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14. 1 The Algebra of Series
There are several simple properties that apply to series that converge. We have
already been using the first of these results.
THEOREM 14. 1. 1. Suppose that we have two convergent series:
Â
n = 0
a
n
= A and
Â
n = 0
b
n
B. Then
(a) If c is a constant, then
Â
n = 0
ca
n
= cA.
(b)
Â
n = 0
a
n
± b
n
= A ± B.
EXAMPLE 14. 1. 2. Determine the sum of the series
Â
n = 0
n
n
if it exists.
Solution. Using the Geometric Series Test (Theorem 13. 2. 2 ) and Theorem 14. 1. 1
we have
Â
n = 0
n
n
Â
n = 0
n
Â
n = 0
n
14. 2 The Divergence Test
The next theorem shows that for a series to converge, the terms of the series must
get small as n gets large.
THEOREM 14. 2. 1 (The nth term Test). If
Â
n = 0
a
n
converges to A, then lim
n! •
a
n
Proof. We make use of a comment we made earlier. Let S
n
be the nth partial sum
of the series. We know that lim
n! •
S
n
= lim
n! •
S
n 1
= A because both sequences are
really just the same set of numbers. But
S
n
= a
+ a
+ · · · + a
n 1
+ a
n
and
S
n 1
= a
+ a
+ · · · + a
n 1
So
S
n
S
n 1
= a
n
Therefore
lim
n! •
a
n
= lim
n! •
( S
n
S
n 1
) = lim
n! •
S
n
lim
n! •
S
n 1
= A A = 0,
which is what we wanted to prove.
The theorem is seldom used in the form as stated above. Rather, if the nth term
of a series does not converge to 0, then series cannot convergence. This is a more
useful way of stating of Theorem 14. 2. 1 and in this form it is a test for divergence.
THEOREM 14. 2. 2 (The nth term test for divergence). If lim
n! •
a
n
6 = 0, then
Â
n = 0
a
n
diverges.
Warning! The nth term test for divergence never allows us to conclude that a
series converges, only that it does not converge. If lim
n! •
a n
= 0, we can’t conclude
anything. The series may or may not converge. The test simply fails to provide us
with any useful information in such a case.
EXAMPLE 14. 2. 3. Determine whether the series
Â
n = 0
n
3 n
converges.
Solution. Notice that
lim
n! •
a
n
= lim
n! •
n
3 n
= lim
n! •
1 +
n
3 +
n
n
=
1
3
6 = 0.
By the nth term test for divergence (Theorem 14. 2. 2 ), the series
Â
n = 0
n
3 n
diverges.
EXAMPLE 14. 2. 4. Determine whether the series
Â
n = 1
✓
1 +
k
n
◆
n
converges.
Solution. This time using using one of our key limits (see Theorem 13. 2 )
lim
n! •
a
n
= lim
n! •
✓
1 +
k
n
◆
n
= e
k
6 = 0.
By the nth term test for divergence (Theorem 14. 2. 2 ), the series
Â
n = 1
✓
1 +
k
n
◆
n
diverges.
EXAMPLE 14. 2. 5. Determine whether the series
Â
n = 1
n
p
n converges.
Solution. Using another of our key limits (see Theorem 13. 2 )
lim
n! •
a
n
= lim
n! •
n
p
n = 1 6 = 0.
By the nth term test for divergence (Theorem 14. 2. 2 ), the series
Â
n = 1
n
p
n diverges.
EXAMPLE 14. 2. 6. Determine whether the series
Â
k = 1
2
n
n
=
2
1
4
4
8
9
Solution. Notice that both numerator and denominator both tend to infinity. So
converting to x and using l’Hôpital’s rule (twice!)
lim
n! •
2
n
n
2
= lim
x! •
2
x
x
2
= lim
x! •
( ln 2 ) 2
x
2 x
= lim
x! •
( ln 2 )( ln 2 ) 2
x
2
= •.
By the nth term test for divergence (Theorem 14. 2. 2 ), the series diverges.
Use the derivative formula
d
dx
( a
x
) =
a
x
ln a, which is valid when a > 0.
EXAMPLE 14. 2. 7. Determine whether the series
Â
k = 1
1
n
and
Â
k = 1
1
n
converge.
Solution. Notice that both
lim
n! •
1
n
= 0.
and
lim
n! •
1
n
2
= 0.
