Nathan Wakefield, Christine Kelley, Marla Williams, Michelle Haver, Lawrence Seminario-Romero, Robert Huben, Aurora Marks, Stephanie Prahl, Based upon Active Calculus by Matthew Boelkins

as a geometric series, we found a way to write \(N\) as a single fraction: \(N =
\frac{4}{33}\text{.}\) In this section, we explore other types of infinite sums. In Example 7.20 we see how one such sum is related to the famous number \(e\) .

Example7.20.

Have you ever wondered how your calculator can produce a numeric approximation for complicated numbers like \(e\text{,}\)\(\pi\) or \(\ln(2)\text{?}\) After all, the only operations a calculator can really perform are addition, subtraction, multiplication, and division, the operations that make up polynomials. This example provides the first steps in understanding how this process works. Throughout the example, let \(f(x) = e^x\text{.}\)

Find the tangent line to \(f\) at \(x=0\) and use this linearization to approximate \(e\text{.}\) That is, find a formula \(L(x)\) for the tangent line, and compute \(L(1)\text{,}\) since \(L(1) \approx f(1) = e\text{.}\)

The linearization of \(e^x\) does not provide a good approximation to \(e\) 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 \(e\text{,}\) we put an appropriate bend in our estimating function to make it better fit the graph of \(e^x\) for \(x\) close to 0. With the linearization, we had both \(f(x)\) and \(f'(x)\) share the same value as the linearization at \(x=0\text{.}\) We will now use a quadratic approximation \(P_2(x)\) to \(f(x) = e^x\) centered at \(x=0\) which has the property that \(P_2(0) = f(0)\text{,}\)\(P'_2(0)
= f'(0)\text{,}\) and \(P''_2(0) = f''(0)\text{.}\)

Let \(P_2(x) = 1+x+\frac{x^2}{2}\text{.}\) Show that \(P_2(0) =
f(0)\text{,}\)\(P'_2(0) = f'(0)\text{,}\) and \(P''_2(0) = f''(0)\text{.}\) Then, use \(P_2(x)\) to approximate \(e\) by observing that \(P_2(1) \approx f(1)\) .

We can continue approximating \(e\) with polynomials of larger degree whose higher derivatives agree with those of \(f\) at 0. This turns out to make the polynomials fit the graph of \(f\) better for more values of \(x\) around 0. For example, let \(P_3(x) =
1+x+\frac{x^2}{2}+\frac{x^3}{6}\text{.}\) Show that \(P_3(0) = f(0)\text{,}\)\(P'_3(0)
= f'(0)\text{,}\)\(P''_3(0) = f''(0)\text{,}\) and \(P'''_3(0) = f'''(0)\text{.}\) Use \(P_3(x)\) to approximate \(e\) in a way similar to how you did so with \(P_2(x)\) above.

Example 7.20 shows that an approximation to \(e\) using a linear polynomial is 2, an approximation to \(e\) using a quadratic polynomial is \(2.5\text{,}\) 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 \(e\text{:}\)

We can calculate this sum using as large an \(n\) as we want, and the larger \(n\) is the more accurate the approximation (7.12) is. This suggests that we can write the number \(e\) as the infinite sum

\begin{equation}
e = \sum_{k=0}^{\infty} \frac{1}{k!}\text{.}\tag{7.13}
\end{equation}

This sum is an example of an infinite series. Note that the series (7.13) is the sum of the terms of the infinite sequence \(\left\{\frac{1}{n!}\right\}\text{.}\) In general, we use the following notation and terminology.

Infinite Series.

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 \(a_1\text{,}\)\(a_2\text{,}\)\(\ldots\text{,}\) are real numbers.

We use summation notation to identify a series. If the series adds the entries of a sequence \(\{a_n\}_{n \geq 1}\text{,}\) we will write the series as

Each of these notations is simply shorthand for the infinite sum \(a_1 + a_2 +
\cdots + a_n + \cdots\text{.}\)

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. The definite integral is the sum of an infinite collections of numbers, and we have already examined the special case of geometric series in the previous section. 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?

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 \(n\)th partial sum (that is, the sum of the first \(n\) terms) enables us to make sense of the infinite sum.

Sum the first two numbers in this series. That is, find a numeric value for

The sums in the table in part c form a sequence whose \(n\)th term is \(S_n = \sum_{k=1}^{n} \frac{1}{k^2}\text{.}\) Based on your calculations in the table, do you think the sequence \(\{S_n\}\) converges or diverges? Explain. How do you think this sequence \(\{S_n\}\) is related to the series \(\sum_{k=1}^{\infty}
\frac{1}{k^2}\text{?}\)

Hint.

