EXERCISES FOR CHAPTER 3: Infinite Seriesdbhs.wvusd.k12.ca.us/.../2011/4/27/57166616/P3notes_Infinite_Serie… · 2011-04-27 · EXERCISES FOR CHAPTER 3: Infinite Series SET 3.1 1

EXERCISES FOR CHAPTER 3: Infinite Series SET 3.1

1. Express 4

k(k + 2) in partial fractions and hence find the sum

4

k(k + 2)k=1

n

.

Determine if limn

4

k(k + 2)k=1

n

exists.

Solution 4

k(k + 2)=A

k+

B

k + 2. By the cover-up method A = 2 and B = 2 . Hence

4

k(k + 2)k=1

n

=2

1

2

3+2

2

2

4+ +

2

n 1

2

n +1+2

n

2

n + 2

= 2 +12

n +1

2

n + 2

= 32n + 3

(n +1)(n + 2)

=n(3n + 5)

(n +1)(n + 2)

Thus, 4

k(k + 2)k=1

= limn

n(3n + 5)

(n +1)(n + 2)= 3 .

2. Show that

5

1 2+

5

2 3+

5

3 4+ exists and find its value.

Solution 5

k(k +1)=A

k+

B

k +1. By the cover-up method A = 5 and B = 5 . Hence

5

k(k +1)k=1

n

=5

1

5

2+5

2

5

3+ +

5

n 1

5

n+5

n

5

n +1

= 55

n +1

=5n

n +1

Thus, 5

k(k +1)k=1

= limn

5n

n +1= 5 .

3. Find the value of the infinite sum

1

1 4+

1

2 5+

1

3 6+ +

1

n(n + 3)+ .

Solution 1

n(n + 3)=A

n+

B

n + 3. By the cover-up method A =

1

3 and B =

1

3. Hence

K. A. Tsokos: Series and D.E.

25

1

n(n + 3)n=1

N

=1

3

1

1

1

4+1

3

1

2

1

5+1

3

1

3

1

6+1

3

1

4

1

7+

+1

3

1

N 2

1

N +1+1

3

1

N 1

1

N + 2+1

3

1

N

1

N + 3

=1

31+

1

2+1

3

1

N +1

1

N + 2

1

N + 3

=N(11N 2

+ 48N + 49)

18(N +1)(N + 2)(N + 3)

Thus 1

n(n + 3)n=1

= limN

N(11N 2+ 48N + 49)

18(N +1)(N + 2)(N + 3)=11

18.

4. Evaluate

1

1 3 5+

1

2 4 6+ +

1

n(n + 2)(n + 4)+ .

Solution 1

n(n + 2)(n + 4)=A

n+

B

n + 2+

C

n + 4. By the cover-up method A =

1

8, B =

1

4 and

C =1

8. We then write

1

n(n + 2)(n + 4)=1

8n

2

8(n + 2)+

1

8(n + 4)

=1

8n

1

8(n + 2)+

1

8(n + 4)

1

8(n + 2)

=1

8

1

n

1

(n + 2)+1

8

1

(n + 4)

1

(n + 2)

so that

1

n(n + 2)(n + 4)n=1

N

=1

8

1

n

1

(n + 2)

1

8

1

(n + 2)

1

(n + 4)1

N

1

N

=1

8(1

1

3) + (

1

2

1

4) + (

1

3

1

5) + + (

1

N 1

1

N +1) + (

1

N

1

N + 2) +

1

8(1

3

1

5) + (

1

4

1

6) + (

1

5

1

7) + + (

1

N +1

1

N + 3) + (

1

N + 2

1

N + 4)

=1

81+

1

2

1

N +1

1

N + 2+

1

8

1

3+1

4

1

N + 3

1

N + 4

=N(N + 5)(11N 2

+ 55N + 62)

96(N +1)(N + 2)(N + 3)(N + 4)


26

Thus, 1

n(n + 2)(n + 4)n=1

= limN

N(N + 5)(11N 2+ 55N + 62)

96(N +1)(N + 2)(N + 3)(N + 4)=11

96.

5. Find

2

1 3 5+

3

3 5 7+ +

n +1

(2n 1)(2n +1)(2n + 3)+ .

Solution n +1

(2n +1)(2n 1)(2n + 3)=

A

2n +1+

B

2n 1+

C

2n + 3. By the cover-up method,

A =1

8, B =

3

16 and C =

1

16. Hence we may write:

n +1

(2n +1)(2n 1)(2n + 3)=

2

16

1

2n +1+3

16

1

2n 1

1

16

1

2n + 3

=2

16

1

2n 1

1

2n +1+1

16

1

2n 1

1

2n + 3

Then,

n +1

(2n 1)(2n +1)(2n + 3)n=1

N

=2

16(1

1

3) + (

1

3

1

5) + (

1

5

1

7) + + (

1

2N 1

1

2N +1)

+1

16(1

1

5) + (

1

3

1

7) + (

1

5

1

9) + + (

1

2N 3

1

2N +1) + (

1

2N 1

1

2N + 3)

n +1

(2n 1)(2n +1)(2n + 3)n=1

N

=2

161

1

2N +1+1

161+

1

3

1

2N +1

1

2N + 3)

=N(5N + 7)

6(2N + 3)(2N +1)

Hence, n +1

(2n 1)(2n +1)(2n + 3)=

n=1

limN

N(5N + 7)

6(2N + 3)(2N +1)=5

24.

6. Express k + 3

k(k2 1) in partial fractions and hence find the sum

k + 3

k(k2 1)k=2

n

.

