1. No category

Strategy for Testing Series
We now have several ways of testing a series for convergence or divergence; the problem
is to decide which test to use on which series. In this respect testing series is similar to integrating functions. Again there are no hard and fast rules about which test to apply to a
given series, but you may find the following advice of some use.
It is not wise to apply a list of the tests in a specific order until one finally works. That
would be a waste of time and effort. Instead, as with integration, the main strategy is to
classify the series according to its form.
1. If the series is of the form
1n p, it is a p-series, which we know to be conver-
gent if p 1 and divergent if p 1.
ar n1 or ar n, it is a geometric series, which converges if r 1 and diverges if r 1. Some preliminary algebraic manipulation may be required to bring the series into this form.
If the series has a form that is similar to a p-series or a geometric series, then
one of the comparison tests should be considered. In particular, if a n is a rational
function or algebraic function of n (involving roots of polynomials), then the
series should be compared with a p-series. (The value of p should be chosen as
in Section 8.3 by keeping only the highest powers of n in the numerator and
denominator.) The comparison tests apply only to series with positive terms, but
if a n has some negative terms, then we can apply the Comparison Test to
a n and test for absolute convergence.
If you can see at a glance that lim n l a n 0, then the Test for Divergence
should be used.
If the series is of the form 1n1bn or 1nbn , then the Alternating Series
Test is an obvious possibility.
Series that involve factorials or other products (including a constant raised to the
n th power) are often conveniently tested using the Ratio Test. Bear in mind that
a n1a n l 1 as n l for all p-series and therefore all rational or algebraic
functions of n. Thus, the Ratio Test should not be used for such series.
If a n f n, where x1 f x dx is easily evaluated, then the Integral Test is effective (assuming the hypotheses of this test are satisfied).
2. If the series has the form
3.
4.
5.
6.
7.
In the following examples we don’t work out all the details but simply indicate which
tests should be used.
EXAMPLE 1
n1
n1
2n 1
Since a n l 12 0 as n l , we should use the Test for Divergence.
EXAMPLE 2
n1
sn 3 1
3n 4n 2 2
3
Since a n is an algebraic function of n, we compare the given series with a p-series.
The comparison series for the Limit Comparison Test is bn , where
bn n 32
1
sn 3
3n 3
3n 3
3n 32
1
2 ■ STRATEGY FOR TESTING SERIES
EXAMPLE 3
ne
n 2
n1
2
Since the integral x1 xex dx is easily evaluated, we use the Integral Test. The Ratio Test
also works.
EXAMPLE 4
1
n
n1
n3
n 1
4
Since the series is alternating, we use the Alternating Series Test.
EXAMPLE 5
k1
2k
k!
Since the series involves k!, we use the Ratio Test.
EXAMPLE 6
n1
1
2 3n
Since the series is closely related to the geometric series 13 n, we use the Comparison
Test.
Exercises
A Click here for answers.
Click here for solutions.
S
17.
1 2
n
1n
18.
n1
1–34
Test the series for convergence or divergence.
1.
n1
3.
n1
5.
n1
7.
n2
9.
n 1
n2 n
2
1
n n
2
2.
1
nsln n
8.
n2
13.
n1
15.
n0
16.
10.
n1
n1
n2 n
2 k k!
k 2!
ne
14.
21.
n1
23.
n1
22n
nn
tan1n
25.
20.
22.
24.
27.
28.
k1
1
33.
5k
3 4k
32.
sin1n
sn
34.
n1
1n1
sn 1
k5
5k
sn 2 1
n 2n 2 5
3
cosn2
n 2 4n
n2 1
5n
e 1n
n2
k
n1
30.
k1
n1
tan n
n sn
31.
n1
k ln k
k 13
n1
26.
n1
k1
n!
2
en
n1
n
n 2 25
ln n
sn
n1
29.
n
n
n2
sin n
n1
2 n 3
1
1
1
n n cos2 n
3nn 2
n!