Compute \(\frac{1}{1^2} + \frac{1}{2^2}\text{.}\)

Do likewise but include the third term, \(\frac{1}{3^2}\text{.}\)

Note that you can use your work in each previous stage and just add one more term to get the next desired sum.

Look for a pattern in the numbers you computed in part (c).

These sums in the table in part (c) seem to indicate that the sequence \(\{S_n\}\) converges to something a bit larger than 1.5. Since the sequence \(\{S_n\}\) is found by adding up the first \(k\) terms of the series, we should expect that the series \(\sum_{k=1}^{n} \frac{1}{k^2}\) is the limit of the sequence \(\{S_n\}\) as \(n\) goes to infinity.

The example in Example 7.21 illustrates how we define the sum of an infinite series. We construct a new sequence of numbers (called the sequence of partial sums) where the \(n\)th term in the sequence consists of the sum of the first \(n\) terms of the series. If this sequence converges, the corresponding infinite series is said to converge, and we say that we can find the sum of the series. More formally, we have the following definition.

Partial Sum.

The \(n\)th partial sum of the series \(\sum_{k=1}^{\infty}
a_k\) is the finite sum \(S_n = \sum_{k=1}^{n} a_k\text{.}\)

In other words, the \(n\)th partial sum \(S_n\) of a series is the sum of the first \(n\) terms in the series,

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 Example 7.21, the series

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.

is determined by what happens to the terms \(a_k\) for very large values of \(k\text{.}\) To see why, suppose that \(m\) is some constant larger than 1. Then

Since \(a_1+a_2+ \cdots + a_m\) is a finite number, the series \(\sum_{k=1}^{\infty}
a_k\) will converge if and only if the series \(\sum_{k=m+1}^{\infty} a_k\) converges. Because the starting index of the series doesn’t affect whether the series converges or diverges, we will often just write

\begin{equation*}
\sum a_k
\end{equation*}

when we are interested in questions of convergence/divergence as opposed to the exact sum of a series.

We summarize this property and a few others in the next remark.

Basic Convergence Properties of Series.

If \(\displaystyle \sum_{n=1}^\infty a_n \) and \(\displaystyle \sum_{n=1}^\infty
b_n \) converge, and \(k \) is a constant, then:

Changing a finite number of terms in a series does not change whether or not the series converges. However, it can change the number that the series will converge to.

If \(\displaystyle \sum_{n=1}^\infty a_n \) and \(\displaystyle
\sum_{n=1}^\infty b_n \) converge, then

If \(\displaystyle \sum_{n=1}^\infty a_n \) does not converge, and \(k \neq 0\) is a constant, then \(\displaystyle \sum_{n=1}^\infty (ka_n) \) also diverges.

In Section 7.2 we encountered the special family of infinite geometric series. Recall that a geometric series has the form \(\sum_{k=0}^{\infty}
ar^k\text{,}\) where \(a\) and \(r\) are real numbers (and \(r \ne 1\)). We found that the \(n\)th partial sum \(S_n\) of a geometric series is given by the convenient formula

and thus a geometric series converges if \(|r|
\lt 1\text{.}\) Geometric series diverge for all other values of \(r\text{.}\)

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.

Subsection7.3.2The 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 example.

Example7.22.

If the series\(\sum a_k\) converges, then an important result necessarily follows regarding the sequence\(\{a_n\}\text{.}\) This example explores this result.

Assume that the series \(\sum_{k=1}^{\infty} a_k\) converges and has sum equal to \(L\text{.}\)

What is the \(n\)th partial sum \(S_n\) of the series \(\sum_{k=1}^{\infty}
a_k\text{?}\)

What is the \((n-1)\)st partial sum \(S_{n-1}\) of the series \(\sum_{k=1}^{\infty}
a_k\text{?}\)

What is the difference between the \(n\)th partial sum and the \((n-1)\)st partial sum of the series \(\sum_{k=1}^{\infty} a_k\text{?}\)

Since we are assuming that \(\sum_{k=1}^{\infty} a_k = L\text{,}\) what does that tell us about \(\lim_{n \to \infty} S_n\text{?}\) Why? What does that tell us about \(\lim_{n
\to \infty} S_{n-1}\text{?}\) Why?

Combine the results of the previous two parts of this example to determine \(\lim_{n \to \infty} a_n = \lim_{n \to \infty} (S_n - S_{n-1})\text{.}\)

Hint.

Recall that the \(n\)th partial sum is the sum of the first \(n\) terms.

Consider the sum with one fewer term.

Subtract your results in (a) and (b).

