Section 11.7 Strategy for Testing Seriesmathcal/download/107/HW/11.7.pdfSection 11.7 Strategy for...

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Section 11.7 Strategy for Testing Series 3. =1 (1) 2 1 3 +1 = =1 (1) . Now = 2 1 3 +1 0 for 2, { } is decreasing for 2, and lim →∞ =0, so the series =1 (1) 2 1 3 +1 converges by the Alternating Series Test. By Exercise 1, =1 2 1 3 +1 diverges, so the series =1 (1) 2 1 3 +1 is conditionally convergent. 7. =1 2 = =1 2 . Using the Ratio Test, we get lim →∞ +1 = lim →∞ ( + 1) 2 +1 · 2 = lim →∞ +1 2 · 1 =1 2 · 1 = 1 1, so the series converges. 13. lim →∞ +1 = lim →∞ 3 +1 ( + 1) 2 ( + 1)! · ! 3 2 = lim →∞ 3( + 1) 2 ( + 1) 2 = 3 lim →∞ +1 2 =0 1, so the series =1 3 2 ! converges by the Ratio Test. 17. lim →∞ +1 = lim →∞ 1 · 3 · 5 ····· (2 1)(2 + 1) 2 · 5 · 8 ····· (3 1)(3 + 2) · 2 · 5 · 8 ····· (3 1) 1 · 3 · 5 ····· (2 1) = lim →∞ 2 +1 3 +2 = lim →∞ 2+1 3+2 = 2 3 1 so the series =1 1 · 3 · 5 ····· (2 1) 2 · 5 · 8 ····· (3 1) converges by the Ratio Test. 24. lim →∞ = lim →∞ sin 1 = lim →∞ sin(1) 1 = lim 0 + sin =1 6=0, so the series =1 sin(1) diverges by the Test for Divergence. 36. Note that (ln ) ln = ln ln ln = ln ln ln = ln ln and ln ln →∞ as →∞, so ln ln 2 for sufficiently large . For these we have (ln ) ln 2 , so 1 (ln ) ln 1 2 . Since =2 1 2 converges [ =2 1], so does =2 1 (ln ) ln by the Comparison Test. 37. lim →∞ || = lim →∞ (2 1 1) = 1 1=0 1, so the series =1 2 1 converges by the Root Test. 1

Transcript of Section 11.7 Strategy for Testing Seriesmathcal/download/107/HW/11.7.pdfSection 11.7 Strategy for...

Page 1: Section 11.7 Strategy for Testing Seriesmathcal/download/107/HW/11.7.pdfSection 11.7 Strategy for Testing Series 1018 ¤ CHAPTER11INFINITESEQUENCESANDSERIES 11.7 Strategy for TestingSeries

Section 11.7 Strategy for Testing Series

1018 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES

11.7 Strategy for Testing Series

1. Use the Limit Comparison Test with =2 − 1

3 + 1and =

1

:

lim→∞

= lim

→∞(2 − 1)

3 + 1= lim

→∞3 −

3 + 1= lim

→∞1− 12

1 + 1 /3= 1 0. Since

∞=1

1

is the divergent harmonic series, the

series∞=1

2 − 1

3 + 1also diverges.

2.− 1

3 + 1

3 + 1

3=

1

2for ≥ 1, so

∞=1

− 1

3 + 1converges by comparison with

∞=1

1

2, which converges because it

is a p-series with = 2 1.

3.∞=1

(−1)2 − 1

3 + 1=

∞=1

(−1). Now =2 − 1

3 + 1 0 for ≥ 2, {} is decreasing for ≥ 2, and lim

→∞ = 0, so

the series∞=1

(−1)2 − 1

3 + 1converges by the Alternating Series Test. By Exercise 1,

∞=1

2 − 1

3 + 1diverges, so the series

∞=1

(−1)2 − 1

3 + 1is conditionally convergent.

4. lim→∞

|| = lim→∞

(−1)2 − 1

2 + 1

= lim→∞

1− 12

1 + 12= 1 6= 0, so the series

∞=1

(−1)2 − 1

2 + 1diverges by the Test for

Divergence.

Note that lim

→∞(−1)

2 − 1

2 + 1does not exist.

5. lim→∞

2

H= lim

→∞

2

H= lim

→∞

2=∞, so lim

→∞

2=∞. Thus, the series

∞=1

2diverges by the Test for Divergence.

6. lim→∞

|| = lim

→∞

2

(1 + )3= lim

→∞2

(1 + )3= lim

→∞1

(1+ 1)3=

0

1= 0 1, so the series

∞=1

2

(1 + )3

converges by the Root Test.

7.∞=1

2− =∞=1

2

. Using the Ratio Test, we get

lim→∞

+1

= lim→∞

( + 1)2

+1·

2

= lim→∞

+ 1

2

· 1

= 12 · 1

=

1

1, so the series converges.