Determine if limn

k + 3

k(k2 1)k=2

n

exists.

Solution k + 3

k(k2 1)=

k + 3

k(k 1)(k +1)=A

k+

B

k 1+

C

k +1. By the cover-up method, A = 3 ,

B = 2 and C = 1 . Hence we may write:

k + 3

k(k2 1)=

3

k+

2

k 1+

1

k +1= 2

1

k 1

1

k

1

k

1

k +1

so that


27

k + 3

k(k2 1)2

N

= 2 (11

2) + (

1

2

1

3) + + (

1

N 1

1

N)

(1

2

1

3) + (

1

3

1

4) + + (

1

N

1

N +1

= 2 11

N

1

2

1

N +1

=2(N 1)

N

N 1

2(N +1)

=(N 1)(3N + 4)

2N(N +1)

and hence k + 3

k(k2 1)2

= limN

(N 1)(3N + 4)

2N(N +1)=3

2.

7. Investigate, by using partial fractions, whether the integrals (a) 1

x2 5x + 65

dx

and (b) x +1

x2 5x + 65

dx exist.

Solution

(a) 1

x2 5x + 6=

1

(x 3)(x 2)=

A

x 3+

B

x 2. By the cover up method, A = 1 and

B = 1 , so that I = lim1

x2 5x + 65

dx = lim (1

x 3

1

x 2)

5

dx . Then,

I = lim (ln x 3 ln x 2 )5

= lim lnx 3

x 2 5

= lim ln3

2ln2

3

= 0 + ln3

2

The integral converges. (Notice that lim ln3

2= ln lim

3

2= ln1 = 0 .)

(b) x +1

x2 5x + 6=

x +1

(x 3)(x 2)=

A

x 3+

B

x 2. By the cover up method, A = 4

and B = 3 , so that I = limx +1

x2 5x + 65

dx = lim (4

x 3

3

x 2)

5

dx . Then,


28

I = lim (4 ln x 3 3ln x 2 )5

= lim(4 ln 3 3ln 2 ) (4 ln2 3ln 3)

= lim(4 ln 3 3ln 2 ) ln16

27

To evaluate the limit, lim(4 ln 3 3ln 2 ) write

lim(4 ln 3 3ln 2 ) = lim(4 ln 3 4 ln 2 ) + lim ln 2

= 4 lim ln3

2+ lim ln 2

= 0 + lim ln 2

=

The integral diverges. (Again, lim 4 ln3

2= 4 ln lim

3

2= 4 ln1 = 0 .)

SET 3.2 Apply the divergence test to the series below to determine those that diverge. If the test is inconclusive you should say so.

1. (a) n +1

n1

(b) n2

n2 +11

(c) n

2n + 51

Solution

(a) limn

n +1

n= 1, so series diverges.

(b) limn

n2 +1

n2= 1 , so series diverges.

(c) ) limn

n

2n + 5=1

2, so series diverges.

2. (a) 1+ ( 1)n( )1

(b) ( e)1

(c) ln(n + 2) lnn( )2

Solution (a) lim

n1+ ( 1)n( ) does not exist so series diverges.

(b) limn

un = e 0 , so series diverges.

(c) limn

ln(n + 2) lnn( ) = limn

lnn + 2

n= ln lim

n

n + 2

n= ln1 = 0 , inconclusive.

3. (a) e n

1

(b) en

1

(c) ne n

1

Solution (a) lim

ne n

= 0 , so test is inconclusive. (Series is geometric and converges.)


29

(b) limn

en = , so series diverges.

(c) limn

ne n= 0 , so test in inconclusive.

4. (a) ( n2 + n1

n) (b) 1

n +1 n

1

(c) 1

2

1

n

Solution

(a) limn

n2 + n n( ) = 12 by the example in the text so series diverges.

(b)

limn

1

n +1 n= lim

n

1

n +1 n

n +1 + n

n +1 + n

= limn

n +1 + n

n +1 n

= limn

n +1 + n

=

so series diverges.

(c) limn

1

2

n


5. (a) (1( 1)n

n)

1

(b) nn

1

(c) en 1

n1

(a) limn

1( 1)n

n= 1 , so series diverges.

(b) limn

nn= 1 , so series diverges.

(c) limn

en 1n


6. (a) n 2

n

n

1

(b) n + 3

n

n

1

(c) n

n + 3

n

1

Solution

(a) limn

n 2

n

n

= e 2 , so series diverges.

(b) limn

n + 3

n

n

= e3 , so series diverges.

(c) limn

n

n + 3

n

= e 3 , so series diverges.

7. (a) 1

lnn2

(b) lnn +1

n

1

(c) lnn

n +1

1

Solution


30

(a) limn

1

lnn= 0 , so test is inconclusive.

(b) limn

ln(n +1

n) = 0 , so test is inconclusive.

(c) limn

ln(n

n +1) = 0 , so test is inconclusive.

8. (a) cos1

n

1

(b) n sin1

n

1

(c) n

n + 21

Solution

(a) limn

cos1

n= 1 , so series diverges.

(b) limn

nsin1

n= lim

n

sin1n1n

= 1 , so series diverges.

(c) limn

n

n + 2= 1, so series diverges.

9. (a) n!

5nn=1

(b) nn

n!n=1

(c) 2n!+7

n10 + n!1

Solution

(a) limn

n!

5n= , so series diverges.

(b) limn

nn

n!= , so series diverges.

(c) limn

2n!+ 7

n10 + n!= 2 , so series diverges.