Remember that the partials sums have to tend to the sum of the series.

Note that \(S_n \to L\) and \(S_{n-1} \to L\text{.}\)

The result of Example 7.22 is the following important conditional statement:

If the series \(\sum_{k = 1}^{\infty} a_k\) converges, then the sequence \(\{a_k\}\) of \(k\)th terms converges to 0.

It is logically equivalent to say that if the sequence \(\{a_k\}\) does not converge to 0, then the series \(\sum_{k = 1}^{\infty} a_k\) cannot converge. This statement is called the Divergence Test.

The Divergence Test.

If the sequence \(\{a_k\}\) does not converge, or if it does converge but \(\lim_{k
\to \infty} a_k \neq 0\text{,}\) then the series \(\sum a_k\) diverges. This is also sometimes called the \(n\)th term test.

Example7.23.

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?

Remember that if the terms of a series go to zero, the Divergence Test doesn’t make any claims.

Answer.

\(\sum \frac{k}{k+1}\) diverges.

\(\sum (-1)^k\) diverges.

The Divergence Test does not apply.

Solution.

Since the sequence \(\frac{k}{k+1}\) converges to 1 and not 0 as \(k\) goes to infinity, the Divergence Test applies here and shows that the series \(\sum
\frac{k}{k+1}\) diverges.

Since the terms of this series don’t even converge to a number, they can’t converge to zero, and thus this series diverges.

Since \(\lim_{k \to \infty} \frac{1}{k} = 0\text{,}\) the Divergence Test does not apply to this series.

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.

Subsection7.3.3The Integral Test

The Divergence Test settles the questions of divergence or convergence of series \(\sum a_k\) in which \(\lim_{k \to \infty} a_k \neq 0\text{.}\) Determining the convergence or divergence of series \(\sum a_k\) in which \(\lim_{k \to \infty} a_k = 0\) turns out to be more complicated. Often, we have to investigate the sequence of partial sums or apply some other technique.

Next, we consider the harmonic series ^{ 7 }

This series is called harmonic because each term in the series after the first is the harmonic mean of the term before it and the term after it. The harmonic mean of two numbers \(a\) and \(b\) is \(\frac{2ab}{a+b}\text{.}\) See “What’s Harmonic about the Harmonic Series”, by David E. Kullman (in the College Mathematics Journal, Vol. 32, No. 3 (May, 2001), 201-203) for an interesting discussion of the harmonic mean.

This information alone doesn’t seem to be enough to tell us if the series \(\sum_{k=1}^{\infty}
\frac{1}{k}\) converges or diverges. The partial sums could eventually level off to some fixed number or continue to grow without bound. Even if we look at larger partial sums, such as \(\sum_{n=1}^{1000} \frac{1}{k} \approx 7.485470861\text{,}\) the result isn’t particularly convincing one way or another. The Integral Test is one way to determine whether or not the harmonic series converges, and we explore this test further in the next example.

Example7.24.

Consider the harmonic series \(\sum_{k=1}^{\infty} \frac{1}{k}\text{.}\) Recall that the harmonic series will converge provided that its sequence of partial sums converges. The \(n\)th partial sum \(S_n\) of the series \(\sum_{k=1}^{\infty}
\frac{1}{k}\) is

Through this last expression for \(S_n\text{,}\) we can visualize this partial sum as a sum of areas of rectangles with heights \(\frac{1}{m}\) and bases of length 1, as shown in Figure 7.25, which uses the 9th partial sum.

The graph of the continuous function \(f\) defined by \(f(x) =
\frac{1}{x}\) is overlaid on this plot.

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 \(S_9\) of the series? Why?

If instead of considering the 9th partial sum, we consider the \(n\)th partial sum, and we let \(n\) go to infinity, we can then compare the series \(\sum_{k=1}^{\infty}
\frac{1}{k}\) to the improper integral \(\int_{1}^{\infty} \frac{1}{x} \ dx\text{.}\) Which of these quantities is larger? Why?

Does the improper integral \(\int_{1}^{\infty} \frac{1}{x} \ dx\) converge or diverge? What does that result, together with your work in (d), tell us about the series \(\sum_{k=1}^{\infty} \frac{1}{k}\text{?}\)

Answer.

The \(n\)th partial sum of the series \(\sum_{k=1}^{\infty}
\frac{1}{k}\) is the left hand Riemann sum of \(f(x)\) on the interval \([1,n]\) .

\(\int_{1}^{\infty} f(x) \ dx = \lim_{t \to \infty} \int_{1}^{t} \frac{1}{x} \ dx
= \infty \) so the series \(\sum_{k=1}^{\infty} \frac{1}{k}\) diverges.

