Testando Séries Infinitas - Exercícios, estratégias e soluções

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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 inte-
grating 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 , it is a -series, which we know to be conver-
gent if and divergent if .
2. If the series has the form or , it is a geometric series, which con-
verges if and diverges if . Some preliminary algebraic manipula-
tion may be required to bring the series into this form.
3. If the series has a form that is similar to a -series or a geometric series, then
one of the comparison tests should be considered. In particular, if is a rational
function or algebraic function of (involving roots of polynomials), then the
series should be compared with a -series. (The value of should be chosen as
in Section 8.3 by keeping only the highest powers of in the numerator and
denominator.) The comparison tests apply only to series with positive terms, but
if has some negative terms, then we can apply the Comparison Test to
and test for absolute convergence.
4. If you can see at a glance that , then the Test for Divergence
should be used.
5. If the series is of the form or , then the Alternating Series
Test is an obvious possibility.
6. Series that involve factorials or other products (including a constant raised to the
th power) are often conveniently tested using the Ratio Test. Bear in mind that
as for all -series and therefore all rational or algebraic
functions of . Thus, the Ratio Test should not be used for such series.
7. If , where is easily evaluated, then the Integral Test is effec-
tive (assuming the hypotheses of this test are satisfied).
In the following examples we don’t work out all the details but simply indicate which
tests should be used.
EXAMPLE 1
Since as , we should use the Test for Divergence.
EXAMPLE 2
Since is an algebraic function of , we compare the given series with a -series.
The comparison series for the Limit Comparison Test is , where
bn �
sn 3
3n 3
�
n 3�2
3n 3
�
1
3n 3�2
�
 bn
pnan
�
�
n�1
 
sn 3 � 1
3n 3 � 4n 2 � 2
n l �an l
1
2 � 0
�
�
n�1
 
n � 1
2n � 1
x�1 f �x� dxan � f �n�
n
pn l �� an�1�an �l 1
n
�
 ��1�nbn� ��1�n�1bn
lim n l � an � 0
�
 � an �
�
 an
n
pp
n
an
p
� r � � 1� r � � 1
�
 arn� arn�1
p � 1p � 1
p� 1�np
1
EXAMPLE 3
Since the integral is easily evaluated, we use the Integral Test. The Ratio Test
also works.
EXAMPLE 4
Since the series is alternating, we use the Alternating Series Test.
EXAMPLE 5
Since the series involves , we use the Ratio Test.
EXAMPLE 6
Since the series is closely related to the geometric series , we use the Comparison
Test.
�
 1�3n
�
�
n�1
 
1
2 � 3n
k!
�
�
k�1
 
2k
k!
�
�
n�1
 ��1�n 
n 3
n 4 � 1
x�1 xe�x2 dx
�
�
n�1
 ne�n
2
2 ■ STRATEGY FOR T ES T ING SER I ES
Exercises
1–34 Test the series for convergence or divergence.
1. 2.
3. 4.
5. 6.
7. 8.
9. 10.
11. 12.
13. 14.
15.
16. �
�
n�1
 
n 2 � 1
n 3 � 1
�
�
n�0
 
n!
2 � 5 � 8 � � � � � �3n � 2�
�
�
n�1
 sin n�
�
n�1
 
3nn 2
n!
�
�
n�1
 ��1�n 
n
n 2 � 25�
�
n�2
 
��1�n�1
n ln n
�
�
n�1
 n 2e�n
3�
�
k�1
 k 2e�k 
�
�
k�1
 
2 k k!
�k � 2�!�
�
n�2
 
1
nsln n
�
�
n�1
 
1
n � n cos2n
�
�
n�1
 
��3�n�1
23n
�
�
n�1
 ��1�n�1 
n � 1
n 2 � n
�
�
n�1
 
1
n 2 � n
�
�
n�1
 
n � 1
n 2 � n
�
�
n�1
 
n 2 � 1
n 2 � n
17. 18.
19. 20.
21. 22.
23. 24.
25. 26.
27. 28.
29. 30.
31. 32.
33. 34. �
�
n�1
 (sn 2 � 1)�
�
n�1
 
