Answers to Selected Exercises in Chapter 7

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Section 7.2 Section 7.3 Section 7.4 Section 7.5 Section 7.6 Section 7.7 Section 7.8

Section 7.1


5.

n
n+1

7.

2n-1
2n

9.

( -1) n27
6n

11. 1

13. 1

15. 1/4

17. ¥

19. ¥

21. 0

23. 0

25. 0
 
 

Section 7.2

5. yn = 4n·3; y0 = 3, y1 = 4·3 = 12, y2 = 42·3 = 48, no limit

7. yn = 3/3n = 31-n; y0 = 3, y1 = 1, y2 = 1/3, y3 = 1/9, limit of 0

9. y0 = 3, y1 = -4. 4, y2 = -5.88,limit is -6.25 , general term is

yn = ( 0.2) n·3-5 æ
ç
è		 1-( 0.2) n
1-0.2		 ö
÷
ø

11. y0 = 3, y1 = 10.7, y2 = 9.93,limit is 10 , general term is

yn = ( -0.1) n·3+11 æ
ç
è		 1-( -0.1) n
1-0.1		 ö
÷
ø

13. y0 = 3, y1 = -7, y2 = 13, no limit , general term is

yn = ( -2) n·3-1 æ
ç
è		 1-( -2) n
1-2		 ö
÷
ø

15. y0 = 3, y1 = 3.1, y2 = 3.18, limit is 3.5 , general term is

yn =  æ
ç
è		 4
5		 ö
÷
ø		 n
 		 ·3+ 7
10		 æ
ç
è		 1-( 4/5) n
1-4/5		 ö
÷
ø

17. 2x2+15x-50 = 0, Solution is x = 2. 5

21. 11x+3 = 0, x = -3/11

23. x-e-x = 0, Solution is x = 0.56714329

25. ex-e-2x = 0, Solution is : x = 0
 


Section 7.3

5. if n is odd, then sn = 1: if n is even, then sn = 1-1/n. Converges to 1.

7. if n is odd, then sn = ( 2n+1) /2n; if n is even, then sn = ( 2n-1) /2n. Converges to 1

9. if n is odd, then sn = 4; If n is even, then sn = 0: Diverges

11. 4/9

13. 23/99

15. 5/111

17. a = 1/2, r = 1/3 < 1. convergent geometric series with sum

1/2
1-1/3		 =  3
4

19. a = 3/2, r = 3/2 > 1, divergent geometric series

21. a = p/e, r = p/e > 1, divergent geometric series

23. series is of the form

1+ 2
3		 + 3
5		 +¼+ n
2n-1		 +¼
individual coefficients have limit of 1 ¹ 0. Divergence test implies divergence.

25. series is of the form

1+0-1-0+1+0-1-0+¼
Coefficents do not have a limit of zero. Divergence test implies divergence.

27. converges when | cos( t) | £ 1, which is when t ¹ np for any integer n. Sum is

1
1-cos( t) 

29. converges for all t. sum is

1
1+tanh( t) 

 


Section 7.4

5. diverges since
ó
õ		 ¥

1

 dx
x+1		 = ln( x+1) |1¥   diverges

7. diverges

9. converges since (by partial fractions)

ó
õ		 ¥

2

 1
x2-1		 dx = - 1
2		 ln æ
ç
è		 x+1
x-1		 ö
÷
ø		 ê
ê
ê		 ¥

2

 = 0.5493061443

11. converges

13. converges

15. converges

17. converges
 


Section 7.5

5. a0 = p/2 and
an =  1
p		 ó
õ		 p

-p

 | x| cos( nx)dx =  2
p		 ó
õ		 p

0

 xcos( nx) dx =  2[( -1) n-1]
pn2

7. a0 = p and

an =  1
p		 ó
õ		 p/2

-p/2

 cos( nx) dx =  sin( np/2)
np/2

9. 0

11. a0 = 2p3/3-2p and

an =  2
p		 ó
õ		 p

0

 ( x2-1) cos(nx) dx =  4( -1) n
n2

13. a0 = 0 and

an =  2
p		 ó
õ		 p

0

 æ
ç
è		 p
2		 -x ö
÷
ø		 cos( nx) dx =  2( 1-( -1) n)
pn2

15. neither: use computer algebra system: a1 = 0,    b1 = 1/2

a0
= 
1
2p		 ó
õ		 p

0

 sin( x) dx =  1
p
an
= 
1
p		 ó
õ		 p

0

 sin( x) cos(nx) dx =  ( -1) n+1
p( n2-1) 		 ,    n ¹ 1 
bn
= 
1
p		 ó
õ		 p

0

 sin( x) sin(nx) dx = 0,    n ¹ 1

17. neither: use computer algebra system: a0 = ( e2p-e-2p) /( 4p)

an =  ( -1) n( 2e2p-2e-2p)
p( n2+4) 		 ,        bn =  ( -1) n(ne-2p-ne2p)
p( n2+4) 

19. a0 = 2/p, a1 = 0 and for n ¹ 1, we have

an =  2
p		 ó
õ		 p/2

0

 cos( x) cos(0x) dx- 2
p		 ó
õ		 p

p/2

 cos( x) cos( 0x) dx =  4cos( np/2)
p(n2-1) 

21.

bn =  2
p		 ó
õ		 p

0

 sinh( 2x) sin(nx) dx =  ( -1) n+1
p		 2nsinh2p
4+n2

23. bn = 2n( -1) nsin( p2) /( p2-n2)

25. function is neither: a0 = p/2, an = 0 and

bn =  1
p		 ó
õ		 0

-p

 ( x+p) sin(nx) dx+ 1
p		 ó
õ		 p

0

 xsin( nx) dx =  ( -1) n+1-1
n

 


Section 7.6


All these problems are graphical.  No answers will be given.
 


Section 7.7


5. converges

7. converges

9. diverges

11. converges

13. converges absolutely

15. does not converge absolutely

17. converges absolutely

19. converges absolutely

21. does not coverge absolutely
 



 

Section 7.8

5. converges uniformly

7. converge uniformly

9. converges uniformly

11. converges uniformly

13. derivative of the series does converge to the derivative of the function

15. derivative of the series does converge to the derivative of the function

17. derivative of the series does converge to the derivative of the function

19.

¥
å
n = 1		 æ
ç
è		 n2p2+6
n3		 ö
÷
ø		 2
 
 		 =  1
p		 ó
õ		 p

-p

 ê
ê
ê		 x3
2		 ê
ê
ê		 2
 
 		 dx =  2
p		 ó
õ		 p

0

 x6
4		 dx =  p6
14

21.

¥
å
n = 1		 n2
( 4n2-1) 2		 =  1
p		 ó
õ		 p

-p

 ê
ê
ê		 p
8		 sin æ
ç
è		 x
2		 ö
÷
ø		 ê
ê
ê		 2
 
 		 dx =  p2
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