Motivating Questions
- What is an alternating series?
- Under what conditions does an alternating series converge? Why?
- How well does the th 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
has and so that every other term alternates in sign. This series converges to
In Preview Activity and our following discussion, we investigate the behavior of similar series where consecutive terms have opposite signs.
Preview Activity
Preview Activity 8.3.1 showed how we can approximate the number with linear, quadratic, and other polynomial approximations. We use a similar approach in this activity to obtain linear and quadratic approximations to 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
- Find the tangent line to at and use this linearization to approximate That is, find the tangent line approximation to and use the fact that to estimate
- The linearization of does not provide a very good approximation to since is not that close to To obtain a better approximation, we alter our approach; instead of using a straight line to approximate we use a quadratic function to account for the concavity of for close to With the linearization, both the function's value and slope agree with the linearization's value and slope at We will now make a quadratic approximation to centered at with the property that and
- Let Show that and Use to approximate by using the fact that
- We can continue approximating with polynomials of larger degree whose derivatives agree with those of at This makes the polynomials fit the graph of better for more values of around For example, let Show that and Taking a similar approach to preceding questions, use to approximate
- If we used a degree or degree polynomial to approximate what approximations of do you think would result? Use the preceding questions to conjecture a pattern that holds, and state the degree and degree approximation.
The Alternating Series Test
Preview Activity gives us several approximations to The linear approximation is and the quadratic approximation is If we continue this process, cubic, quartic (degree ), quintic (degree ), and higher degree polynomials give us the approximations to in Table .
| linear | ||
|---|---|---|
| quadratic | ||
| cubic | ||
| quartic | ||
| quintic |
The pattern here shows that can be approximated by the partial sums of the infinite series
where the alternating signs are indicated by the factor
Using computational technology, we find that the sum of the first 100 terms in this series is 0.6881721793. As a comparison,
Definition 8.4.1 8.4.1
An alternating series is a series of the form
where
We have some flexibility in how we write an alternating series; for example, the series
whose index starts at
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 8.4.2 8.4.2
Remember that, by definition, a series converges if and only if its corresponding sequence of partial sums converges.
- Calculate the first few partial sums (to 10 decimal places) of the alternating series
∑ k = 1 ∞ ( − 1 ) k + 1 1 k . \sum_{k=1}^{\infty} (-1)^{k+1}\frac{1}{k}\text{.} Label each partial sum with the notation
for an appropriate choice ofS n = ∑ k = 1 n ( − 1 ) k + 1 1 k S_n = \sum_{k=1}^{n} (-1)^{k+1}\frac{1}{k} n . n\text{.} - Plot the sequence of partial sums from part (a). What do you notice about this sequence?
Activity
Recall that if
in which the sequence
Notice that
and sinceS 2 = a 1 − a 2 , S_2 = a_1 - a_2\text{,} we havea 1 > a 2 a_1 \gt a_2 0 < S 2 < S 1 . 0 \lt S_2 \lt S_1 \text{.} and soS 3 = S 2 + a 3 S_3 = S_2+a_3 ButS 2 < S 3 . S_2 \lt S_3\text{.} soa 3 < a 2 , a_3 \lt a_2\text{,} Thus,S 3 < S 1 . S_3 \lt S_1\text{.} 0 < S 2 < S 3 < S 1 . 0 \lt S_2 \lt S_3 \lt S_1 \text{.} and soS 4 = S 3 − a 4 S_4 = S_3-a_4 ButS 4 < S 3 . S_4 \lt S_3\text{.} soa 4 < a 3 , a_4 \lt a_3\text{,} Thus,S 2 < S 4 . S_2 \lt S_4\text{.} 0 < S 2 < S 4 < S 3 < S 1 . 0 \lt S_2 \lt S_4 \lt S_3 \lt S_1 \text{.} and soS 5 = S 4 + a 5 S_5 = S_4+a_5 ButS 4 < S 5 . S_4 \lt S_5\text{.} soa 5 < a 4 , a_5 \lt a_4\text{,} Thus,S 5 < S 3 . S_5 \lt S_3\text{.} 0 < S 2 < S 4 < S 5 < S 3 < S 1 . 0 \lt S_2 \lt S_4 \lt S_5 \lt S_3 \lt S_1 \text{.}
This pattern continues as illustrated in Figure

Note further that the absolute value of the difference between the
Because the sequence
The preceding discussion has demonstrated the truth of the Alternating Series Test.