8. lim→∞

+1

= lim→∞

(+ 1)4

4+1· 4

4

= lim→∞

(+ 1)4

44=

1

4lim→∞

1 +

1

4

=1

4(1) =

1

4 1, so the series

∞=1

(−1)−14

4is absolutely convergent (and therefore convergent) by the Ratio Test.

c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

1018 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES

11.7 Strategy for Testing Series

1. Use the Limit Comparison Test with =2 − 1

3 + 1and =

1

:

lim→∞

= lim

→∞(2 − 1)

3 + 1= lim

→∞3 −

3 + 1= lim

→∞1− 12

1 + 1 /3= 1 0. Since

∞=1

1

is the divergent harmonic series, the

series∞=1

2 − 1

3 + 1also diverges.

2.− 1

3 + 1

3 + 1

3=

1

2for ≥ 1, so

∞=1

− 1

3 + 1converges by comparison with

∞=1

1

2, which converges because it

is a p-series with = 2 1.

3.∞=1

(−1)2 − 1

3 + 1=

∞=1

(−1). Now =2 − 1

3 + 1 0 for ≥ 2, {} is decreasing for ≥ 2, and lim

→∞ = 0, so

the series∞=1

(−1)2 − 1

3 + 1converges by the Alternating Series Test. By Exercise 1,

∞=1

2 − 1

3 + 1diverges, so the series

∞=1

(−1)2 − 1

3 + 1is conditionally convergent.

4. lim→∞

|| = lim→∞

(−1)2 − 1

2 + 1

= lim→∞

1− 12

1 + 12= 1 6= 0, so the series

∞=1

(−1)2 − 1

2 + 1diverges by the Test for

Divergence.

Note that lim

→∞(−1)

2 − 1

2 + 1does not exist.

5. lim→∞

2

H= lim

→∞

2

H= lim

→∞

2=∞, so lim

→∞

2=∞. Thus, the series

∞=1

2diverges by the Test for Divergence.

6. lim→∞

|| = lim

→∞

2

(1 + )3= lim

→∞2

(1 + )3= lim

→∞1

(1+ 1)3=

0

1= 0 1, so the series

∞=1

2

(1 + )3

converges by the Root Test.

7.∞=1

2− =∞=1

2

. Using the Ratio Test, we get

lim→∞

+1

= lim→∞

( + 1)2

+1·

2

= lim→∞

+ 1

2

· 1

= 12 · 1

=

1

1, so the series converges.

8. lim→∞

+1

= lim→∞

(+ 1)4

4+1· 4

4

= lim→∞

(+ 1)4

44=

1

4lim→∞

1 +

1

4

=1

4(1) =

1

4 1, so the series

∞=1

(−1)−14

4is absolutely convergent (and therefore convergent) by the Ratio Test.

c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.7 STRATEGY FOR TESTING SERIES ¤ 1019

9. lim→∞

+1

= lim→∞

2+2

(2+ 2)!· (2)!

2

= lim→∞

2

(2+ 2)(2+ 1)= 0 1, so the series

∞=0

(−1)2

(2)!is absolutely

convergent (and therefore convergent) by the Ratio Test.

10. lim→∞

|| = lim

→∞

(2+ 1)

2

= lim→∞

2+ 1

2= lim

→∞

2

+

1

2

= 0 1, so the series

∞=1

(2+ 1)

2

converges by the Root Test.

11.∞=1

1

3+

1

3

=

∞=1

1

3+

∞=1

1

3

. The first series converges since it is a -series with = 3 1 and the second

series converges since it is geometric with || = 13 1. The sum of two convergent series is convergent.

12.∞=1

2!

( + 2)!=

∞=1

2

( + 1)( + 2). Using the Ratio Test, we get

lim→∞

+1

= lim→∞

2+1

( + 2)( + 3)· ( + 1)( + 2)

2

= lim→∞

2 · + 1

+ 3

= 2 1, so the series diverges.

Or: Use the Test for Divergence.

13. lim→∞

+1

= lim→∞

3+1 (+ 1)2

(+ 1)!· !

32

= lim→∞

3(+ 1)2

(+ 1)2= 3 lim

→∞+ 1

2= 0 1, so the series

∞=1

32

!

converges by the Ratio Test.

14.

sin 2

1 + 2

≤ 1

1 + 2

1

2=

1

2

, so the series

∞=1

sin 2

1 + 2

converges by comparison with the geometric series∞=1

1

2

with || = 1

2 1. Thus, the series

∞=1

sin 2

1 + 2converges absolutely, implying convergence.