10. 2n + 7

3n 21

n

Solution

limn

2n + 7

3n 2

n

= limn

(2n)n (1+72n)n

(3n)n (123n)n

=2

3

ne7 /2

e 2 /3= 0 , test is inconclusive.

SET 3.3 Apply, if possible, the integral test on the following series. If the test cannot be applied, state why.


31

1. (a) n

n +11

(b) n

n2 +11

(c) n4

4n5 + 21

Solution

(a) Terms are increasing because d

dn

n

n +1=(n +1) n

(n +1)2=

1

(n +1)2> 0 , test cannot be

applied.

(b) Terms are decreasing because d

dn

n

n2 +1=(n2 +1) n 2n

(n2 +1)2=1 n2

(n2 +1)2< 0 and

n

n2 +11

dn = limn

n2 +11

dn =1

2lim ln(n2 +1)

1= so series diverges.

(c) d

dn

n4

4n5 + 2=4n3(4n5 + 2) n4 20n4

(4n5 + 2)2=8n3 4n8

(4n5 + 2)2=4n3(2 n5 )

(4n5 + 2)2< 0 for n > 1 so

terms are decreasing. limn4

4n5 + 21

=1

20lim ln(4 5

+ 2)1

20ln6 = , so series

diverges.

2. (a) 1

n lnn2

(b) lnn

n21

(c) lnn

n1

Solution

(a) Terms are clearly decreasing. The integral is lim1

n lnndn

1

. Make the substitution

u = lnn du =dn

n so that

lim1

n lnndn

2

= lim1

udu

ln 2

ln

= lnuln 2

ln= so series diverges.

(b) d

dn

lnn

n2=

1nn2 2n lnn

n4=1 2 lnn

n3< 0 so terms are decreasing. The integral is

limlnn

n2dn

2

= limlnn

n 2

+ lim1

n2dn

2

= limlnn

n 2

lim1

n 2

=ln2

2+1

2

and so series converges.

(c) d

dn

lnn

n=

n

nlnn

1

2 nn

=2 lnn

2n3/2< 0 for n > e2 . The integral is lim

lnn

ndn

1

.

Using integration by parts,


32

limlnn

ndn

1

= 2 lim n lnn1

2 limn

ndn

1

= 2 lim n lnn1

4 lim n1

= lim 2 n( 2 + lnn)1=

and the series diverges.

3. (a) en

(en +1)21

(b) n1

e n (c) 1

2n1

Solution

(a) d

dn

en

(en +1)2=en (en +1)2 en2(en +1)en

(en +1)4=en (1 en )

(en +1)3< 0 so terms are decreasing.

limen

(en +1)21

dn = limen

(en +1)21

dn

= lim1

en +1 1

=1

e +1

so series converges.

(b) d

dnne n

=en nen

e2n=en (1 n)

e2n< 0 for n > 1 so terms are decreasing.

lim ne n

1

dn = lim ne n

1+ lim e n

1

dn = lim ne n

1lim e n

1

= 1+ e 1

so series converges. (c) The terms are obviously decreasing. The integral gives

lim 2 n

1

dn = lim2 n

ln21

=1

2 ln2 and so series converges.

4. (a) 1

2n +11

(b) 1

n + 51

(c) ne n 2

1

Solution (a) The terms are obviously decreasing. The integral gives

lim1

2n +11

dn =1

2lim ln(2n +1)

1= and so series diverges.

(b) The terms are obviously decreasing. The integral gives

lim1

n + 51

dn = lim ln(n + 5)1= and so series diverges.


33

(c) d

dnne n2

=en

2

nen2

(2n)

e2n2 =

1 2n2

en2 < 0 and so terms are decreasing.

lim ne n2

1

dn =1

2lim e n2

1=1

2 so series converges.

5. (a) lnn

n2

(b) n

lnn2

(c) e n

e n+11

Solution

(a) d

dn

lnn

n=1 lnn

n2< 0 for n > 2 so terms are decreasing. Making the substitution

u = lnn du =dn

n the integral is lim

lnn

n2

dn = lim uln 2

ln

du = lim(lnn)2

2)2

= and

the series diverges.

(b) d

dn

n

lnn=lnn 1

(lnn)2> 0 for n > 2 so test cannot be applied.

(c) The general term can be re-written as en

ene n

e n+1

=1

1+ en and is obviously

decreasing. The integral is

lime n

e n+11

dn = lim ln(e n+1)

1= lim ln(e +1)+ ln(e 1

+1) = ln(e 1+1) and so

series converges.

6. (a) 1

en +11

(b) 1

(n +1)(n + 2)1

(c) 1

n2 +11

Solution (a) Terms are clearly decreasing.

lim1

en +11

dn = lime n

e n+11

dn = lim ln(e n+1)

1= lim ln(e +1) + ln(e 1

+1)

= ln(e 1+1)

and so series converges. (b) Terms are clearly decreasing.

lim1

(n +1)(n + 2)1

dn = lim1

n2 + 2n + 21

dn = lim1

(n +1)2 +11

dn

= lim arctan(n +1)1=2

arctan2

and so series converges. (c) Terms are clearly decreasing.

lim1

n2 +11

dn = lim arctan(n)1=2

arctan1 =4



34

7. (a) 1

n2 + 4n + 51

(b) sin1

n

2 (c)

n2

n4 +11

Solution (a) Terms are clearly decreasing.

lim1

n2 + 4n + 51

dn = lim1

(n + 2)2 +11

dn = lim arctan(n + 2)1=2

arctan 3

and so series converges. (b) Test cannot be applied. The terms are not positive and are not decreasing.

