Calculus AB and Calculus BC
CHAPTER 10 Sequences and Series
B. INFINITE SERIES
B1. Definitions.
Infinite series
If an is a sequence of real numbers, then an infinite series is an expression of the form
![]()
The elements in the sum are called terms; an is the nth or general term of the series.
EXAMPLE 7
A series of the form
is called a p-series.
The p-series for p = 2 is ![]()
EXAMPLE 8
The p-series with p = 1 is called the harmonic series:
![]()
EXAMPLE 9
A geometric series has a first term, a, and common ratio of terms, r:
![]()
If there is a finite number S such that
![]()
then we say that infinite series is convergent, or converges to S, or has the sum S, and we write, in this case,
![]()
When there is no source of confusion, the infinite series (1) may be indicated simply by
![]()
EXAMPLE 10
Show that the geometric series
converges to 2.
SOLUTION: Let S represent the sum of the series; then:

EXAMPLE 11
Show that the harmonic series
diverges.
SOLUTION The terms in the series can be grouped as follows:

This sum clearly exceeds
![]()
which equals
![]()
Since that sum is not bounded, it follows that
diverges to ∞.
B2. Theorems About Convergence or Divergence of Infinite Series.
The following theorems are important.
THEOREM 2a. If
converges, then ![]()
This provides a convenient and useful test for divergence, since it is equivalent to the statement: If an does not approach zero, then the series
diverges. Note, however, particularly that the converse of Theorem 2a is not true. The condition that an approach zero is necessary but not sufficient for the convergence of the series. The harmonic series
is an excellent example of a series whose nth term goes to zero but that diverges (see Example 11 above). The series
diverges because
not zero; the series
does not converge (as will be shown shortly) even though ![]()
THEOREM 2b. A finite number of terms may be added to or deleted from a series without affecting its convergence or divergence; thus
![]()
(where m is any positive integer) both converge or both diverge. (Note that the sums most likely will differ.)
THEOREM 2c. The terms of a series may be multiplied by a nonzero constant without affecting the convergence or divergence; thus
![]()
both converge or both diverge. (Again, the sums will usually differ.)
THEOREM 2d. If
both converge, so does ![]()
THEOREM 2e. If the terms of a convergent series are regrouped, the new series converges.
B3. Tests for Convergence of Infinite Series.
THE nth TERM TEST
If
diverges.
NOTE: When working with series, it’s a good idea to start by checking the nth Term Test. If the terms don’t approach 0, the series cannot converge. This is often the quickest and easiest way to identify a divergent series.
(Because this is the contrapositive of Theorem 2a, it’s always true. But beware of the converse! Seeing that the terms do approach 0 does not guarantee that the series must converge. It just means that you need to try other tests.)
EXAMPLE 12
Does
converge or diverge?
SOLUTION: Since
the series
diverges by the nth Term Test.
THE GEOMETRIC SERIES TEST
A geometric series
converges if and only if |r| < 1.
If |r| < 1, the sum is ![]()
The series cannot converge unless it passes the nth Term Test;
only if |r| < 1. As noted earlier, this is a necessary condition for convergence, but may not be sufficient. We now examine the sum using the same technique we employed in Example 10:

EXAMPLE 13
Does 0.3 + 0.03 + 0.003 + · · · converge or diverge?
SOLUTION: The series 0.3 + 0.03 + 0.003 + · · · is geometric with a = 0.3 and r = 0.1. Since |r| < 1, the series converges, and its sum is
![]()
NOTE:
= 0.333 …, which is the given series.
B4. Tests for Convergence of Nonnegative Series.
The series
is called a nonnegative series if an ≥ 0 for all n.
THE INTEGRAL TEST
If f (x) is a continuous, positive, decreasing function and f (n) = an, then
converges if and only if the improper integral
converges.
EXAMPLE 14
Does
converge?
SOLUTION: The associated improper integral is
![]()
which equals
![]()
The improper integral and the infinite series both diverge.
EXAMPLE 15
Test the series
for convergence.
SOLUTION: 
by an application of L’Hôpital’s Rule. Thus
converges.
THE p-SERIES TEST
A p-series
converges if p > 1, but diverges if p ≤ 1.
This follows immediately from the Integral Test and the behavior of improper integrals of the form ![]()
EXAMPLE 16
Does the series
converge or diverge?
SOLUTION: The series
is a p-series with p = 3;
hence the series converges by the p-Series Test.
EXAMPLE 17
Does the series
converge or diverge?
SOLUTION:
diverges, because it is a p-series with ![]()
THE COMPARISON TEST
We compare the general term of
the nonnegative series we are investigating, with the general term of a series known to converge or diverge.
(1) If
converges and an
un, then
converges.
(2) If
diverges and an
un, then
diverges.
Any known series can be used for comparison. Particularly useful are p-series, which converge if p > 1 but diverge if p
1, and geometric series, which converge if |r| < 1 but diverge if |r|
1.
EXAMPLE 18
Does
converge or diverge?
SOLUTION: Since
and the p-series
converges,
converges by the Comparison Test.
EXAMPLE 19
Does the series
converge or diverge?
SOLUTION:
diverges, since
![]()
the latter is the general term of the divergent p-series
where
and ![]()
Remember in using the Comparison Test that you may either discard a finite number of terms or multiply each term by a nonzero constant without affecting the convergence of the series you are testing.
EXAMPLE 20
Show that
converges.
SOLUTION: For
is a convergent geometric series with ![]()
THE LIMIT COMPARISON TEST
If
is finite and nonzero, then
and
both converge or both diverge.
This test is useful when the direct comparisons required by the Comparison Test are difficult to establish. Note that, if the limit is zero or infinite, the test is inconclusive and some other approach must be used.
EXAMPLE 21
Does
converge or diverge?
SOLUTION: This series seems to be related to the divergent harmonic series, but
so the comparison fails. However, the Limit Comparison Test yields:

