Motivating Questions
- What is an infinite series?
- What is the th partial sum of an infinite series?
- How do we add up an infinite number of numbers? In other words, what does it mean for an infinite series of real numbers to converge?
- What does it mean for an infinite series of real numbers to diverge?
In Section 8.2, we encountered infinite geometric series. For example, by writing
as a geometric series, we found a way to write as a single fraction: In this section, we explore other types of infinite sums. In Preview Activity we see how one such sum is related to the famous number
Preview Activity
Have you ever wondered how your calculator can produce a numeric approximation for complicated numbers like or After all, the only operations a calculator can really perform are addition, subtraction, multiplication, and division, the operations that make up polynomials. This activity provides the first steps in understanding how this process works. Throughout the activity, let
- Find the tangent line to at and use this linearization to approximate That is, find a formula for the tangent line, and compute since
- The linearization of does not provide a good approximation to since 1 is not very close to 0. To obtain a better approximation, we alter our approach a bit. Instead of using a straight line to approximate we put an appropriate bend in our estimating function to make it better fit the graph of for close to 0. With the linearization, we had both and share the same value as the linearization at We will now use a quadratic approximation to centered at which has the property that and
- Let Show that and Then, use to approximate by observing that
- We can continue approximating with polynomials of larger degree whose higher derivatives agree with those of at 0. This turns out to make the polynomials fit the graph of better for more values of around 0. For example, let Show that and Use to approximate in a way similar to how you did so with above.
Infinite Series
Preview Activity shows that an approximation to using a linear polynomial is 2, an approximation to using a quadratic polynomial is and an approximation using a cubic polynomial is 2.6667. If we continue this process we can obtain approximations from quartic (degree 4) and quintic (degree 5) polynomials, giving us the following approximations to
| linear | ||
|---|---|---|
| quadratic | ||
| cubic | ||
| quartic | ||
| quintic |
We see an interesting pattern here. The number is being approximated by the sum
for increasing values of
We can use summation notation as a shorthand 1 for writing the sum in Equation (
We can calculate this sum using as large an
This sum is an example of an infinite series. Note that the series (
Definition 8.3.1 8.3.1
An infinite series of real numbers is the sum of the entries in an infinite sequence of real numbers. In other words, an infinite series is sum of the form
where
We use summation notation to identify a series. If the series adds the entries of a sequence
or
Each of these notations is simply shorthand for the infinite sum
Is it even possible to sum an infinite list of numbers? That doing so is possible in some situations shouldn't come as a surprise. We have already examined the special case of geometric series in the previous section. Moreover, the definite integral is defineed as the limit of a sequence of finite sums, which provides insight into how we'll think of infinite sums generally. As we investigate other infinite sums, there are two key questions we seek to answer: (1) is the sum finite? and (2) if yes, what is its value?
Activity 8.3.2 8.3.2
Consider the series
While it is physically impossible to add an infinite collection of numbers, we can, of course, add any finite collection of them. In what follows, we investigate how understanding how to find the
- Sum the first two numbers in this series. That is, find a numeric value for
∑ k = 1 2 1 k 2 \sum_{k=1}^2 \frac{1}{k^2} - Next, add the first three numbers in the series.
- Continue adding terms in this series to complete the list below. Carry each sum to at least 8 decimal places.
∑ k = 1 1 1 k 2 = 1 \displaystyle \displaystyle\sum_{k=1}^{1} \frac{1}{k^2}=1 ∑ k = 1 2 1 k 2 = \displaystyle \displaystyle\sum_{k=1}^{2} \frac{1}{k^2}= ∑ k = 1 3 1 k 2 = \displaystyle \displaystyle\sum_{k=1}^{3} \frac{1}{k^2}= ∑ k = 1 4 1 k 2 = \displaystyle \displaystyle\sum_{k=1}^{4} \frac{1}{k^2}= ∑ k = 1 5 1 k 2 = 1.46361111 \displaystyle \displaystyle\sum_{k=1}^{5} \frac{1}{k^2}=\phantom{1.46361111} ∑ k = 1 6 1 k 2 = \displaystyle \displaystyle\sum_{k=1}^{6} \frac{1}{k^2}= ∑ k = 1 7 1 k 2 = \displaystyle \displaystyle\sum_{k=1}^{7} \frac{1}{k^2}= ∑ k = 1 8 1 k 2 = \displaystyle \displaystyle\sum_{k=1}^{8} \frac{1}{k^2}= ∑ k = 1 9 1 k 2 = \displaystyle \displaystyle\sum_{k=1}^{9} \frac{1}{k^2}= ∑ k = 1 10 1 k 2 = 1.549767731 \displaystyle \displaystyle\sum_{k=1}^{10} \frac{1}{k^2} = \phantom{1.549767731}
- The sums in the table in part c form a sequence whose
th term isn n Based on your calculations in the table, do you think the sequenceS n = ∑ k = 1 n 1 k 2 . S_n = \sum_{k=1}^{n} \frac{1}{k^2}\text{.} converges or diverges? Explain. How do you think this sequence{ S n } \{S_n\} is related to the series{ S n } \{S_n\} ∑ k = 1 ∞ 1 k 2 ? \sum_{k=1}^{\infty} \frac{1}{k^2}\text{?}
The example in Activity
Definition 8.3.2 8.3.2
The
In other words, the
We investigate the behavior of the series by examining the sequence
of its partial sums. If the sequence of partial sums converges to some finite number, we say that the corresponding series converges. Otherwise, we say the series diverges. From our work in Activity
appears to converge to some number near 1.54977. We formalize the concept of convergence and divergence of an infinite series in the following definition.
