−1)k+1ak\sum_{k=1}^{\infty} (-1)^{k+1}a_k has positive terms aka_k that decrease to zero as k→∞,k \to \infty\text{,} and Sn=∑k=1n(−1)k+1akS_n = \sum_{k=1}^{n} (-1)^{k+1}a_k is the nnth partial sum of the alternating series, then

∣∑k=1∞(−1)k+1ak−Sn∣≤an+1. \left\lvert \sum_{k=1}^{\infty} (-1)^{k+1}a_k - S_n \right\rvert \leq a_{n+1}\text{.}

Example 8.4.18.4.1

Determine how well the 100100th partial sum S100S_{100} of

∑k=1∞(−1)k+1k \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}

approximates the sum of the series.

Answer

If we let SS be the sum of the series ∑k=1∞(−1)k+1k,\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}\text{,} then we know that

∣S100−S∣<a101. \left| S_{100} - S \right| \lt a_{101}\text{.}

Now

a101=1101≈0.0099, a_{101} = \frac{1}{101} \approx 0.0099\text{,}

so the 100th partial sum is within 0.0099 of the sum of the series. We have discussed the fact (and will later verify) that

S=∑k=1∞(−1)k+1k=ln⁡(2), S = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} = \ln(2)\text{,}

and so S≈0.693147S \approx 0.693147 while

S100=∑k=1100(−1)k+1k≈0.6881721793. S_{100} = \sum_{k=1}^{100} \frac{(-1)^{k+1}}{k} \approx 0.6881721793\text{.}

We see that the actual difference between SS and S100S_{100} is approximately 0.0049750013,0.0049750013\text{,} which is indeed less than 0.0099.0.0099\text{.}

Activity 8.4.48.4.4

Determine the number of terms it takes to approximate the sum of the convergent alternating series

∑k=1∞(−1)k+1k4 \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^4}

to within 0.0001.

Absolute and Conditional Convergence

A series such as

1−14−19+116+125+136−149−164−181−1100+⋯(8.4.2) 1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \cdots\tag{8.4.28.4.2}

whose terms are neither all nonnegative nor alternating is different from any series that we have considered so far. The behavior of such a series can be rather complicated, but there is an important connection between a series with some negative terms and series with all positive terms.

Activity 8.4.58.4.5
  1. Explain why the series
    1−14−19+116+125+136−149−164−181−1100+⋯ 1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \cdots

    must have a sum that is less than the series

    ∑k=1∞1k2. \sum_{k=1}^{\infty} \frac{1}{k^2}\text{.}
  2. Explain why the series
    1−14−19+116+125+136−149−164−181−1100+⋯ 1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \cdots

    must have a sum that is greater than the series

    ∑k=1∞−1k2. \sum_{k=1}^{\infty} -\frac{1}{k^2}\text{.}
  3. Given that the terms in the series
    1−14−19+116+125+136−149−164−181−1100+⋯ 1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \cdots

    converge to 0, what do you think the previous two results tell us about the convergence status of this series?

As the example in Activity 8.4.58.4.5 suggests, if a series ∑ak\sum a_k has some negative terms but ∑∣ak∣\sum |a_k| converges, then the original series, ∑ak,\sum a_k\text{,} must also converge. That is, if ∑∣ak∣\sum | a_k | converges, then so must ∑ak.\sum a_k\text{.}

As we just observed, this is the case for the series (8.4.28.4.2), because the corresponding series of the absolute values of its terms is the convergent pp-series ∑1k2.\sum \frac{1}{k^2}\text{.} But there are series, such as the alternating harmonic series ∑(−1)k+11k,\sum (-1)^{k+1} \frac{1}{k}\text{,} that converge while the corresponding series of absolute values, ∑1k,\sum \frac{1}{k}\text{,} diverges. We distinguish between these behaviors by introducing the following language.

Definition 8.4.68.4.6

Consider a series ∑ak.\sum a_k\text{.}

  1. The series ∑ak\sum a_k converges absolutely (or is absolutely convergent) provided that ∑∣ak∣\sum | a_k | converges.
  2. The series ∑ak\sum a_k converges conditionally (or is conditionally convergent) provided that ∑∣ak∣\sum | a_k | diverges and ∑ak\sum a_k converges.

In this terminology, the series (8.4.28.4.2) converges absolutely while the alternating harmonic series is conditionally convergent.