Solution.

Notice that the \(n\)th partial sum of the series \(\sum_{k=1}^{\infty}
\frac{1}{k}\) is equal to the left hand Riemann sum of \(f(x)\) on the interval \([1,n]\text{.}\)

Since \(f\) is a decreasing function, it follows that

Since \(f\) is decreasing, the improper integral \(\int_{1}^{\infty}
\frac{1}{x} \ dx\) is smaller than the limit of the Riemann sums as \(n\) goes to infinity. So we wind up with a comparison between the series \(\sum_{k=1}^{\infty}
\frac{1}{n}\) and the improper integral:

Since the value of the series \(\sum_{k=1}^{\infty}
\frac{1}{k}\) exceeds the value of the infinite improper integral, we must conclude that the series \(\sum_{k=1}^{\infty} \frac{1}{k}\) diverges.

The ideas from Example 7.24 hold more generally. Suppose that \(f\) is a continuous decreasing function and that \(a_k = f(k)\) for each value of \(k\text{.}\) Consider the corresponding series \(\sum_{k=1}^{\infty} a_k\text{.}\) The partial sum

can always be viewed as a left hand Riemann sum of \(f(x)\text{,}\) using rectangles with bases of width 1 and heights given by the values \(a_k\text{.}\) A representative picture is shown at left in Figure 7.26. Since \(f\) is a decreasing function, we have that

So if \(\int_{1}^{\infty} f(x) \ dx\) converges, then so does \(\sum_{k=2}^{\infty}
a_k\text{,}\) which also means that the series \(\sum_{k=1}^{\infty} a_k\) also converges. Our preceding discussion has demonstrated the truth of the Integral Test.

The Integral Test.

Let \(f\) be a real valued function and assume \(f\) is decreasing and positive for all \(x\) larger than some number \(c\text{.}\) Let \(a_k = f(k)\) for each positive integer \(k\text{.}\)

If the improper integral \(\int_{c}^{\infty} f(x) \, dx\) converges, then the series \(\sum_{k=1}^{\infty} a_k\) converges.

If the improper integral \(\int_{c}^{\infty} f(x) \, dx\) diverges, then the series \(\sum_{k=1}^{\infty} a_k\) diverges.

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 example, we apply the Integral Test to determine the convergence or divergence of a class of important series.

Example7.27.

The series \(\sum \frac{1}{k^p}\) are special series called \(p\)-series. We have already seen that the \(p\)-series with \(p=1\) (the harmonic series) diverges. We investigate the behavior of other \(p\)-series in this example.

Evaluate the improper integral \(\int_1^{\infty} \frac{1}{x^2} \ dx\text{.}\) Does the series \(\sum_{k=1}^{\infty} \frac{1}{k^2}\) converge or diverge? Explain.

Evaluate the improper integral \(\int_1^{\infty} \frac{1}{x^p} \ dx\) where \(p \gt 1\text{.}\) For which values of \(p\) can we conclude that the series \(\sum_{k=1}^{\infty}
\frac{1}{k^p}\) converges?

Evaluate the improper integral \(\int_1^{\infty} \frac{1}{x^p} \ dx\) where \(p \lt 1\text{.}\) What does this tell us about the corresponding \(p\)-series \(\sum_{k=1}^{\infty}
\frac{1}{k^p}\text{?}\)

Summarize your work in this example by completing the following statement.

The \(p\)-series \(\sum_{k=1}^{\infty} \frac{1}{k^p}\) converges if and only if .

Answer.

\(\sum_{k=1}^{\infty} \frac{1}{k^2}\) converges.

\(\sum_{k=1}^{\infty} \frac{1}{k^p}\) converges when \(p \gt 1\text{.}\)

\(\sum_{k=1}^{\infty} \frac{1}{k^p}\) diverges when \(p \lt 1\text{.}\)

The \(p\)-series \(\sum_{k=1}^{\infty} \frac{1}{k^p}\) converges if and only if \(p \gt 1\text{.}\)

So the improper integral \(\int_1^{\infty}
\frac{1}{x^p} \ dx\) converges when \(p \gt 1\text{.}\) The Integral Test then shows that the series \(\sum_{k=1}^{\infty} \frac{1}{k^p}\) converges when \(p \gt 1\) .