15. =2−13+1

=

22−1331

=

3

2

2 · 3

. By the Root Test, lim

→∞

6

= lim

→∞6

= 0 1, so the series

∞=1

6

converges. It follows from Theorem 8(i) in Section 11.2 that the given series,

∞=1

2−13+1

=

∞=1

3

2

6

,

also converges.

16. Use the Limit Comparison Test with =

√4 + 1

3 + and =

1

:

lim→∞

= lim

→∞√4 + 1

(2 + 1)= lim

→∞

√4 + 12

(2 + 1)2= lim

→∞

1 + 14

1 + 12= 1 0. Since

∞=1

1

is the divergent harmonic

series, the series∞=1

√4 + 1

3 + also diverges.

17. lim→∞

+1

= lim→∞

1 · 3 · 5 · · · · · (2− 1)(2+ 1)

2 · 5 · 8 · · · · · (3− 1)(3+ 2)· 2 · 5 · 8 · · · · · (3− 1)

1 · 3 · 5 · · · · · (2− 1)

= lim→∞

2+ 1

3+ 2

= lim→∞

2 + 1

3 + 2=

2

3 1

so the series∞=1

1 · 3 · 5 · · · · · (2− 1)

2 · 5 · 8 · · · · · (3− 1)converges by the Ratio Test.

c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

SECTION 11.7 STRATEGY FOR TESTING SERIES ¤ 1019

9. lim→∞

+1

= lim→∞

2+2

(2+ 2)!· (2)!

2

= lim→∞

2

(2+ 2)(2+ 1)= 0 1, so the series

∞=0

(−1)2

(2)!is absolutely

convergent (and therefore convergent) by the Ratio Test.

10. lim→∞

|| = lim

→∞

(2+ 1)

2

= lim→∞

2+ 1

2= lim

→∞

2

+

1

2

= 0 1, so the series

∞=1

(2+ 1)

2

converges by the Root Test.

11.∞=1

1

3+

1

3

=

∞=1

1

3+

∞=1

1

3

. The first series converges since it is a -series with = 3 1 and the second

series converges since it is geometric with || = 13 1. The sum of two convergent series is convergent.

12.∞=1

2!

( + 2)!=

∞=1

2

( + 1)( + 2). Using the Ratio Test, we get

lim→∞

+1

= lim→∞

2+1

( + 2)( + 3)· ( + 1)( + 2)

2

= lim→∞

2 · + 1

+ 3

= 2 1, so the series diverges.

Or: Use the Test for Divergence.

13. lim→∞

+1

= lim→∞

3+1 (+ 1)2

(+ 1)!· !

32

= lim→∞

3(+ 1)2

(+ 1)2= 3 lim

→∞+ 1

2= 0 1, so the series

∞=1

32

!

converges by the Ratio Test.

14.

sin 2

1 + 2

≤ 1

1 + 2

1

2=

1

2

, so the series

∞=1

sin 2

1 + 2

converges by comparison with the geometric series∞=1

1

2

with || = 1

2 1. Thus, the series

∞=1

sin 2

1 + 2converges absolutely, implying convergence.

15. =2−13+1

=

22−1331

=

3

2

2 · 3

. By the Root Test, lim

→∞

6

= lim

→∞6

= 0 1, so the series

∞=1

6

converges. It follows from Theorem 8(i) in Section 11.2 that the given series,

∞=1

2−13+1

=

∞=1

3

2

6

,

also converges.

16. Use the Limit Comparison Test with =

√4 + 1

3 + and =

1

:

lim→∞

= lim

→∞√4 + 1

(2 + 1)= lim

→∞

√4 + 12

(2 + 1)2= lim

→∞

1 + 14

1 + 12= 1 0. Since

∞=1

1

is the divergent harmonic

series, the series∞=1

√4 + 1

3 + also diverges.

17. lim→∞

+1

= lim→∞

1 · 3 · 5 · · · · · (2− 1)(2+ 1)

2 · 5 · 8 · · · · · (3− 1)(3+ 2)· 2 · 5 · 8 · · · · · (3− 1)

1 · 3 · 5 · · · · · (2− 1)

= lim→∞

2+ 1

3+ 2

= lim→∞

2 + 1

3 + 2=

2

3 1

so the series∞=1

1 · 3 · 5 · · · · · (2− 1)

2 · 5 · 8 · · · · · (3− 1)converges by the Ratio Test.

c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

1020 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES

18. Using the Limit Comparison Test with =2 + 1

3 + 1and =

1

, we have

lim→∞

= lim

→∞

2 + 1

3 + 1·

1

= lim

→∞3 +

3 + 1= lim

→∞1 + 12

1 + 13= 1 0. Since

∞=1

is the divergent harmonic

series,∞=1

is also divergent.