sin�1�n�
sn
�
�
n�2
 
1
�ln n�ln n�
�
k�1
 
5 k
3 k � 4 k
�
�
j�1
 ��1� j sj
 
j � 5�
�
n�1
 
tan�1n
nsn
�
�
n�1
 
e 1�n
n 2
�
�
k�1
 
k ln k
�k � 1�3
�
�
n�1
 
n 2 � 1
5n�
�
n�1
 
n!
e n
2
�
�
n�1
 
cos�n�2�
n 2 � 4n�
�
n�1
 tan�1�n�
�
�
n�1
 
sn 2 � 1
n 3 � 2n 2 � 5�
�
n�1
 
��2�2n
n n
�
�
k�1
 
k � 5
5k�
�
n�1
 ��1�n 
ln n
sn
�
�
n�2
 
��1�n�1
sn � 1�
�
n�1
 ��1�n21�nClick here for answers.A Click here for solutions.S
STRATEGY FOR T ES T ING SER I ES ■ 3
Answers
1. D 3. C 5. C 7. D 9. C 11. C 13. C
15. C 17. D 19. C 21. C 23. D 25. C
27. C 29. C 31. D 33. C
Click here for solutions.S
1. lim
n→∞
an = lim
n→∞
n2 − 1
n2 + 1
= lim
n→∞
1− 1/n2
1 + 1/n
= 1 6= 0, so the series
∞X
n=1
n2 − 1
n2 + 1
diverges by the Test for
Divergence.
3.
1
n2 + n
<
1
n2
for all n ≥ 1, so
∞P
n=1
1
n2 + n
converges by the Comparison Test with
∞P
n=1
1
n2
, a p-series that
converges because p = 2 > 1.
5. lim
n→∞
¯¯¯¯
an+1
an
¯¯¯¯
= lim
n→∞
¯¯¯¯
(−3)n+2
23(n+1)
· 2
3n
(−3)n+1
¯¯¯¯
= lim
n→∞
¯¯¯¯
−3 · 23n
23n · 23
¯¯¯¯
= lim
n→∞
3
23
=
3
8
< 1, so the series
∞X
n=1
(−3)n+1
23n
is absolutely convergent by the Ratio Test.
7. Let f(x) = 1
x
√
lnx
. Then f is positive, continuous, and decreasing on [2,∞), so we can apply the Integral Test.
Since
Z
1
x
√
ln x
dx
"
u = lnx,
du = dx/x
#
=
Z
u−1/2 du = 2u1/2 + C = 2
√
ln x+ C, we findZ ∞
2
dx
x
√
ln x
= lim
t→∞
Z t
2
dx
x
√
ln x
= lim
t→∞
h
2
√
lnx
it
2
= lim
t→∞
³
2
√
ln t− 2
√
ln 2
´
=∞. Since the integral
diverges, the given series
∞X
n=2
1
n
√
lnn
diverges.
9.
∞X
k=1
k2e−k =
∞X
k=1
k2
ek
. Using the Ratio Test, we get
lim
k→∞
¯¯¯¯
ak+1
ak
¯¯¯¯
= lim
k→∞
¯¯¯¯
(k + 1)2
ek+1
· e
k
k2
¯¯¯¯
= lim
k→∞
"µ
k + 1
k
¶2
· 1
e
#
= 12 · 1
e
=
1
e
< 1, so the series converges.
11. bn =
1
n lnn
> 0 for n ≥ 2, {bn} is decreasing, and lim
n→∞
bn = 0, so the given series
∞P
n=2
(−1)n+1
n lnn
converges by
the Alternating Series Test.
13. lim
n→∞
¯¯¯¯
an+1
an
¯¯¯¯
= lim
n→∞
¯¯¯¯
3n+1 (n+ 1)2
(n+ 1)!
· n!
3nn2
¯¯¯¯
= lim
n→∞
·
3(n+ 1)2
(n+ 1)n2
¸
= 3 lim
n→∞
n+ 1
n2
= 0 < 1, so the series
∞X
n=1
3nn2
n!
converges by the Ratio Test.
15. lim
n→∞
¯¯¯¯
an+1
an
¯¯¯¯
= lim
n→∞
¯¯¯¯
(n+ 1)!
2 · 5 · 8 · · · · · (3n+ 2)[3(n+ 1) + 2] ·