Activity 8.4.68.4.6
  1. Consider the series ∑(−1)kln⁡(k)k.\sum (-1)^k \frac{\ln(k)}{k}\text{.}
    1. Does this series converge? Explain.
    2. Does this series converge absolutely? Explain what test you use to determine your answer.
  2. Consider the series ∑(−1)kln⁡(k)k2.\sum (-1)^k \frac{\ln(k)}{k^2}\text{.}
    1. Does this series converge? Explain.
    2. Does this series converge absolutely? Hint: Use the fact that ln⁡(k)<k\ln(k) \lt \sqrt{k} for large values of kk and then compare to an appropriate pp-series.

Conditionally convergent series turn out to be very interesting. If the sequence {an}\{a_n\} decreases to 0, but the series ∑ak\sum a_k diverges, the conditionally convergent series ∑(−1)kak\sum (-1)^k a_k is right on the borderline of being a divergent series. As a result, any conditionally convergent series converges very slowly. Furthermore, some very strange things can happen with conditionally convergent series, as illustrated in some of the exercises.

Summary of Tests for Convergence of Series

We have discussed several tests for convergence/divergence of series in our sections and in exercises. We close this section of the text with a summary of all the tests we have encountered, followed by an activity that challenges you to decide which convergence test to apply to several different series.

Geometric Series

The geometric series ∑ark\sum ar^k with ratio rr converges for −1<r<1-1 \lt r \lt 1 and diverges for ∣r∣≥1.|r| \geq 1\text{.}

The sum of the convergent geometric series ∑k=0∞ark\displaystyle \sum_{k=0}^{\infty} ar^k is a1−r.\frac{a}{1-r}\text{.}

Divergence Test

If the sequence ana_n does not converge to 0, then the series ∑ak\sum a_k diverges.

This is the first test to apply because the conclusion is simple. However, if lim⁡n→∞an=0,\lim_{n \to \infty} a_n = 0\text{,} no conclusion can be drawn.

Integral Test

Let ff be a positive, decreasing function on an interval [c,∞)[c,\infty) and let ak=f(k)a_k = f(k) for each positive integer k≥c.k \geq c\text{.}

  • If ∫c∞f(t) dt\int_c^{\infty} f(t) \ dt converges, then ∑ak\sum a_k converges.
  • If ∫c∞f(t) dt\int_c^{\infty} f(t) \ dt diverges, then ∑ak\sum a_k diverges.

Use this test when f(x)f(x) is easy to integrate.

Direct Comp. Test

(see Ex 4 in Section 8.3)

Let 0≤ak≤bk0 \leq a_k \leq b_k for each positive integer k.k\text{.}

  • If ∑bk\sum b_k converges, then ∑ak\sum a_k converges.
  • If ∑ak\sum a_k diverges, then ∑bk\sum b_k diverges.

Use this test when you have a series with known behavior that you can compare to — this test can be difficult to apply.

Limit Comp. Test

Let ana_n and bnb_n be sequences of positive terms. If

lim⁡k→∞akbk=L \displaystyle \lim_{k \to \infty} \frac{a_k}{b_k} = L

for some positive finite number L,L\text{,} then the two series ∑ak\sum a_k and ∑bk\sum b_k either both converge or both diverge.

Easier to apply in general than the comparison test, but you must have a series with known behavior to compare. Useful to apply to series of rational functions.

Ratio Test

Let ak≠0a_k \neq 0 for each kk and suppose

lim⁡k→∞∣ak+1∣∣ak∣=r. \displaystyle \lim_{k \to \infty} \frac{|a_{k+1}|}{|a_k|} = r\text{.}
  • If r<1,r \lt 1\text{,} then the series ∑ak\sum a_k converges absolutely.
  • If r>1,r \gt 1\text{,} then the series ∑ak\sum a_k diverges.
  • If r=1,r=1\text{,} then test is inconclusive.

This test is useful when a series involves factorials and powers.

Root Test

(see Exercise 2 in Section 8.3)

Let ak≥0a_k \geq 0 for each kk and suppose

lim⁡k→∞akk=r. \displaystyle \lim_{k \to \infty} \sqrt[k]{a_k} = r\text{.}
  • If r<1,r \lt 1\text{,} then the series ∑ak\sum a_k converges.
  • If r>1,r \gt 1\text{,} then the series ∑ak\sum a_k diverges.
  • If r=1,r=1\text{,} then test is inconclusive.

In general, the Ratio Test can usually be used in place of the Root Test. However, the Root Test can be quick to use when aka_k involves kkth powers.

Alt. Series Test

If ana_n is a positive, decreasing sequence so that lim⁡n→∞an=0,\displaystyle \lim_{n \to \infty} a_n = 0\text{,} then the alternating series ∑(−1)k+1ak\sum (-1)^{k+1} a_k converges.

This test applies only to alternating series — we assume that the terms ana_n are all positive and that the sequence {an}\{a_n\} is decreasing.