19.
n1
n1
12.
n2 1
n3 1
k1
1 n1
n ln n
n!
2 5 8 3n 2
n1
k1
11.
n1
6.
e
1
n1
3
2 3n
k
n1
n2 n
4.
n1
2 k
n1
1
j1
n2
j
sj
j5
1
ln nln n
(s2 1)
n
n1
STRATEGY FOR TESTING SERIES ■ 3
Answers
S
1. D
15. C
27. C
Click here for solutions.
3. C
17. D
29. C
5. C
7. D
9. C
11. C
13. C
19. C
21. C
23. D
25. C
31. D
33. C
4 ■ STRATEGY FOR TESTING SERIES
Solutions: Strategy for Testing Series
∞
X
n2 − 1
1 − 1/n2
n2 − 1
=
lim
=
1
=
6
0,
so
the
series
diverges by the Test for
2
n→∞ n + 1
n→∞ 1 + 1/n
n2 + 1
n=1
1. lim an = lim
n→∞
Divergence.
∞
∞ 1
P
P
1
1
1
, a p-series that
< 2 for all n ≥ 1, so
converges by the Comparison Test with
2 +n
2
n2 + n
n
n
n=1
n=1 n
converges because p = 2 > 1.
¯
¯
¯
¯
¯
¯
n+2
¯ an+1 ¯
¯
¯ −3 · 23n ¯
23n ¯¯
¯ = lim ¯ (−3)
¯
¯ = lim 3 = 3 < 1, so the series
=
lim
·
5. lim ¯¯
n→∞
an ¯ n→∞ ¯ 23(n+1) (−3)n+1 ¯ n→∞ ¯ 23n · 23 ¯ n→∞ 23
8
∞
n+1
X
(−3)
is absolutely convergent by the Ratio Test.
23n
n=1
3.
1
√
. Then f is positive, continuous, and decreasing on [2, ∞), so we can apply the Integral Test.
x ln x
"
# Z
Z
√
u = ln x,
1
√
Since
dx
=
u−1/2 du = 2u1/2 + C = 2 ln x + C, we find
x ln x
du = dx/x
Z t
Z ∞
it
´
h √
³ √
√
dx
dx
√
√
= lim
= lim 2 ln x = lim 2 ln t − 2 ln 2 = ∞. Since the integral
t→∞ 2 x
t→∞
2
x ln x
ln x t→∞
2
∞
X
1
√
diverges.
diverges, the given series
ln n
n=2 n
7. Let f (x) =
9.
∞
X
∞
X
k2
. Using the Ratio Test, we get
ek
k=1
k=1
"µ
#
¯
¯
¯
¯
¶2
¯ ak+1 ¯
¯ (k + 1)2 ek ¯
k+1
1
1
1
¯
¯
¯
¯
= lim ¯ k+1 · 2 ¯ = lim
lim
·
= 12 · = < 1, so the series converges.
k→∞ ¯ ak ¯
k→∞
k→∞
e
k
k
e
e
e
k2 e−k =
∞ (−1)n+1
P
1
> 0 for n ≥ 2, {bn } is decreasing, and lim bn = 0, so the given series
converges by
n→∞
n ln n
n=2 n ln n
the Alternating Series Test.
¯
¯
¯ n+1
¯
·
¸
¯ an+1 ¯
¯
(n + 1)2
n+1
n! ¯¯
3(n + 1)2
¯ = lim ¯ 3
=
lim
·
13. lim ¯¯
= 3 lim
= 0 < 1, so the series
n→∞
n→∞ n2
an ¯ n→∞ ¯ (n + 1)!
3n n2 ¯ n→∞ (n + 1)n2
∞
X
3n n2
converges by the Ratio Test.
n!
n=1
¯
¯
¯
¯
¯ an+1 ¯
¯
(n + 1)!
