Thomas Calculus 13th [Solutions]

762 Chapter 10 Infinite Sequences and Series
51.
3 3 1 1 n
g( x) x 5 ,
x 2 3 [ ( x 5)] x 5 3
1
n 0
3
x
3
5
1 or 2 x 8.
n
which converges for
52. (a) We can write the given series as
1 x
2 4
(b) The function represented by the series in (a) is
n
0
n
which shows that the interval of convergence is 4 x 4.
2
4 x
2
we can represent it by the geometric series
1 ( x 3)
x 3 1or 2 x 4.
for 4 x 4. If we rewrite this function as
n
0
2( x 3) n which will converge only for
53.
n 1
( x 3) n
2
n 1
2 ( x 3)
lim 1 x 3 2 1 x 5; when x 1 we have
n
n
n
(1) n
0
which diverges; when x 5
we have
n
( 1) n
1
geometric series is
x 3
which also diverges; the interval of convergence is 1 x 5; the sum of this convergent
1
1 2
x 1
2
. If
1 1 1
n
2 2 2
f x is
2
,
2
( x 1)
1 1 2 1
n n 2
2 4 2 x 1
f ( x) 1 ( x 3) ( x 3) ( x 3)
then
n 1
f ( x) ( x 3) n( x 3) is convergent when 1 x 5, and diverges when x 1 or
5. The sum for ( )
the derivative of
21 . x
54. If
1 1 2 1
n n 2
2 4 2 x 1
f ( x) 1 ( x 3) ( x 3) ( x 3)
then
( x 3.) ( x 3) 1
n ( x 3)
4 12 2 n 1
2 3 n 1
f ( x) dx x . At x 1 the series
2
series
n
( 1) 2
n 1
n 1
2 ln x 1 (3 ln 4), since
2
1
n 1 n
converges. Therefore the interval of convergence is 1 x 5 and the sum is
x 1
dx 2 ln x 1 C , where C 3 ln 4 when x 3.
2 4 6 8 10
3 5 7 9 11
3! 5! 7! 9! 11!
55. (a) Differentiate the series for sin x to get cos x 1
x x x x x
(b)
(c)
n
2 4 6 8 10
1
x x x x x
. The series converges for all values of x since
2! 4! 6! 8! 10!
2n
2
x (2 n)! 2
x
1
for all x.
(2n 2)! 2n
x n
(2n 1)(2n
2)
x x
2 x 2 x 2 x 2 x 2 x
x
8x 32x 128x 512x 2048x
3! 5! 7! 9! 11! 3! 5! 7! 9! 11!
1 2 1 1 3
x x x x x
2 2 3!
1 1 1 4 1 1 1 1 5
0 1 0 0 0 0 1 x 0 0 1 0 0 0 0 1 x
4! 2 3! 4! 2 3! 5!
lim lim 0 1
3 3 5 5 7 7 9 9 11 11 3 5 7 9 11
sin 2 2 2
2sin cos 2 (0 1) (0 0 1 1) 0 1 0 0 1 0 0 1 0 0 1
diverges; at x 5 the
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