Since
also diverges by the Limit Comparison Test.
THE RATIO TEST
Let
if it exists. Then
converges if L < 1 and diverges if L > 1.
If L = 1, this test is inconclusive; apply one of the other tests.
EXAMPLE 22
Does
converge or diverge?
SOLUTION: 
Therefore this series converges by the Ratio Test.
EXAMPLE 23
Does
converge or diverge?
SOLUTION: ![]()
and
![]()
(See §E2.) Since e > 1,
diverges by the Ratio Test.
EXAMPLE 24
If the Ratio Test is applied to any p-series,
then

But if p > 1 then
converges, while if p
1 then
diverges. This illustrates the failure of the Ratio Test to resolve the question of convergence when the limit of the ratio is 1.
THE nth ROOT TEST
Let
if it exists. Then
converges if L < 1 and diverges if L > 1.
If L = 1 this test is inconclusive; try one of the other tests.
Note that the decision rule for this test is the same as that for the Ratio Test.
EXAMPLE 25
The series
converges by the nth Root Test, since

B5. Alternating Series and Absolute Convergence.
Any test that can be applied to a nonnegative series can be used for a series all of whose terms are negative. We consider here only one type of series with mixed signs, the so-called alternating series. This has the form:
![]()
where ak > 0. The series
![]()
is the alternating harmonic series.
THE ALTERNATING SERIES TEST
An alternating series converges if:
(1) an + 1 < an for all n, and
(2) ![]()
EXAMPLE 26
Does the series
converge or diverge?
SOLUTION: The alternating harmonic series
converges, since
(1)
for all n and
(2) ![]()
EXAMPLE 27
Does the series
converge or diverge?
SOLUTION: The series
diverges, since we see that
is 1, not 0. (By the nth Term Test, if an does not approach 0, then
does not converge.)
DEFINITION
Absolute convergence
A series with mixed signs is said to converge absolutely (or to be absolutely convergent) if the series obtained by taking the absolute values of its terms converges; that is,
converges absolutely if
converges.
A series that converges but not absolutely is said to converge conditionally (or to be conditionally convergent). The alternating harmonic series converges conditionally since it converges, but does not converge absolutely. (The harmonic series diverges.)
When asked to determine whether an alternating series is absolutely convergent, conditionally convergent, or divergent, it is often advisable to first consider the series of absolute values. Check first for divergence, using the nth Term Test. If that test shows that the series may converge, investigate further, using the tests for nonnegative series. If you find that the series of absolute values converges, then the alternating series is absolutely convergent. If, however, you find that the series of absolute values diverges, then you’ll need to use the Alternating Series Test to see whether the series is conditionally convergent.
EXAMPLE 28
Determine whether
converges absolutely, converges conditionally, or diverges.
SOLUTION: We see that
not 0, so by the nth Term Test the series
is divergent.
EXAMPLE 29
Determine whether
converges absolutely, converges conditionally, or diverges.
SOLUTION: Note that, since
the series passes the nth Term Test.

But
is the general term of a convergent p-series (p = 2), so by the Comparison Test the nonnegative series converges, and therefore the alternating series converges absolutely.
EXAMPLE 30
Determine whether
converges absolutely, converges conditionally, or diverges.
SOLUTION:
is a p-series with
so the nonnegative series diverges.
We see that ![]()
so the alternating series converges; hence
is conditionally convergent.
APPROXIMATING THE LIMIT OF AN ALTERNATING SERIES
Evaluating the sum of the first n terms of an alternating series, given by
yields an approximation of the limit, L. The error (the difference between the approximation and the true limit) is called the remainder after n terms and is denoted by Rn. When an alternating series is first shown to pass the Alternating Series Test, it’s easy to place an upper bound on this remainder. Because the terms alternate in sign and become progressively smaller in magnitude, an alternating series converges on its limit by oscillation, as shown in Figure N10–1.

FIGURE N10–1
Error bound
Because carrying out the approximation one more term would once more carry us beyond L, we see that the error is always less than that next term. Since |Rn | < an + 1, the error bound for an alternating series is the first term omitted or dropped.
EXAMPLE 31
The series
passes the Alternating Series Test; hence its sum differs from the sum
![]()
by less than
which is the error bound.
EXAMPLE 32
How many terms must be summed to approximate to three decimal places the value of ![]()
SOLUTION: Since
the series converges by the Alternating Series Test; therefore after summing a number of terms the remainder (error) will be less than the first omitted term.
We seek n such that
Thus n must satisfy (n + 1)2 > 1000, or n > 30.623. Therefore 31 terms are needed for the desired accuracy.