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Thomas Calculus 13th [Solutions]

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798 Chapter 10 Infinite Sequences and Series<br />

22. A polynomial has only a finite number of nonzero terms in its Taylor series, but the functions sin x, ln<br />

x<br />

e have infinitely many nonzero terms in their Taylor expansions.<br />

x and<br />

23.<br />

3 3 3<br />

ax a x x<br />

x<br />

x<br />

sin( ax) sin x x 3! 3!<br />

3 5<br />

a 2 a 1 a 1 2<br />

3 3 2<br />

lim lim lim<br />

x 0 x x 0 x x 0 x<br />

3<br />

if a 2 0 a 2; lim sin 2x sin x x 2 1 7<br />

3<br />

x 0 x<br />

3! 3! 6<br />

3! 3! 5! 5!<br />

x is finite<br />

24.<br />

a<br />

2<br />

x<br />

2<br />

a<br />

4<br />

x<br />

4<br />

b<br />

2 4!<br />

1<br />

cos ax b 2 2 2<br />

lim 1 lim 1 lim 1 b a a x 1 b 1 and a 2<br />

2 2 2<br />

x 0 2x x 0 2x x 0 2x<br />

4 48<br />

25. (a)<br />

u<br />

u<br />

n<br />

n 1<br />

2<br />

2 2<br />

( n 1)<br />

1 2 1 C 2 1 and 1<br />

n n<br />

2<br />

n<br />

n 1 n<br />

un<br />

(b) n 1 1 1 0 C 1 1 and 1<br />

u 2<br />

n 1 n n n<br />

n 1 n diverges<br />

converges<br />

26.<br />

5<br />

2<br />

6 3<br />

n<br />

2<br />

4 2 4n<br />

2<br />

4n<br />

1<br />

1 1<br />

2 2 2 2<br />

un<br />

2 n(2n 1) 4n 2n<br />

5<br />

un<br />

1 (2n 1) 4n 4n 1 n 4n 4n 1 n n<br />

after long division C 3 1 and<br />

2<br />

2<br />

f ( n) 5n<br />

5 5 u<br />

2<br />

n converges by Raabes Test<br />

4n<br />

4n<br />

1 4 n 1<br />

4 1<br />

n<br />

n<br />

2<br />

27. (a)<br />

2 2<br />

an L an an an anL a n converges by the Direct Comparison Test<br />

n 1 n 1 n 1<br />

(b) converges by the Limit Comparison Test:<br />

lim an<br />

0<br />

n<br />

a n<br />

1 a n<br />

1<br />

lim lim 1 since<br />

n<br />

an<br />

n<br />

1 an<br />

n<br />

a n converges and therefore<br />

1<br />

2 3<br />

an an 2 3 an<br />

28. If 0 a n 1 then ln 1 an ln 1 an an a ,<br />

2 3 n an a n a positive term of a<br />

1 an<br />

convergent series, by the Limit Comparison Test and Exercise 27b<br />

29.<br />

1<br />

n<br />

(1 x) 1 x where 1<br />

1 n 1<br />

x 1 d (1 x)<br />

nx and when x 1 we have<br />

2<br />

(1 x)<br />

dx<br />

2<br />

n 1<br />

n 1<br />

1 1<br />

2<br />

1<br />

3<br />

1<br />

n 1<br />

4 1 2 3 4 n<br />

2 2 2 2<br />

30. (a)<br />

2 2<br />

n 1 n 2 n 1 2<br />

n<br />

x x ( n 1) x x x n( n 1) x n( n 1) x 2x<br />

1 x (1 x) (1 x) (1 x)<br />

n 1 n 1 n 1 n 1<br />

2 3 3<br />

n<br />

1<br />

2<br />

x<br />

2<br />

x<br />

3<br />

1 ( x<br />

3<br />

1)<br />

x<br />

n( n 1) 2<br />

x<br />

n<br />

1<br />

, x 1<br />

Copyright<br />

2014 Pearson Education, Inc.

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