In this situation the nth term test for divergence (Theorem 14. 2. 2 ) fails to provide
us with any information. The series may or may not converge. In fact, we will soon
see that one of these series converges while the other diverges.
ht = a
= f ( 1 )
area = a
= a
area = a
area = a
3 area = a
4 area = a
5 area = a
Dx = 1
Figure 14. 1 : Each • indicates a point
of the sequence { a
n
} with the graph
of the corresponding function f ( x ) for
x 1. Notice that f ( x ) is positive,
decreasing, and continuous. Because
f ( x ) is decreasing , the left-hand sum
is an overestimate of the integral. And
the function is integrable because it is
continuous.
Notice that because f ( x ) is decreasing , the left-hand sum is an overestimate of
the integral. And the function is integrable because it is continuous. So
S
n
n
Â
k = 1
a
k
= Left ( n )
Z
n + 1
f ( x ) dx.
Taking the limit as n! • , we get the improper integral which we assumed di-
verged:
Â
n = 1
a
n
= lim
n! •
S
n
lim
n! •
Z
n + 1
f ( x ) dx =
Z
f ( x ) dx! •.
Since the series is at least as big as the corresponding improper integral:
Take-home message: If the improper integral diverges to infinity, so does the series!!
The Same Idea with a Twist
Assume, as before, that we have
Â
n = 1
a
n
and a function f ( x ) that corresponds to the
terms a
n
and that f ( x ) is positive , continuous , and decreasing for x 1. This time
assume that
Â
n = 1
a
n
diverges. If we toss out the first term,
Â
n = 2
a
n
still diverges.
This time approximate the area under f ( x ) on the interval [ 1, n + 1 ] by using a
right -hand Riemann sum, Right ( n ) sum. As before Dx = 1. The height of the first
rectangle is f ( 2 ) = a
. The second rectangle has height f ( 3 ) = a
. In general, the
kth rectangle which lies above the interval [ k, k + 1 ] has height f ( k ) = a
k + 1
. So the
Riemann sum is
Right ( n ) = a
· Dx + a
· Dx + · · · + a
n + 1
· Dx = a
· 1 + a
· 1 + · · · + a
n
· 1 + a
n + 1
n + 1
Â
k = 2
a
k
= T
n
which is the n + 1st partial sum of the series
Â
n = 2
a
n
area = a
= a
area = a
area = a
4 area = a
5 area = a
6 area = a
Dx = 1
Because f ( x ) is decreasing , the right-hand sum is an underestimate of the integral.
So
T
n
n + 1
Â
k = 2
a
k
= Right ( n )
Z
n + 1
f ( x ) dx.
Taking the limit as n! • , we get the entire series, which we assumed diverged,
and the improper integral:
• = lim
n! •
n + 1
Â
k = 2
a
k
lim
n! •
Z
n + 1
f ( x ) dx =
Z
f ( x ) dx.
Since the improper integral is at least as big as the diverging series, the integral
must diverge. Consequently, if the integral con verges, the series cannot diverge. In
other words,
Take-home message 2 : Therefore, if the improper integral converges , so does the
series!! We can combine the two take-home messages into the following neat
theorem.
THEOREM 14. 3. 1 (The Integral Test). Given
Â
n= 1
a
n
and a positive , continuous , and decreas-
ing function f (x) such that f (n) = a
n
Â
n= 1
a
n
diverges if and only if diverges
Z
f (x) dx diverges. This is the same as saying
Â
n= 1
a
n
converges if and only if
Z
f (x) dx converges.
Note: It is sufficient if f (x) is positive and decreasing on some interval of the form [a, • )
where a > 1. It is the infinite tail of the the improper integral or the series that determines
convergence or divergence, not the first few terms.
EXAMPLE 14. 3. 2. Does
Â
n = 1
n
+ · · · converge?
Solution. The series is not geometric and the Divergence Test is inconclusive. We
use the integral test. The corresponding function is f ( x ) =
x
and is clearly pos-
itive and continuous on [ 1, • ). Notice that f
( x ) =
2 x
( 4 + x
< 0 for all x in [ 1, • ).
So f is decreasing. The improper integral that we must evaluate is
Z
x
dx We could also say that f is decreasing
because as x increases, the denomi-
nator increases, but the numerator is
constant—which makes the function
values smaller.
which you should recognize as an inverse tangent integral.
Z
x
dx = lim
b! •
Z
b
x
dx
= lim
b! •
arctan (
x
b
= lim
b! •
arctan (
b
arctan (
[
p
arctan (
)].
Since the integral converges, by Theorem 14. 3. 1 the series
Â
n = 1
n
also con-
verges.
14. 4 Applying the Integral Test: p-Series
The integral test shows why improper integrals are important.