Definition 8.3.3 8.3.3
The infinite series
converges (or is convergent) if the sequence
If
If the sequence of partial sums does not converge, then the series
diverges (or is divergent).
The early terms in a series do not influence whether or not the series converges or diverges. Rather, the convergence or divergence of a series
is determined by what happens to the terms
Since
when we are interested in questions of convergence/divergence as opposed to the exact sum of a series.
In Section 8.2 we encountered the special family of infinite geometric series. Recall that a geometric series has the form
and thus a geometric series converges if
It is generally a difficult question to determine if a given nongeometric series converges or diverges. There are several tests we can use that we will consider in the following sections.
The Divergence Test
The first question we ask about any infinite series is usually “Does the series converge or diverge?” There is a straightforward way to check that certain series diverge, and we explore this test in the next activity.
Activity 8.3.3 8.3.3
If the series
Assume that the series
- What is the
th partial sumn n of the seriesS n S_n ∑ k = 1 ∞ a k ? \sum_{k=1}^{\infty} a_k\text{?} - What is the
st partial sum( n − 1 ) (n-1) of the seriesS n − 1 S_{n-1} ∑ k = 1 ∞ a k ? \sum_{k=1}^{\infty} a_k\text{?} - What is the difference between the
th partial sum and then n st partial sum of the series( n − 1 ) (n-1) ∑ k = 1 ∞ a k ? \sum_{k=1}^{\infty} a_k\text{?} - Since we are assuming that
what does that tell us about∑ k = 1 ∞ a k = L , \sum_{k=1}^{\infty} a_k = L\text{,} Why? What does that tell us aboutlim n → ∞ S n ? \lim_{n \to \infty} S_n\text{?} Why?lim n → ∞ S n − 1 ? \lim_{n \to \infty} S_{n-1}\text{?} - Combine the results of the previous two parts of this activity to determine
lim n → ∞ a n = lim n → ∞ ( S n − S n − 1 ) . \lim_{n \to \infty} a_n = \lim_{n \to \infty} (S_n - S_{n-1})\text{.}
The result of Activity
If the series
converges, then the sequence ∑ k = 1 ∞ a k \sum_{k = 1}^{\infty} a_k of { a k } \{a_k\} th terms converges to 0. k k
It is logically equivalent to say that if the sequence
The Divergence Test
If
Activity 8.3.4 8.3.4
Determine if the Divergence Test applies to the following series. If the test does not apply, explain why. If the test does apply, what does it tell us about the series?
∑ k k + 1 \displaystyle \sum \frac{k}{k+1} ∑ ( − 1 ) k \displaystyle \sum (-1)^k ∑ 1 k \displaystyle \sum \frac{1}{k}
Note well: be very careful with the Divergence Test. This test only tells us what happens to a series if the terms of the corresponding sequence do not converge to 0. If the sequence of the terms of the series does converge to 0, the Divergence Test does not apply: indeed, as we will soon see, a series whose terms go to zero may either converge or diverge.
The Integral Test
The Divergence Test settles the questions of divergence or convergence of series
Next, we consider the harmonic series 2
The first 9 partial sums of this series are shown following.
This information alone doesn't seem to be enough to tell us if the series
Activity 8.3.5 8.3.5
Consider the harmonic series
Through this last expression for

The graph of the continuous function
- Explain how this picture represents a particular Riemann sum.
- What is the definite integral that corresponds to the Riemann sum you considered in (a)?