Assume \(p \lt 1\text{.}\) Then \(1-p \gt 0\) and so

So the improper integral \(\int_1^{\infty}
\frac{1}{x^p} \ dx\) diverges when \(p \lt 1\text{.}\) The Integral Test then shows that the series \(\sum_{k=1}^{\infty} \frac{1}{k^p}\) also diverges when \(p \lt
1\text{.}\)

We complete the statement as

The \(p\)-series \(\sum_{k=1}^{\infty} \frac{1}{k^p}\) converges if and only if \(p \gt 1\text{.}\)

\(p \) -Series.

Series of the form \(\sum \frac{1}{k^p}\) are called \(p \) -series. The harmonic series \(\sum \frac{1}{k}\) is the special case when \(p =1\text{.}\)

The \(p\)-series \(\sum_{k=1}^{\infty} \frac{1}{k^p}\) converges for \(p \gt 1\text{,}\) and diverges for \(p \leq 1 \text{.}\)

Subsection7.3.4Summary

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

The sequence \(\{S_n\}\) of partial sums of a series \(\sum_{k=1}^{\infty}
a_k\) tells us about the convergence or divergence of the series. In particular

The series \(\sum_{k=1}^{\infty} a_k\) converges if the sequence \(\{S_n\}\) of partial sums converges. In this case we say that the series is the limit of the sequence of partial sums and write

For the following answer blanks, decide whether the given sequence or series is convergent or divergent. If convergent, enter the limit (for a sequence) or the sum (for a series). If divergent, enter ’infinity’ if it diverges to \(\infty\text{,}\) ’-infinity’ if it diverges to \(- \infty\) or ’DNE’ otherwise.

(a) The series \(\displaystyle
\sum_{n=1}^\infty \frac{9n}{8n+5}\text{.}\)

(b) The sequence \(\displaystyle
\biggl \lbrace \frac{9n}{8n+5} \biggr \rbrace\text{.}\)

4.Convergence of an integral and a related series.

Compute the value of the following improper integral. If it converges, enter its value. Enter infinity if it diverges to \(\infty\text{,}\) and -infinity if it diverges to \(-\infty\text{.}\) Otherwise, enter diverges.

Does the series \(\displaystyle \sum_{n=1}^{\infty} \frac{9}{n^2 + 1}\) converge or diverge?

converges

diverges to +infinity

diverges to -infinity

diverges

5.Adding two series together.

The associative and commutative laws of addition allow us to add finite sums in any order we want. That is, if \(\sum_{k=0}^n a_k\) and \(\sum_{k=0}^n b_k\) are finite sums of real numbers, then

This shows that it is possible to have to two divergent series \(\sum_{k=0}^{\infty} a_k\) and \(\sum_{k=0}^{\infty} b_k\) but yet have the series \(\sum_{k=0}^{\infty}
(a_k+b_k)\) converge.

While part (a) shows that we cannot add series term by term in general, we can under reasonable conditions. The problem in part (a) is that we tried to add divergent series. In this exercise we will show that if \(\sum a_k\) and \(\sum
b_k\) are convergent series, then \(\sum (a_k+b_k)\) is a convergent series and

Let \(A_n\) and \(B_n\) be the \(n\)th partial sums of the series \(\sum_{k=1}^{\infty} a_k\) and \(\sum_{k=1}^{\infty} b_k\text{,}\) respectively. Explain why

(Note that the starting point of the sum is irrelevant in this problem, so it doesn’t matter where we begin the sum.)

Use the prior result to find the sum of the series \(\sum_{k=0}^{\infty}
\frac{2^k+3^k}{5^k}\text{.}\)

6.Using the integral test on a series involving a logarithm.

Find the value of the integral

\begin{equation*}
\int_2^{\infty} \frac{dx}{2 x (\ln(2 x))^2}.
\end{equation*}

Determine whether the series

\begin{equation*}
\sum_{n=2}^\infty \frac{1}{2 n (\ln(2 n))^2}
\end{equation*}

is convergent. Enter C if the series is convergent, D if the series is divergent.

7.Using the integral test on a series involving an exponential.

Compute the value of the following improper integral. If it converges, enter its value. Enter infinity if it diverges to \(\infty\text{,}\) and -infinity if it diverges to \(-\infty\text{.}\) Otherwise, enter diverges.

Does the series \(\displaystyle \sum_{n=1}^{\infty} 3 n^2 e^{-n^3}\) converge or diverge?

converges

diverges to +infinity

diverges to -infinity

diverges

8.Deciding when the integral test applies and using it.

Test each of the following series for convergence by the Integral Test. If the Integral Test can be applied to the series, enter CONV if it converges or DIV if it diverges. If the integral test cannot be applied to the series, enter NA. (Note: this means that even if you know a given series converges by some other test, but the Integral Test cannot be applied to it, then you must enter NA rather than CONV.)