(c) d

dn

n2

n4 +1=2n(n4 +1) n2 4n3

(n4 +1)2=2n(1 n4 )

(n4 +1)2< 0 so terms are decreasing. The

integral is limn2

n4 +11

dn < limn2

n41

dn = lim1

n21

dn = lim1

n 1

= 1 and so the series

converges.

8. (a) 1

1+ n1

(b) n

2n1

(c) 1

n3 +11

Solution

(a) Terms are clearly decreasing. The integral is lim1

1+ n1

dn . Make the substitution

u = 1+ n du =1

2 ndn . Then

lim1

1+ n1

dn = 2 limu 1

u2

du = 2 lim ( 2) ln2

= , and so series diverges.

(b)

lim n2 n

1

dn = limn2 n

ln21

+1

ln2lim 2 n

1

dn = limn2 n

ln21

+1

ln2lim

2 n

ln21

=1

ln2lim

2

1

(ln2)2lim

2+1+ ln2

2(ln2)2

Both limits are clearly zero and so series converges.

(c) d

dn

1

n3 +1=

3n2

(n3 +1)2< 0 so the terms are decreasing. The integral is

lim1

n3 +11

dn < lim1

n31

dn =1

2lim

1

n2 1

=1

2 and so the series converges.

9. For what values of p does the following series converge: 1

n(lnn)p2

?

Solution


35

Terms are obviously decreasing. dn

n(lnn)p2

= limdn

n(lnn)p2

. Call u = lnn du =dn

n

and so the integral is dn

n(lnn)p2

= limdu

u pln 2

ln

= limu1 p

1 pln 2

ln

= lim(ln )1 p

1 p

ln21 p

1 p.

The limit is zero if p > 1 and infinite if p < 1 . If p = 1 the integral must be redone to give

dn

n(lnn)2

= limdu

uln 2

ln

= lim lnuln 2

ln= lim(ln ln ) ln ln2 .

Hence the series converges for p > 1 and diverges for p 1 . SET 3.4 Apply the ratio test to these series. If the ratio test is inconclusive apply a different test.

1. (a) 5

3n1

(b) 5n

3n1

(c) en

n1

Solution

(a) = limn

53n+1

53n

=1

3<1 so series converges.

(b) = limn

5n+1

3n+1

5n

3n

=5

3>1 so series diverges.

(c) = limn

en+1

n+1

en

n

=e<1 so series converges.

2. (a) n +1

n

1

n

(b) 7n

4n +11

(c) e nn!1

Solution

(a) = limn

n + 2n +1

n+1

n +1n

n = limn

1+1

n +1

n+1

1+1n

n =e

e= 1 so test is inconclusive. (Series

diverges by the divergence test.)


36

(b) = limn

7n+1

4n+1 +17n

4n +1

= limn

7n+1

7n4n +1

4n+1 +1= 7 lim

n

4n (1+14n)

4n+1(1+14n+1

)=7

4> 1 , series diverges.

(c) = limn

(n +1)!en+1

n!en

= limn

(n +1)

e= , series diverges.

3. (a) 2n

n!1

(b) 1+ n2

n21

(c) 2n

3n +11

Solution

(a) = limn

2n+1

(n +1)!2n

n!

= limn

2

n +1= 0 , series converges.

(b)

= limn

1+ (n +1)2

(n +1)2

1+ n2

n2

= limn

n

n +1

21+ (n +1)2

1+ n2= lim

n

n

n +1

2(n +1)2

n2

1+1

(n +1)2

1+1n2

= 1

so test is inconclusive. (Series diverges by the divergence test.)

(c) = limn

2n+1

3n+1 +12n

3n +1

= limn

2n+1

2n3n +1

3n+1 +1== 2 lim

n

3n (1+13n)

3n+1(1+13n+1

)=2

3< 1 , series converges.

4. (a) lnn

n2

(b) lnn

n2

(c) ne n

1

Solution

(a) = limn

ln(n +1)(n +1)lnnn

= limn

ln(n +1)

lnn

n

n +1= 1 by using L’ Hôpital’s rule on each

fraction. Hence test is inconclusive. (Series diverges by the integral test.)

(b) = limn

ln(n +1)

n +1lnn

n

= limn

ln(n +1)

lnn

n

n +1= 1 1 = 1 by using L’ Hôpital’s rule on

each fraction. Hence test is inconclusive. (Series diverges by the integral test.)

(c) = limn

n +1en+1

n

en

= limn

en

en+1n +1

n=1

e< 1 , series converges.


37

5. (a) 1+ n4

n!1

(b) (2n)!

3n1

(c) e n ln n2

Solution

(a) = limn

1+ (n +1)4

(n +1)!1+ n4

n!

= limn

1

n +1

1+ (n +1)4

1+ n4= 0 1= 0 <1 , series converges.

(b) = limn

(2n + 2)!3n+1

(2n)!3n

= limn

(2n + 2)(2n +1)

3, series diverges.

(c) = limn

ln(n +1)en+1

lnnen

= limn

1

e

ln(n +1)

lnn=1

e< 1 (used L’ Hôpital’s rule), so series

converges.