19. Let () =ln√. Then 0() =

2− ln

232 0 when ln 2 or 2, so

ln√is decreasing for 2.

By l’Hospital’s Rule, lim→∞

ln√

= lim→∞

1

12√ = lim

→∞2√

= 0, so the series∞=1

(−1)ln√converges by the

Alternating Series Test.

20. =3√ − 1

(√ + 1)

3√

(√ + 1)

3√

=13

32=

1

76, so the series

∞=1

3√ − 1

(√ + 1)

converges by comparison with the

convergent -series∞=1

1

76

= 7

6 1

.

21. lim→∞

|| = lim→∞

(−1) cos(12) = lim

→∞

cos(12) = cos 0 = 1, so the series

∞=1

(−1) cos(12) diverges by the

Test for Divergence.

22.

√2 − 1

3 + 22 + 5

3 + 22 + 5

3=

1

2for ≥ 1, so

∞=1

√2 − 1

3 + 22 + 5converges by the Comparison Test with the

convergent -series∞=1

12 [ = 2 1].

23. Using the Limit Comparison Test with = tan

1

and =

1

, we have

lim→∞

= lim

→∞tan(1)

1= lim

→∞tan(1)

1

H= lim

→∞sec2(1) · (−12)

−12= lim

→∞sec2(1) = 12 = 1 0. Since

∞=1

is the divergent harmonic series,∞=1

is also divergent.

24. lim→∞

= lim→∞

sin

1

= lim

→∞sin(1)

1= lim

→0+

sin

= 1 6= 0, so the series

∞=1

sin(1) diverges by the

Test for Divergence.

25. Use the Ratio Test. lim→∞

+1

= lim→∞

(+ 1)!

(+1)2·

2

!

= lim→∞

(+ 1)! · 2

2+2+1!= lim

→∞+ 1

2+1= 0 1, so

∞=1

!

2

converges.

26. lim→∞

+1

= lim→∞

+1

= lim

→∞

2 + 2+ 2

5+1· 5

2 + 1

= lim

→∞

1 + 2+ 22

1 + 12· 1

5

=

1

5 1, so

∞=1

2 + 1

5

converges by the Ratio Test.

c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

1022 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES

36. Note that (ln)ln

=ln ln

ln=ln

ln ln= ln ln and ln ln→∞ as →∞, so ln ln 2 for sufficiently

large . For these we have (ln)ln

2, so1

(ln)ln

1

2. Since

∞=2

1

2converges [ = 2 1], so does

∞=2

1

(ln)ln

by the Comparison Test.

37. lim→∞

|| = lim

→∞(21 − 1) = 1− 1 = 0 1, so the series

∞=1

2− 1

converges by the Root Test.

38. Use the Limit Comparison Test with =√

2− 1 and = 1. Then

lim→∞

= lim

→∞21 − 1

1= lim

→∞21 − 1

1

H= lim

→∞21 · ln 2 · (−12)

−12= lim

→∞(21 · ln 2) = 1 · ln 2 = ln 2 0.

So since∞=1

diverges (harmonic series), so does∞=1

2− 1.

Alternate solution: √

2− 1 =1

2(−1) + 2(−2) + 2(−3) + · · ·+ 21 + 1[rationalize the numerator] ≥ 1

2,

and since∞=1

1

2=

1

2

∞=1

1

diverges (harmonic series), so does

∞=1

2− 1by the Comparison Test.

11.8 Power Series

1. A power series is a series of the form∞

=0 = 0 + 1+ 2

2 + 33 + · · · , where is a variable and the ’s are

constants called the coefficients of the series.

More generally, a series of the form∞

=0 (− ) = 0 + 1(− ) + 2(− )2 + · · · is called a power series in(− ) or a power series centered at or a power series about , where is a constant.

2. (a) Given the power series∞

=0 (− ), the radius of convergence is:

(i) 0 if the series converges only when =

(ii) ∞ if the series converges for all , or

(iii) a positive number such that the series converges if |− | and diverges if |− | .

In most cases, can be found by using the Ratio Test.

(b) The interval of convergence of a power series is the interval that consists of all values of for which the series converges.

Corresponding to the cases in part (a), the interval of convergence is: (i) the single point {}, (ii) all real numbers; that is,the real number line (−∞∞), or (iii) an interval with endpoints − and + which can contain neither, either, or

both of the endpoints. In this case, we must test the series for convergence at each endpoint to determine the interval of

convergence.

3. If = (−1), then

lim→∞

+1

= lim→∞

(−1)+1(+ 1)+1

(−1)

= lim→∞

(−1)+ 1

= lim→∞

1 +

1

||

= ||. By the Ratio Test, the

series∞=1

(−1) converges when || 1, so the radius of convergence = 1. Now we’ll check the endpoints, that is,

c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

1