EXAMPLE 14. 4. 1. Let’s look back at the two series in Example 14. 2. 7. The first was
Â
n = 1
n
which is called the harmonic series. (You should know it by name!) Does the harmonic
series converge?
The second series was
Â
n = 1
n
. Does it converge? In Example 14. 2. 7 we saw that the nth
term test provided no help since the nth terms go to 0 in both cases.
EXAMPLE 14. 4. 5. Does
Â
n = 1
( n + 1 ) ln ( n + 1 )
converge?
Solution. We use the integral test. The corresponding function is f ( x ) =
( x + 1 ) ln ( x + 1 )
and is positive, decreasing (as x gets bigger, so does the denominator), and contin-
uous. The improper integral that we must evaluate is
Z
( x + 1 ) ln ( x + 1 )
dx.
Using a u-substitution with u = ln ( x + 1 ) and du =
x + 1
dx we find
Z
( x + 1 ) ln ( x + 1 )
dx = lim
b! •
Z
b
( x + 1 ) ln ( x + 1 )
dx
= lim
b! •
ln | ln ( x + 1 )|
b
= lim
b! •
ln | ln ( b + 1 )| ln ( ln ( 2 )) = + •.
By Theorem 14. 3. 1 the series
Â
n = 1
( n + 1 ) ln ( n + 1 )
also diverges.
EXAMPLE 14. 4. 6. Does
Â
n = 1
n
+ n
converge?
Solution. This is one that we could do as a telescoping series as in Example 13. 1. 3.
Instead we use the integral test. The corresponding function is f ( x ) =
x
+ x
and is
clearly positive, decreasing, and continuous. The improper integral that we must
evaluate is
Z
x
+ x
dx =
Z
x
x + 1
dx (check that the partial fractions are
correct).
Z
x
x + 1
dx = lim
b! •
Z
b
x
x + 1
dx
= lim
b! •
( ln | x | ln | x + 1 |)
b
= lim
b! •
ln
x
x + 1
b
= lim
b! •
ln
b
b + 1
ln
= ln 1 ln
= ln 2.
Since the integral converges, by Theorem 14. 3. 1 the series
Â
n = 1
n
+ n
also con-
verges. Notice that unlike in Example 13. 1. 3 , we do not know what
Â
n = 1
n
+ n
converges to, only that it does converge.
EXAMPLE 14. 4. 7. Does
Â
n = 1
ln n
n
converge?
Solution. The corresponding function is f ( x ) =
ln x
x
. Notice that f ( 1 ) = 0 so it is
not always positive on [ 1, • ). Further, we know that a function is decreasing when
its derivative is negative. But using the quotient rule,
f
( x ) =
1 ln x
x
< 0 () 1 < ln x () x > e.
So f is positive, decreasing, and continuous only if x > e. That’s good enough! It
is the infinite tail of the series that matters, not the first few terms. Applying the
integral test
Z
ln x
x
dx = lim
b! •
Z
b
ln x
x
dx = lim
b! •
( ln x )
b
= lim
b! •
( ln b )
Since the integral diverges, by Theorem 14. 3. 1 the series
Â
n = 1
ln n
n
also diverges.
EXAMPLE 14. 4. 8. Does
Â
n = 1
n
e
n
converge?
Solution. The nth term test fails and it is not a geometric or p-series. We use the
integral test. The corresponding function is f ( x ) =
x
e
x
and is clearly positive and
continuous. It is decreasing since f
( x ) =
e
x
xe
x
( e
x
1 x
e
x
0 when x 1. We
evaluate (using integration by parts with u = x and dv = e
x
Z
x
e
x
dx = lim
b! •
Z
b
xe
x
dx = lim
b! •
xe
x
Z
b
e
x
dx
= lim
b! •
xe
x
e
x
b
= lim
b! •
b
e
b
e
b
e
e
l’Ho
= lim
b! •
e
b
e
e
e
Since the integral converges, by Theorem 14. 3. 1 the series
Â
n = 1
n
e
n
also converges.
webwork: Click to try Problems 149 through 152. Use guest login, if not in my course.
YOU TRY IT 14. 6. Use the integral test, geometric series test, or the p-series test to determine
whether the following series converge or diverge. While you could use the integral test on
all of these, it is possible to use one of the other tests for four of the seven.
(a)
Â
n = 1
p
n
(b)
Â
n = 1
e
2 n
(c)
Â
n = 1
n
(d)
Â
n = 1
9 + n
(e)
Â
n = 1
n
+ 7 n + 10
(f )
Â
n = 1
n
p
n
(g)
Â
n = 1
3 n
n
(h)
Â
n = 1
n