6. (a) 5n

3n1

(b) n

n2 +11

(c) 2n + 1

5n 3

1

n

Solution

(a) = limn

5(n +1)3n+1

5n3n

= limn

3n

3n+1n +1

n=1


(b) = limn

n +1(n +1)2 +1

n

n2 +1

= limn

n2 +1

(n +1)2 +1

n +1

n= lim

n

n2 +1

(n +1)2 +1limn

n +1

n= 1 by using L’

Hôpital’s rule, so test is inconclusive. (Series diverges by the integral test.) (c)

= limn

2n + 35n + 2

n+1

2n +15n 3

n = limn

2n + 35n + 2

n+1

2n +15n 3

n+1

2n +1

5n 3

= limn

(2n)n+1(1+32n)n+1

(5n)n+1(1+25n)n+1

(5n)n+1(135n)n+1

(2n)n+1(1+12n)n+1

2n +1

5n 3

=e3/2

e2 /5e 3/5

e1/22

5

=2

5< 1

hence series converges.

7. (a) 2n+1n3

3n1

(b) 5

9n +1

1

n

(c) 5n +3

2n 1

1

n


38

Solution

(a) = limn

2n+2 (n +1)3

3n+1

2n+1n3

3n

=2

3limn

n +1

n

3

=2

3limn

n +1

n

3

=2

3< 1 so series converges.

(b)

= limn

59n +10

n+1

59n +1

n = 5 limn

9n +1

9n +10

n1

9n +10= 5 lim

n

(1+19n)n

(1+109n)n

1

9n +10

= 5e1/9

e10 /9limn

1

9n +10= 0 < 1


(c) Working as in 6 (c) the limit is =5

2> 1 so series diverges.

8. (a) 2n +1

3n1

(b) 3n

2n +11

(c) n3

n!1

Solution

(a) = limn

2n+1 +13n+1

2n +13n

=1

3limn

2n+1 +1

2n +1=1

3limn

2n (1+12n)

2n+1(1+12n+1

)=2


(b) = limn

3n+1

2n+1 +13n

2n +1

= 3limn

2n +1

2n+1 +1= 3lim

n

2n (1+12n)

2n+1(1+12n+1

)=3

2> 1, series diverges.

(c) = limn

(n +1)3

(n +1)!n3

n!

= limn

1

n +1

n +1

n

3

= 0 1 = 0 < 1 , series converges.

9. (a) nn

n!1

(b) n!

(2n)n1

(c) nn

n!( )21

Solution

(a) = limn

(n +1)n+1

(n +1)!nn

n!

= limn

n!

(n +1)!

n +1

n

n+1

n = limn

n

n +11+

1

n

n+1

= 1 e = e > 1 ,

series diverges.


39

(b) = lim

n

(n +1)!(2n + 2)n+1

n!(2n)n

= limn

(n +1)!

n!

2n

2n + 2

n+11

2n= lim

n

n +1

2n1

1

n +1

n+1

=1

2e 1

< 1

and the series converges. (c)

= limn

(n +1)n+1

((n +1)!)2

nn

(n!)2

= limn

n!

(n +1)!

2n +1

n

n+1

n = limn

n

(n +1)21+

1

n

n+1

= 0 e = 0 < 1 ,

series converges.

10. (a) 5n +1

3n1

(b) 3n +1

n5n1

(c) n

3n1

(a) = limn

5n+1 +13n+1

5n +13n

=1

3limn

5n+1 +1

5n +1=1

3limn

5n+1(1+15n+1

)

5n (1+15n)

=1

35 =

5

3> 1

and so the series diverges. (b)

= limn

3n+1 +1(n +1)5n+1

3n +1n5n

=1

5limn

n

n +1

3n+1 +1

3n +1=1

5limn

n

n +1

3n+1(1+13n+1

)

3n (1+13n)

=1

51 3 =

3

5< 1 ,

series converges.

(c) = limn

n +13n+1

n

3n

=1

3limn

n +1

n=1

31 =

1

3< 1 and the series converges.

SET 3.5 Determine the convergence or divergence of the series by using one of the comparison tests.

1. (a) 1

n(n +1)1

(b) 10

3n n!1

(c) 2

lnn2

Solution

(a) Series behaves like the divergent 1

n1

. Using the limit comparison test we have that


40

limn

1

n(n +1)1n

= limn

n

n(n +1)= 1 and so the series diverges.

(b) Since 10

3n n!<10

3n and the series

10

3n1

converges being geometric, the other series

converges as well.

(c) Since 2

lnn>1

lnn>1

n and the series

1

n2

diverges, the other series diverges as well.

2. (a) 1

3n +11

(b) n

n2 +11

(c) lnn

n

2

2

Solution

(a) 3n +1 > 3n1

3n +1<1

3n. Hence series converges because

1

3n1

does.

(b) n

n2 +1<

n

n2=1

n3/2. Hence series converges because

1

n3/21

does.

(c) For large enough n, lnn < n1/8 (lnn)2 < n1/4lnn

n

2

<n1/4

n2=1

n3/4. Hence series

converges.

3. (a) 5

n4 12

(b) 2n 1

3n 11

(c) 2

3n +11

Solution

(a) Use the limit comparison test with the convergent series 5

n41

.

limn

5n4 15n4

= limn

n4

n4 1= 1 so the series converges.

(b) Use the direct comparison test with the convergent series 2n

2.5n1

.

2n 1

3n 1<

2n

3n 1<2n

2.5n (because 3n 1 > 2.5n for n > 1) and so the series converges.

(c) 2

3n +1>

2

3n + n=1

2n and so series diverges because

1

2n1

does.

4. (a) lnn

n3/22

(b) 7

n2 + n +11

(c) 2n + 3

n3n1

Solution

(a) lnn

n3/2<n1/4

n3/2=1

n5 /4 and so series converges since

1

n5 /42

does.