The Alternating Series Test
Given an alternating series
Note that if the limit of the sequence
Activity 8.4.3 8.4.3
Which series converge and which diverge? Justify your answers.
∑ k = 1 ∞ ( − 1 ) k k 2 + 2 \displaystyle \displaystyle\sum_{k=1}^{\infty} \frac{(-1)^k}{k^2+2} ∑ k = 1 ∞ ( − 1 ) k + 1 2 k k + 5 \displaystyle \displaystyle\sum_{k=1}^{\infty} \frac{(-1)^{k+1}2k}{k+5} ∑ k = 2 ∞ ( − 1 ) k ln ( 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
Recall that the sequence of partial sums oscillates around the sum
Therefore, the value of the term
Alternating Series Estimation Theorem
If the alternating series
Example 8.4.1 8.4.1
Determine how well the
approximates the sum of the series.
- Answer
-
If we let
be the sum of the seriesS S then we know that∑ k = 1 ∞ ( − 1 ) k + 1 k , \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}\text{,} ∣ S 100 − S ∣ < a 101 . \left| S_{100} - S \right| \lt a_{101}\text{.} Now
a 101 = 1 101 ≈ 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 + 1 k = ln ( 2 ) , S = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} = \ln(2)\text{,} and so
whileS ≈ 0.693147 S \approx 0.693147 S 100 = ∑ k = 1 100 ( − 1 ) k + 1 k ≈ 0.6881721793 . S_{100} = \sum_{k=1}^{100} \frac{(-1)^{k+1}}{k} \approx 0.6881721793\text{.} We see that the actual difference between
andS S is approximatelyS 100 S_{100} which is indeed less than0.0049750013 , 0.0049750013\text{,} 0.0099 . 0.0099\text{.}
Activity 8.4.4 8.4.4
Determine the number of terms it takes to approximate the sum of the convergent alternating series
to within 0.0001.
Absolute and Conditional Convergence
A series such as
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 8.4.5 8.4.5
- Explain why the series
1 − 1 4 − 1 9 + 1 16 + 1 25 + 1 36 − 1 49 − 1 64 − 1 81 − 1 100 + ⋯ 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 ∞ 1 k 2 . \sum_{k=1}^{\infty} \frac{1}{k^2}\text{.} - Explain why the series
1 − 1 4 − 1 9 + 1 16 + 1 25 + 1 36 − 1 49 − 1 64 − 1 81 − 1 100 + ⋯ 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 ∞ − 1 k 2 . \sum_{k=1}^{\infty} -\frac{1}{k^2}\text{.} - Given that the terms in the series
1 − 1 4 − 1 9 + 1 16 + 1 25 + 1 36 − 1 49 − 1 64 − 1 81 − 1 100 + ⋯ 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
As we just observed, this is the case for the series (
Definition 8.4.6 8.4.6
Consider a series
- The series
converges absolutely (or is absolutely convergent) provided that∑ a k \sum a_k converges.∑ ∣ a k ∣ \sum | a_k | - The series
converges conditionally (or is conditionally convergent) provided that∑ a k \sum a_k diverges and∑ ∣ a k ∣ \sum | a_k | converges.∑ a k \sum a_k
In this terminology, the series (
Activity 8.4.6 8.4.6
- Consider the series
∑ ( − 1 ) k ln ( k ) k . \sum (-1)^k \frac{\ln(k)}{k}\text{.} - Does this series converge? Explain.
- Does this series converge absolutely? Explain what test you use to determine your answer.
- Consider the series
∑ ( − 1 ) k ln ( k ) k 2 . \sum (-1)^k \frac{\ln(k)}{k^2}\text{.} - Does this series converge? Explain.