- Which is larger, the definite integral in (b), or the corresponding partial sum
of the series? Why?S 9 S_9 - If instead of considering the 9th partial sum, we consider the
th partial sum, and we letn n go to infinity, we can then compare the seriesn n to the improper integral∑ k = 1 ∞ 1 k \sum_{k=1}^{\infty} \frac{1}{k} Which of these quantities is larger? Why?∫ 1 ∞ 1 x d x . \int_{1}^{\infty} \frac{1}{x} \ dx\text{.} - Does the improper integral
converge or diverge? What does that result, together with your work in (d), tell us about the series∫ 1 ∞ 1 x d x \int_{1}^{\infty} \frac{1}{x} \ dx ∑ k = 1 ∞ 1 k ? \sum_{k=1}^{\infty} \frac{1}{k}\text{?}
The ideas from Activity
can always be viewed as a left hand Riemann sum of
Taking the limit as
Therefore, if the improper integral

What's more, if we look at the right hand Riemann sums of
So if
The Integral Test
Let
- If the improper integral
converges, then the series∫ c ∞ f ( x ) d x \int_{c}^{\infty} f(x) \, dx converges.∑ k = 1 ∞ a k \sum_{k=1}^{\infty} a_k - If the improper integral
diverges, then the series∫ c ∞ f ( x ) d x \int_{c}^{\infty} f(x) \, dx diverges.∑ k = 1 ∞ a k \sum_{k=1}^{\infty} a_k
The Integral Test compares a given infinite series to a natural, corresponding improper integral and says that the infinite series and corresponding improper integral both have the same convergence status. In the next activity, we apply the Integral Test to determine the convergence or divergence of a class of important series.
Activity 8.3.6 8.3.6
The series
- Evaluate the improper integral
Does the series∫ 1 ∞ 1 x 2 d x . \int_1^{\infty} \frac{1}{x^2} \ dx\text{.} converge or diverge? Explain.∑ k = 1 ∞ 1 k 2 \sum_{k=1}^{\infty} \frac{1}{k^2} - Evaluate the improper integral
where∫ 1 ∞ 1 x p d x \int_1^{\infty} \frac{1}{x^p} \ dx For which values ofp > 1 . p \gt 1\text{.} can we conclude that the seriesp p converges?∑ k = 1 ∞ 1 k p \sum_{k=1}^{\infty} \frac{1}{k^p} - Evaluate the improper integral
where∫ 1 ∞ 1 x p d x \int_1^{\infty} \frac{1}{x^p} \ dx What does this tell us about the correspondingp < 1 . p \lt 1\text{.} -seriesp p ∑ k = 1 ∞ 1 k p ? \sum_{k=1}^{\infty} \frac{1}{k^p}\text{?} - Summarize your work in this activity by completing the following statement.
The
-seriesp p converges if and only if .∑ k = 1 ∞ 1 k p \sum_{k=1}^{\infty} \frac{1}{k^p}
The Limit Comparison Test
The Integral Test allows us to determine the convergence of an entire family of series: the
Activity 8.3.7 8.3.7
Consider the series
- By computing the value of
fork + 1 k 3 + 2 \frac{k+1}{k^3+2} andk = 100 k = 100 explain why the termsk = 1000 , k = 1000\text{,} are essentiallyk + 1 k 3 + 2 \frac{k+1}{k^3+2} whenk k 3 \frac{k}{k^3} is large.k k - Let's formalize our observations in (a) a bit more. Let
anda k = k + 1 k 3 + 2 a_k = \frac{k+1}{k^3+2} Calculateb k = k k 3 . b_k = \frac{k}{k^3}\text{.} lim k → ∞ a k b k . \lim_{k \to \infty} \frac{a_k}{b_k}\text{.} What does the value of the limit tell you about
anda k a_k for large values ofb k b_k Compare your response from part (a).k ? k\text{?} - Does the series
converge or diverge? Why? What do you think that tells us about the convergence or divergence of the series∑ k k 3 \sum \frac{k}{k^3} Explain.∑ k + 1 k 3 + 2 ? \sum \frac{k+1}{k^3+2}\text{?}
Activity
then
Since multiplying by a nonzero constant does not affect the convergence or divergence of a series, it follows that the series
The Limit Comparison Test
Let
for some positive (finite) constant
The Limit Comparison Test shows that if we have a series
Activity 8.3.8 8.3.8
Use the Limit Comparison Test to determine the convergence or divergence of the series
by comparing it to the series
The Ratio Test
The Limit Comparison Test works well if we can find a series with known behavior to compare. But such series are not always easy to find. In this section we will examine a test that allows us to examine the behavior of a series by comparing it to a geometric series, without knowing in advance which geometric series we need.