2 · 5 · 8 · · · · · (3n+ 2)
n!
¯¯¯¯
= lim
n→∞
n+ 1
3n+ 5
=
1
3
< 1
so the series
∞X
n=0
n!
2 · 5 · 8 · · · · · (3n+ 2) converges by the Ratio Test.
17. lim
n→∞
21/n = 20 = 1, so lim
n→∞
(−1)n 21/n does not exist and the series
∞X
n=1
(−1)n21/n diverges by the
Test for Divergence.
4 ■ S TRATEGY FOR T ES T ING SER I ES
Solutions: Strategy for Testing Series
19. Let f(x) = ln x√
x
. Then f 0(x) = 2− ln x
2x3/2
< 0 when lnx > 2 or x > e2, so lnn√
n
is decreasing for n > e2.
By l’Hospital’s Rule, lim
n→∞
lnn√
n
= lim
n→∞
1/n
1/ (2
√
n)
= lim
n→∞
2√
n
= 0, so the series
∞X
n=1
(−1)n lnn√
n
converges by
the Alternating Series Test.
21.
∞P
n=1
(−2)2n
nn
=
∞P
n=1
µ
4
n
¶n
. lim
n→∞
n
p|an| = lim
n→∞
4
n
= 0 < 1, so the given series is absolutelyconvergent by the
Root Test.
23. Using the Limit Comparison Test with an = tan
µ
1
n
¶
and bn =
1
n
, we have
lim
n→∞
an
bn
= lim
n→∞
tan(1/n)
1/n
= lim
x→∞
tan(1/x)
1/x
H
= lim
x→∞
sec2(1/x) · (−1/x2)
−1/x2 = limx→∞ sec
2(1/x) = 12 = 1 > 0.
Since
P∞
n=1 bn is the divergent harmonic series,
P∞
n=1 an is also divergent.
25. Use the Ratio Test. lim
n→∞
¯¯¯¯
an+1
an
¯¯¯¯
= lim
n→∞
¯¯¯¯
¯ (n+ 1)!e(n+1)2 · en
2
n!
¯¯¯¯
¯ = limn→∞ (n+ 1)n! · en
2
en2+2n+1n!
= lim
n→∞
n+ 1
e2n+1
= 0 < 1, so
∞X
n=1
n!
en2
converges.
27.
Z ∞
2
lnx
x2
dx = lim
t→∞
·
− ln x
x
− 1
x
¸t
1
(using integration by parts) H= 1. So
∞P
n=1
lnn
n2
converges by the Integral Test,
and since k ln k
(k + 1)3
<
k ln k
k3
=
ln k
k2
, the given series
∞P
k=1
k ln k
(k + 1)3
converges by the Comparison Test.
29. 0 <
tan−1 n
n3/2
<
π/2
n3/2
.
∞P
n=1
π/2
n3/2
=
π
2
∞P
n=1
1
n3/2
which is a convergent p-series (p = 3
2
> 1), so
∞P
n=1
tan−1 n
n3/2
converges by the Comparison Test.
31. lim
k→∞
ak = lim
k→∞
5k
3k + 4k
= [divide by 4k] lim
k→∞
(5/4)k
(3/4)k + 1
=∞ since lim
k→∞
µ
3
4
¶k
= 0 and lim
k→∞
µ
5
4
¶k
=∞.
Thus,
∞X
k=1
5k
3k + 4k
diverges by the Test for Divergence.
33. Let an =
sin(1/n)√
n
and bn =
1
n
√
n
. Then lim
n→∞
an
bn
= lim
n→∞
sin(1/n)
1/n
= 1 > 0, so
∞X
n=1
sin(1/n)√
n
converges by
limit comparison with the convergent p-series
∞X
n=1
1
n3/2
(p = 3/2 > 1).
STRATEGY FOR T ES T ING SER I ES ■ 5