- Does this series converge absolutely? Hint: Use the fact that
for large values ofln ( k ) < k \ln(k) \lt \sqrt{k} and then compare to an appropriatek k -series.p p
Conditionally convergent series turn out to be very interesting. If the sequence
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
with ratio∑ a r k \sum ar^k converges forr r and diverges for− 1 < r < 1 -1 \lt r \lt 1 ∣ r ∣ ≥ 1 . |r| \geq 1\text{.} The sum of the convergent geometric series
is∑ k = 0 ∞ a r k \displaystyle \sum_{k=0}^{\infty} ar^k a 1 − r . \frac{a}{1-r}\text{.} - Divergence Test
-
If the sequence
does not converge to 0, then the seriesa n a_n diverges.∑ a k \sum a_k This is the first test to apply because the conclusion is simple. However, if
no conclusion can be drawn.lim n → ∞ a n = 0 , \lim_{n \to \infty} a_n = 0\text{,} - Integral Test
-
Let
be a positive, decreasing function on an intervalf f and let[ c , ∞ ) [c,\infty) for each positive integera k = f ( k ) a_k = f(k) k ≥ c . k \geq c\text{.} - If
converges, then∫ c ∞ f ( t ) d t \int_c^{\infty} f(t) \ dt converges.∑ a k \sum a_k - If
diverges, then∫ c ∞ f ( t ) d t \int_c^{\infty} f(t) \ dt diverges.∑ a k \sum a_k
Use this test when
is easy to integrate.f ( x ) f(x) - If
- Direct Comp. Test
-
(see Ex 4 in Section 8.3)
Let
for each positive integer0 ≤ a k ≤ b k 0 \leq a_k \leq b_k k . k\text{.} - If
converges, then∑ b k \sum b_k converges.∑ a k \sum a_k - If
diverges, then∑ a k \sum a_k diverges.∑ b k \sum b_k
Use this test when you have a series with known behavior that you can compare to — this test can be difficult to apply.
- If
- Limit Comp. Test
-
Let
anda n a_n be sequences of positive terms. Ifb n b_n lim k → ∞ a k b k = L \displaystyle \lim_{k \to \infty} \frac{a_k}{b_k} = L for some positive finite number
then the two seriesL , L\text{,} and∑ a k \sum a_k either both converge or both diverge.∑ b k \sum b_k 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
for eacha k ≠ 0 a_k \neq 0 and supposek k lim k → ∞ ∣ a k + 1 ∣ ∣ a k ∣ = r . \displaystyle \lim_{k \to \infty} \frac{|a_{k+1}|}{|a_k|} = r\text{.} - If
then the seriesr < 1 , r \lt 1\text{,} converges absolutely.∑ a k \sum a_k - If
then the seriesr > 1 , r \gt 1\text{,} diverges.∑ a k \sum a_k - If
then test is inconclusive.r = 1 , r=1\text{,}
This test is useful when a series involves factorials and powers.
- If
- Root Test
-
(see Exercise 2 in Section 8.3)
Let
for eacha k ≥ 0 a_k \geq 0 and supposek k lim k → ∞ a k k = r . \displaystyle \lim_{k \to \infty} \sqrt[k]{a_k} = r\text{.} - If
then the seriesr < 1 , r \lt 1\text{,} converges.∑ a k \sum a_k - If
then the seriesr > 1 , r \gt 1\text{,} diverges.∑ a k \sum a_k - If
then test is inconclusive.r = 1 , r=1\text{,}
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
involvesa k a_k th powers.k k - If
- Alt. Series Test
-
If
is a positive, decreasing sequence so thata n a_n then the alternating serieslim n → ∞ a n = 0 , \displaystyle \lim_{n \to \infty} a_n = 0\text{,} converges.∑ ( − 1 ) k + 1 a k \sum (-1)^{k+1} a_k This test applies only to alternating series — we assume that the terms
are all positive and that the sequencea n a_n is decreasing.{ a n } \{a_n\} - Alt. Series Est.