Alt. Series Est.

Let Sn=∑k=1n(−1)k+1akS_n = \displaystyle \sum_{k=1}^n (-1)^{k+1} a_k be the nnth partial sum of the alternating series ∑k=1∞(−1)k+1ak.\displaystyle \sum_{k=1}^{\infty} (-1)^{k+1} a_k\text{.} Assume an>0a_n \gt 0 for each positive integer n,n\text{,} the sequence ana_n decreases to 0 and lim⁡n→∞Sn=S.\displaystyle \lim_{n \to \infty} S_n = S\text{.} Then it follows that ∣S−Sn∣<an+1.|S - S_n| \lt a_{n+1}\text{.}

This bound can be used to determine the accuracy of the partial sum SnS_n as an approximation of the sum of a convergent alternating series.

Activity 8.4.78.4.7

For (a)-(j), use appropriate tests to determine the convergence or divergence of the following series. Throughout, if a series is a convergent geometric series, find its sum.

  1. ∑k=3∞ 2k−2\displaystyle \displaystyle\sum_{k=3}^{\infty} \ \frac{2}{\sqrt{k-2}}
  2. ∑k=1∞ k1+2k\displaystyle \displaystyle\sum_{k=1}^{\infty} \ \frac{k}{1+2k}
  3. ∑k=0∞ 2k2+1k3+k+1\displaystyle \displaystyle\sum_{k=0}^{\infty} \ \frac{2k^2+1}{k^3+k+1}
  4. ∑k=0∞ 100kk!\displaystyle \displaystyle\sum_{k=0}^{\infty} \ \frac{100^k}{k!}
  5. ∑k=1∞ 2k5k\displaystyle \displaystyle\sum_{k=1}^{\infty} \ \frac{2^k}{5^k}
  6. ∑k=1∞ k3−1k5+1\displaystyle \displaystyle\sum_{k=1}^{\infty} \ \frac{k^3-1}{k^5+1}
  7. ∑k=2∞ 3k−17k\displaystyle \displaystyle\sum_{k=2}^{\infty} \ \frac{3^{k-1}}{7^k}
  8. ∑k=2∞ 1kk\displaystyle \displaystyle\sum_{k=2}^{\infty} \ \frac{1}{k^k}
  9. ∑k=1∞ (−1)k+1k+1\displaystyle \displaystyle\sum_{k=1}^{\infty} \ \frac{(-1)^{k+1}}{\sqrt{k+1}}
  10. ∑k=2∞ 1kln⁡(k)\displaystyle \displaystyle\sum_{k=2}^{\infty} \ \frac{1}{k \ln(k)}
  11. Determine a value of nn so that the nnth partial sum SnS_n of the alternating series ∑n=2∞(−1)nln⁡(n)\displaystyle\sum_{n=2}^{\infty} \frac{(-1)^n}{\ln(n)} approximates the sum to within 0.001.

Summary

  • An alternating series is a series whose terms alternate in sign. It has the form
    ∑(−1)kak \sum (-1)^ka_k

    where aka_k is a positive real number for each k.k\text{.}

  • The sequence of partial sums of a convergent alternating series oscillates around the sum of the series if the sequence of nnth terms converges to 0. That is why the Alternating Series Test shows that the alternating series ∑k=1∞(−1)kak\sum_{k=1}^{\infty} (-1)^ka_k converges whenever the sequence {an}\{a_n\} of nnth terms decreases to 0.
  • The difference between the n−1n-1st partial sum Sn−1S_{n-1} and the nnth partial sum SnS_n of a convergent alternating series ∑k=1∞(−1)kak\sum_{k=1}^{\infty} (-1)^ka_k is ∣Sn−Sn−1∣=an.|S_n - S_{n-1}| = a_n\text{.} Since the partial sums oscillate around the sum SS of the series, it follows that
    ∣S−Sn∣<an. |S - S_n| \lt a_n\text{.}

    So the nnth partial sum of a convergent alternating series ∑k=1∞(−1)kak\sum_{k=1}^{\infty} (-1)^ka_k approximates the actual sum of the series to within an.a_n\text{.}

Credit and licence

This page titled 8.4: Alternating Series was last modified on Fri, 29 Sep 2023 01:03:48 GMT and is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Matthew Boelkins, David Austin, and Steven Schlicker ( ScholarWorks @Grand Valley State University ) via source content that was edited to the style and standards of the LibreTexts platform.

Source: https://math.libretexts.org/Bookshelves/Calculus/Active_Calculus_(Boelkins_et_al.)/08%3A_Sequences_and_Series/8.04%3A_Alternating_Series

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