(b) 7

n2 + n +1<7

n2 and so series converges since

7

n21

does.


41

(c) Compare with the convergent 1

3n1

using the limit comparison test:

limn

2n + 3n3n

13n

= limn

2n + 2

n= 2 and so series converges.

5. (a) lnn

n22

(b) sin21

n1

(c) n + 5

2n2 12

Solution

(a) lnn

n2<n1/4

n2=1

n7 /4 and so the series converges because

1

n7 /41

does.

(b) Compare with the convergent series 1

n21

:

limn

sin21n

1n2

= limn

sin1n1n

2

= 1

and so the series converges.

(c) The series behaves as the divergent 1

n2

and since limn

n + 52n2 11n

= limn

n2 + 5n

2n2 1=1

2

the series diverges.

6. (a) 1

2n 11

(b) 1

n(n + 1)(n + 2)1

(c) n

n3 12

Solution

(a) Use the limit comparison test with the convergent series 1

2n1

:

limn

12n 112n

= 1 and so the series converges.

(b) Use the limit comparison test with the convergent series 1

n3/21

:

limn

1

n(n +1)(n + 2)1n3/2

= limn

n3/2

n(n +1)(n + 2)= 1 and so the series converges.

(c) Use the limit comparison test with the divergent series 1

n1/21

:


42

limn

n

n3 11n1/2

= limn

n3/2

n3 1= 1 and so the series diverges.

7. (a) n +1

n2 +11

(b) lnn

n2 +12

(c) 1

ln lnn2

Solution

(a) Use the limit comparison test with the divergent series 1

n1/21

:

limn

n +1

n2 +11n1/2

= limn

n2 + n

n2 +1= 1 and so the series diverges.

(b) lnn

n2 +1<n1/4

n2=1

n7 /4 and so series converges since

1

n7 /41

does.

(c) ln(lnn) < (lnn)1/2 < n1/2( )1/2

= n1/4 . Hence 1

ln(lnn)>1

n1/4 and so the series diverges.

8. (a) sinn2

(b) 21/n

n1/21

(c) 1

n!1

Solution

(a) Compare with the divergent 1

n2

: limn

sinn1n

= and so the series diverges.

(b) Compare with the divergent series 1

n1/21

:

limn

21/n

n1/2

1n1/2

= limn

21/n = 1 and so the series diverges.

(c) 1

n!<1

n2 for n > 3 and so the series converges.

9. (a) 1+ cosn

4n + n41

(b)4n

n41

(c) n4

4n1

Solution

(a) We have that 1+ cosn

4n + n4<

2

4n + n4. Compare the series

1

4n + n41

with 1

4n1

:

limn

24n + n4

14n

= limn

2 4n

4n + n4= lim

n

2 4n

4n (1+n4

4n)= 2 . Hence

1+ cosn

4n + n41

converges.


43

(b) Because n4 < 2n for n > 16 (proof by induction), 4n

n4>4n

2n. Since

4n

2n1

diverges so

does 4n

n41

.

(c) n4

4n<2n

4n for n > 16 . Since

2n

4n1

converges so does n4

4n1

.

10. (a) 3n

n!1

(b) 1

2n n!1

(c) n!

5n1

Solution

(a) 3n <n

2! for n 44 . Hence

3n

n!<

n

2!

n!<1

n2 and so series converges.

(b) limn

12n n!1n!

= 0 . Since 1

n!1

converges so does 1

2n n!1

.

(c) 5n <n

2! for n 130 . Hence

n!

5n>

n!n

2!> n2 . Hence series diverges.

11. (a)1

nn1

(b) sin2 n

n!1

(c) 3

n + 31

Solution

(a) Since nn > n2 we have that 1

nn<1

n2. But

1

n21

converges and thus so does 1

nn1

.

(b) Since sin2 n

n!<1

n!, series converges.

(c) Using the limit comparison test with the diverging series 1

n1

: we have

limn

3n + 31n

= 3 and so series diverges.

SET 3.6 Apply any test to determine the convergence or divergence of the following infinite series.

1. (a) ne n2

1

(b) 3k

4 k +11

(c) k!

5k1

Solution

(a) d

dnne n2

=en

2

nen2

(2n)

e2n2 =

1 2n2

en2 < 0 and so terms are decreasing.


44

lim ne n2

1

dn =1

2lim e n2

1=1

2 so series converges.

(b) By the ratio test, = limk

3k+1

4 k+1 +13k

4 k +1

= limk

3k+1

3klimk

4 k +1

4 k+1 +1=3

4< 1 so series converges.

(c) By the ratio test, = limk

(k +1)!5k+1

k!5k

= limk

5k

5k+1limk(k +1) = so series diverges.

2. (a) 1

ek1

(b) k 3

k 4 + k +11

(c) 3k + 2k

5k1

Solution

(a) By the ratio test, = limk

1ek+1

1ek

=1

e< 1 so series converges.

(b) Series behaves like the divergent 1

k1

: limk

k 3

k 4 + k +11k

= limk

k 4

k 4 + k +1= 1 so series

diverges.

(c) By the ratio test, = limk

3k+1 + 2k+1

5k+1

3k + 2k

5k

= limk

5k

5k+1limk

3k+1 + 2k+1

3k + 2k=3

5< 1 so series

converges.

3. (a) arctan k

k1

(b) 1

2k k!1

(c) 3k

k 31

Solution

(a) Compare with 1

k1

: limk

arctan kk1k

= limk

arctan k =2

and so series diverges.

(b) By the ratio test = limk

12k+1(k +1)!