-
Let
be theS n = ∑ k = 1 n ( − 1 ) k + 1 a k S_n = \displaystyle \sum_{k=1}^n (-1)^{k+1} a_k th partial sum of the alternating seriesn n Assume∑ k = 1 ∞ ( − 1 ) k + 1 a k . \displaystyle \sum_{k=1}^{\infty} (-1)^{k+1} a_k\text{.} for each positive integera n > 0 a_n \gt 0 the sequencen , n\text{,} decreases to 0 anda n a_n Then it follows thatlim n → ∞ S n = S . \displaystyle \lim_{n \to \infty} S_n = S\text{.} ∣ S − S n ∣ < a n + 1 . |S - S_n| \lt a_{n+1}\text{.} This bound can be used to determine the accuracy of the partial sum
as an approximation of the sum of a convergent alternating series.S n S_n
Activity 8.4.7 8.4.7
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.
∑ k = 3 ∞ 2 k − 2 \displaystyle \displaystyle\sum_{k=3}^{\infty} \ \frac{2}{\sqrt{k-2}} ∑ k = 1 ∞ k 1 + 2 k \displaystyle \displaystyle\sum_{k=1}^{\infty} \ \frac{k}{1+2k} ∑ k = 0 ∞ 2 k 2 + 1 k 3 + k + 1 \displaystyle \displaystyle\sum_{k=0}^{\infty} \ \frac{2k^2+1}{k^3+k+1} ∑ k = 0 ∞ 100 k k ! \displaystyle \displaystyle\sum_{k=0}^{\infty} \ \frac{100^k}{k!} ∑ k = 1 ∞ 2 k 5 k \displaystyle \displaystyle\sum_{k=1}^{\infty} \ \frac{2^k}{5^k} ∑ k = 1 ∞ k 3 − 1 k 5 + 1 \displaystyle \displaystyle\sum_{k=1}^{\infty} \ \frac{k^3-1}{k^5+1} ∑ k = 2 ∞ 3 k − 1 7 k \displaystyle \displaystyle\sum_{k=2}^{\infty} \ \frac{3^{k-1}}{7^k} ∑ k = 2 ∞ 1 k k \displaystyle \displaystyle\sum_{k=2}^{\infty} \ \frac{1}{k^k} ∑ k = 1 ∞ ( − 1 ) k + 1 k + 1 \displaystyle \displaystyle\sum_{k=1}^{\infty} \ \frac{(-1)^{k+1}}{\sqrt{k+1}} ∑ k = 2 ∞ 1 k ln ( k ) \displaystyle \displaystyle\sum_{k=2}^{\infty} \ \frac{1}{k \ln(k)} - Determine a value of
so that then n th partial sumn n of the alternating seriesS n S_n approximates the sum to within 0.001.∑ n = 2 ∞ ( − 1 ) n ln ( n ) \displaystyle\sum_{n=2}^{\infty} \frac{(-1)^n}{\ln(n)}
Summary
- An alternating series is a series whose terms alternate in sign. It has the form
∑ ( − 1 ) k a k \sum (-1)^ka_k where
is a positive real number for eacha k a_k k . k\text{.} - The sequence of partial sums of a convergent alternating series oscillates around the sum of the series if the sequence of
th terms converges to 0. That is why the Alternating Series Test shows that the alternating seriesn n converges whenever the sequence∑ k = 1 ∞ ( − 1 ) k a k \sum_{k=1}^{\infty} (-1)^ka_k of{ a n } \{a_n\} th terms decreases to 0.n n - The difference between the
st partial sumn − 1 n-1 and theS n − 1 S_{n-1} th partial sumn n of a convergent alternating seriesS n S_n is∑ k = 1 ∞ ( − 1 ) k a k \sum_{k=1}^{\infty} (-1)^ka_k Since the partial sums oscillate around the sum∣ S n − S n − 1 ∣ = a n . |S_n - S_{n-1}| = a_n\text{.} of the series, it follows thatS S ∣ S − S n ∣ < a n . |S - S_n| \lt a_n\text{.} So the
th partial sum of a convergent alternating seriesn n approximates the actual sum of the series to within∑ k = 1 ∞ ( − 1 ) k a k \sum_{k=1}^{\infty} (-1)^ka_k a n . a_n\text{.}