Activity 8.3.9 8.3.9
Consider the series defined by
This series is not a geometric series, but this activity will illustrate how we might compare this series to a geometric one. Recall that a series
- To see if
is comparable to a geometric series, we analyze the ratios of successive terms in the series. Complete Table∑ 2 k 3 k − k \sum \frac{2^k}{3^k-k} , listing your calculations to at least 8 decimal places.8.3.6 8.3.6 Table . Ratios of successive terms in the series8.3.1 8.3.1 ∑ 2 k 3 k − k \sum \frac{2^k}{3^k-k} k k 5 5 10 10 20 20 21 21 22 22 23 23 24 24 25 25 a k + 1 a k \dfrac{a_{k+1}}{a_k} - Based on your calculations in Table
, what can we say about the ratio8.3.1 8.3.1 ifa k + 1 a k \frac{a_{k+1}}{a_k} is large?k k - Do you agree or disagree with the statement: “the series
is approximately geometric when∑ 2 k 3 k − k \sum \frac{2^k}{3^k-k} is large”? If not, why not? If so, do you think the seriesk k converges or diverges? Explain.∑ 2 k 3 k − k \sum \frac{2^k}{3^k-k}
We can generalize the argument in Activity
for large values of
for a value of
The Ratio Test
Let
- If
then the series0 ≤ r < 1 , 0 \leq r \lt 1\text{,} converges.∑ a k \sum a_k - If
then the series1 < r , 1 \lt r\text{,} diverges.∑ a k \sum a_k - If
then the test is inconclusive.r = 1 , r = 1\text{,}
Note well: The Ratio Test looks at the limit of the ratio of consecutive terms of a given series; in so doing, the test is asking, “is this series approximately geometric?” If so, the test uses the limit of the ratio of consecutive terms to determine if the given series converges.
We have now encountered several tests for determining convergence or divergence of series.
- The Divergence Test can be used to show that a series diverges, but never to prove that a series converges.
- We used the Integral Test to determine the convergence status of an entire class of series, the
-series.p p - The Limit Comparison Test works well for series that involve rational functions and which can therefore by compared to
-series.p p - Finally, the Ratio Test allows us to compare our series to a geometric series; it is particularly useful for series that involve
th powers and factorials.n n - Two other tests, the Direct Comparison Test and the Root Test, are discussed in the exercises.
One of the challenges of determining whether a series converges or diverges is finding which test answers that question.
Activity 8.3.10 8.3.10
Determine whether each of the following series converges or diverges. Explicitly state which test you use.
∑ k 2 k \displaystyle \sum \frac{k}{2^k} ∑ k 3 + 2 k 2 + 1 \displaystyle \sum \frac{k^3+2}{k^2+1} ∑ 10 k k ! \displaystyle \sum \frac{10^k}{k!} ∑ k 3 − 2 k 2 + 1 k 6 + 4 \displaystyle \sum \frac{k^3-2k^2+1}{k^6+4}
Summary
- An infinite series is a sum of the elements in an infinite sequence. In other words, an infinite series is a sum of the form
a 1 + a 2 + ⋯ + a n + ⋯ = ∑ k = 1 ∞ a k a_1 + a_2 + \cdots + a_n + \cdots = \sum_{k=1}^{\infty} a_k where
is a real number for each positive integera k a_k k . k\text{.} - The
th partial sumn n of the seriesS n S_n is the sum of the first∑ k = 1 ∞ a k \sum_{k=1}^{\infty} a_k terms of the series. That is,n n S n = a 1 + a 2 + ⋯ + a n = ∑ k = 1 n a k . S_n = a_1+a_2+ \cdots + a_n = \sum_{k=1}^{n} a_k\text{.} - The sequence
of partial sums of a series{ S n } \{S_n\} tells us about the convergence or divergence of the series. In particular∑ k = 1 ∞ a k \sum_{k=1}^{\infty} a_k - The series
converges if the sequence∑ k = 1 ∞ a k \sum_{k=1}^{\infty} a_k of partial sums converges. In this case we say that the series is the limit of the sequence of partial sums and write{ S n } \{S_n\} ∑ k = 1 ∞ a k = lim n → ∞ S n . \sum_{k=1}^{\infty} a_k =\lim_{n \to \infty} S_n\text{.} - The series
diverges if the sequence∑ k = 1 ∞ a k \sum_{k=1}^{\infty} a_k of partial sums diverges.{ S n } \{S_n\}
- The series