12k k!

=1

2limk

1

k +1= 0 < 1 and the series

converges.

(c) By the ratio test = limk

3k+1

(k +1)3

3k

k 3

= 3 limk

k

k +1

3

= 3 > 1 and the series diverges.


45

4. (a) 2k

3k k1

(b) 1+ cosk

k1

(c) k 4

4 k1

Solution

(a) By the ratio test = limk

2k+1

3k+1 k +12k

3k k

=2


(b) 1+ cosk

k>1

k and so the series diverges by the direct comparison test.

(c) By the ratio test = limk

(k +1)4

4 k+1

k 4

4 k

=1


5. (a) 1

k1+1

k1

(b) 1

2k +11

(c) k 2

k

k

1

Solution

(a) Compare with 1

k1

: limk

1k1+1/k

1k

= limk

k

k1+1/k= lim

k

1

k1

k

= 1 and so series diverges by

the limit comparison test.

(b) 1

2k +1<1

2k and so series converges since

1

2k1

does.

(c) limk

k 2

k

k

= e 2 0 so series diverges by the divergence test.

6. (a) 1

k2 12

(b) 1

ln k2

(c) 1

ln(ln k)2

Solution

(a) Series behaves like 1

k2

: limk

1

k2 11k

= limk

k

k2 1= lim

k

k

k 11k2

= 1 and so

series diverges.

(b) ln k < k1/21

ln k>1

k1/2 and so series diverges.

(c) ln(ln k) < ln k < k1

ln(ln k)>1

ln k>1

k and so series diverges since

1

k2

does.

7. (a) k

7k1

(b) k +1

k +11

(c) (2k)!

(k!)22

Solution


46

(a) = limk

k +17k+1

k

7k

= limk

7k

7k+1limk

k +1

k!=1

71 =

1


(b) Compare with 1

k1

: limk

k +1k +11

k

= limk

k + k

k +1= 1 and so series diverges since

1

k1

does.

(c) By the ratio test series diverges because:

= limk

(2k + 2)!((k +1)!)2

(2k)!(k!)2

= limk

(2k + 2)!

(2k)!limk

(k!)2

((k +1)!)2= lim

k

(2k + 2)(2k +1)

(k +1)2= 4 1 > 1

8. (a) k!

ek1

(b) k

(k +1)(k + 2)(k + 3)1

(c) 1

k + k1

Solution (a) By the ratio test the series diverges because:

= limk

(k +1)!ek+1

k!ek

= limk

ek

ek+1limk

(k +1)!

k!=1

e=

(b) k

(k +1)(k + 2)(k + 3)<k

k 3=1

k2 and so the series converges by the direct comparison

test.

(c) Compare with 1

k1

: limk

1

k + k1k

= limk

k

k + k= 1 and so series diverges since

1

k1

does.

9. (a) k100

k!1

(b) 1

ln k + k1

(c) 4 k

5k + 2k1

Solution (a) By the ratio test the series converges because:

= limk

(k +1)100

(k +1)!k100

k!

= limk

k +1

k

100

limk

k!

(k +1)!= 1 0 = 0


47

(b) Comparing with the series 1

k1

: = limk

1ln k + k1k

= limn

k

ln k + k= 1 and so the series

diverges. (c) By the ratio test the series converges because:

= limk

4 k+1

5k+1 + 2k+1

4 k

5k + 2k

= limk

4 k+1

4 klimk

5k + 2k

5k+1 + 2k+1= 4

1

5=4

5< 1

10. (a)

1

1+ 2 + + k1

(b)

1

12 + 22 + + k21

(c)

1

1+12+ +

1k

1

Solution

(a) Since

1+ 2 + + k =k(k +1)

2>k2

2, we have that

1

1+ 2 + + k<2

k2. But

2

k21

converges so by the direct comparison test so does

1

1+ 2 + + k1

.

(b) Since 12+ 22 + + k2 > k2 , we have that

1

12 + 22 + + k2<1

k2 and so series

converges since 1

k21

converges.

(c) From Chapter 2 we know that

1+1

2+ +

1

k<1+ ln k <1+ k . Hence

1

1+12+ +

1k

>1

k +1. But

1

1+ k1

diverges (using the integral test for example) and

thus so does our series.

11. (a) k1

2

k

1

(b) k!2k

3k1

(c) (k + 2)!

k!2k1

Solution (a) Using the ratio test series converges because

= limn

(k +1)12

k+1

k12

k =1

2

(b) = limn

(k +1)!2k+1

3k+1

k!2k

3k

=2

3limn

(k +1)!

k!=2

3limn(k +1) = and so series diverges.


48

(c) = limn

(k + 3)!2k+1(k +1)!(k + 2)!2k k!

=1

2limn

k!(k + 3)!

(k +1)!(k + 2)!=1

2limn

(k + 3)

(k +1)=1

2< 1 and so series

converges.

12. (a) k +1 k 1

k1

(b) k2 +1 k2 1

k1

(c) (k2 +1)4

(k 3 + 2)31

Solution

(a) Using the limit comparison test and comparing with the divergent series 1

k1

we

have that:

limk

k +1 k 1

k1k

= limk

k ( k +1 k 1)

= limk

k ( k +1 k 1)( k +1 + k 1)

k +1 + k 1

= limk

2 k

k +1 + k 1= lim

k

2 k

k ( 1+1 / k + 1 1 / k )

= 1

and so the series diverges.