2 · 5 · 8 · · · · · (3n + 2) ¯¯
¯= lim ¯
15. lim ¯¯
·
¯
n→∞
an ¯ n→∞ ¯ 2 · 5 · 8 · · · · · (3n + 2)[3(n + 1) + 2]
n!
n+1
1
= <1
= lim
n→∞ 3n + 5
3
∞
X
n!
so the series
converges by the Ratio Test.
2 · 5 · 8 · · · · · (3n + 2)
n=0
11. bn =
17. lim 21/n = 20 = 1, so lim (−1)n 21/n does not exist and the series
n→∞
Test for Divergence.
n→∞
∞
X
n=1
(−1)n 21/n diverges by the
STRATEGY FOR TESTING SERIES ■ 5
ln x
2 − ln x
ln n
19. Let f (x) = √ . Then f 0 (x) =
< 0 when ln x > 2 or x > e2 , so √ is decreasing for n > e2 .
2x3/2
x
n
∞
X
ln n
1/n
2
ln n
√ = lim √ = 0, so the series
(−1)n √ converges by
By l’Hospital’s Rule, lim √ = lim
n→∞
n→∞ 1/ (2 n)
n→∞
n
n
n
n=1
the Alternating Series Test.
µ ¶n
∞ (−2)2n
∞
p
P
P
4
4
= 0 < 1, so the given series is absolutely convergent by the
=
. lim n |an | = lim
21.
n→∞
n→∞ n
nn
n=1
n=1 n
Root Test.
µ ¶
1
1
23. Using the Limit Comparison Test with an = tan
and bn = , we have
n
n
an
tan(1/n)
tan(1/x) H
sec2 (1/x) · (−1/x2 )
= lim
= lim
= lim sec2 (1/x) = 12 = 1 > 0.
= lim
n→∞ bn
n→∞
x→∞
x→∞
x→∞
1/n
1/x
−1/x2
P∞
P∞
Since n=1 bn is the divergent harmonic series, n=1 an is also divergent.
¯
¯
¯
¯
2
¯
n2 ¯
¯ an+1 ¯
(n + 1)n! · en
n+1
¯ (n + 1)! e ¯
¯
¯
=
lim
·
25. Use the Ratio Test. lim ¯
= lim 2n+1 = 0 < 1, so
=
lim
¯
¯
2
2
n→∞
n→∞ e
an ¯ n→∞ ¯ e(n+1)
n! ¯ n→∞ en +2n+1 n!
lim
∞
X
n!
converges.
n2
e
n=1
·
¸t
∞ ln n
P
ln x
ln x
1
H
dx = lim −
(using integration by parts) = 1. So
converges by the Integral Test,
−
27.
2
2
t→∞
x
x
x 1
n=1 n
2
∞
P
k ln k
k ln k
ln k
k ln k
= 2 , the given series
and since
3 <
3 converges by the Comparison Test.
3
k
k
(k + 1)
k=1 (k + 1)
Z
∞
29. 0 <
∞ π/2
∞
tan−1 n
1
π/2 P
π P
< 3/2 .
=
which is a convergent p-series (p =
3/2
3/2
n
n
2 n=1 n3/2
n=1 n
3
2
> 1), so
∞ tan−1 n
P
converges by the Comparison Test.
n3/2
n=1
31. lim ak = lim
k→∞
Thus,
k→∞
∞
X
k=1
33. Let an =
3k
µ ¶k
µ ¶k
3
5
5k
(5/4)k
k
=
[divide
by
4
]
lim
=
0
and
lim
= ∞.
=
∞
since
lim
k→∞ (3/4)k + 1
k→∞ 4
k→∞ 4
3k + 4k
5k
diverges by the Test for Divergence.
+ 4k
∞
X
sin(1/n)
sin(1/n)
sin(1/n)
an
1
√
√
= lim
= 1 > 0, so
and bn = √ . Then lim
converges by
n→∞ bn
n→∞
1/n
n
n n
n
n=1
limit comparison with the convergent p-series
∞
X
1
(p = 3/2 > 1).
3/2
n
n=1