(b) Compare with the convergent series 1

k21

.

limk

k2 +1 k2 1k1k2

= limk

k( k2 +1 k2 1)

= limk

k( k2 +1 k2 1)( k2 +1 + k2 1)

k2 +1 + k2 1

= limk

2k

k2 +1 + k2 1= lim

k

2k

k( 1+1 / k2 + 1 1 / k2 )

= 1


(c) Compare with the divergent series 1

k1

.


49

limk

(k2 +1)4

(k 3 + 2)3

1k

= limk

k(k2 +1)4

(k 3 + 2)3

= limk

k9 (1+1 / k2 )4

k9 (1+ 2 / k 3)3

= 1

and so the series diverges.

13. (a) (5k)!

10k (3k)!(2k)!1

(b) 9k (3k)!k!

(4k)!1

(c) 4 k (k!)4

(4k)!1

Solution (a) By the ratio test

= limk

(5k + 5)!10k+1(3k + 3)!(2k + 2)!

(5k)!10k (3k)!(2k)!

=1

10limk

(5k + 5)(5k + 4)(5k + 3)(5k + 2)(5k +1)

(3k + 3)(3k + 2)(3k +1)(2k + 2)(2k +1)

=1

10

55

3322> 1

and so series diverges. (b) By the ratio test,

= limk

9k+1(3k + 3)!(k +1)!(4k + 4)!9k (3k)!k!(4k)!

= 9 limk

(3k + 3)(3k + 2)(3k +1)(k +1)

(4k + 4)(4k + 3)(4k + 2)(4k +1)

= 933

44< 1

so series converges. (c) By the ratio test,


50

= limk

4 k+1((k +1)!)4

(4k + 4)!4 k (k!)4

(4k)!

= 4 limk

(k +1)4

(4k + 4)(4k + 3)(4k + 2)(4k +1)

=1

43< 1


14. (a) sin2n21

(b) 1 cosn1

(c) cos2(n 1)

n1

Solution

(a) For large n the general term behaves as 2n2

so we compare with the convergent

series 1

n21

. By the limit comparison test and using L’ Hôpital’s rule,

limn

sin2n2

1n2

= limn

22n3

cos2n2

2n3

=2

,

hence the series converges. (b) We will be able to see when we study MacLaurin series that the general term will

behave as 2n2

so we compare with the convergent series 1

n21

. By the limit

comparison test and using L’ Hôpital’s rule,

limn

1 cosn

1n2

= limn

n2sin

n2n3

=2limn

sinn1n

=2limn

n2cos

n1n2

=

2

2,

hence series converges.

(c) limn

cos2(n 1)

n= cos2 = 1 0 and so series diverges by the divergence test.

15. The series of positive terms un1

converges. Deduce that the series (a) un2

1

and

(b) un1

n1

2

where >1

2 also converge. (c) Hence or otherwise prove that

the series unn1

( >1

2) converges.

Solution


51

(a) Since un1

converges limn

un = 0 and so for sufficiently large n, un < 1 . Then

un2< un and so un

2

1

converges by the direct comparison test.

(b) Since (un1

n)2 = un

2+1

n22unn

< un2+1

n2 (since

2unn

> 0 ) and un2

1

converges

and 1

n21

also converges being a p-series with p = 2 > 1 we have that

un1

n1

2

converges as well by the comparison test.

(c) un1

n1

2

= (un2

1

+1

n22unn) . All the series converge and thus so must

unn1

provided >1

2.

16. For what value(s) of p does the series 1

np lnn2

converge?

Solution

For p > 1 , 1

np lnn<1

np and the series converges by comparison with

1

np2

. For p = 1

the series is 1

n lnn2

that diverges as shown earlier (Set 3.3 Ex. 2(a)) by the integral

test. For p < 1 , np< n and so

1

np lnn>

1

n lnn that diverges. Hence series converges for

p > 1 and diverges for p 1 .

17. Examine the convergence or divergence of the two series (a) 1

(lnn)p2

and (b)

1

(lnn)n2

.

Solution (a) The series clearly diverges for p 0 so we may assume that p is positive. In the text

we stated that for n sufficiently large, lnn < n . This can be proved by examining the

function f (x) = ln x x and seeing that it has a maximum at x =1

1/

. Then,

1

lnn>1

n and so

1

(lnn)p>1

n p . Choose such that p < 1. Then series diverges by

the direct comparison test since 1

n p2

diverges when p < 1.


52

(b) Since lnn > 2 for n > e2 7.4 , then 1

lnn<1

2 and so

1

(lnn)n<1

2n. Hence series

converges by the direct comparison test since 1

2n2

converges.

18. Using Euler’s result 1

n21

=

2

6 find the sum

1

(2n 1)21

.

Solution

1

n21

=

2

6=

1

n2n even

+1

n2n odd

=1

(2n)2+

1

1

(2n 1)21

=1

4

1

n2+

1

1

(2n 1)21

2

6=1

4

2

6+

1

(2n 1)21

1

(2n 1)21

=

2

6

1

4

2

6

=

2

8

19. Euler has shown that the series

1

2+1

3+1

5+1

7+1

11+1

13+ i.e. the sum of the

reciprocals of the prime numbers diverges. Use this fact to explain why the number of prime numbers is infinite.

Solution If the number of primes were finite the infinite series would converge. It diverges, however, and so the number of primes is infinite.

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EXERCISES FOR CHAPTER 3: Infinite Seriesdbhs.wvusd.k12.ca.us/.../2011/4/27/57166616/P3notes_Infinite_Serie… · 2011-04-27 · EXERCISES FOR CHAPTER 3: Infinite Series SET 3.1 1