Thomas Calculus 13th [Solutions]

Section 10.7 Power Series 755
17.
lim 1
u 1
1 lim ( 1)( 3) n n
n
n x 5 1 x 3 1
3
1
5 lim n
1 x
5
1 x 3 5 5 x
u
3 5
n
n
n
n
n n 5 n( x 3)
n
8 x 2; when x 8 we have
n
n
n( 5)
n
5
1 n 1
n
( 1) n , a divergent series; when x 2 we have
n
n5
n 1
n
5
n 1
n , a divergent series
(a) the radius is 5; the interval of convergence is 8 x 2
(b) the interval of absolute convergence is 8 x 2
(c) there are no values for which the series converges conditionally
18.
n 2 2
lim 1
1
1 lim n
u ( 1) 4 1 ( 1) 1
n
n x n 1 x
1 2 4
lim n n
u
1 4 4 4;
n
n
n
2
n n 4 n 2n 2 nx
n n n 2n
2
x x when
n( 1)
x 4 we have
2
n 1
n
n
1
,
a conditionally convergent series; when x 4 we have ,
n
2 1
series
(a) the radius is 4; the interval of convergence is 4 x 4
(b) the interval of absolute convergence is 4 x 4
(c) the series converges conditionally at x 4
n 1
n
a divergent
19.
u 1
1
1 n n
n
n x 3 x n 1 x
n 1
n
n
un
n 3 nx 3
n
n
3
lim 1 lim 1 lim 1 1 x 3 3 x 3; when x 3
we have
n 1
n
( 1) n , a divergent series; when x 3 we have
n 1
(a) the radius is 3; the interval of convergence is 3 x 3
(b) the interval of absolute convergence is 3 x 3
(c) there are no values for which the series converges conditionally
n , a divergent series
20.
lim t
1 1
1
lim
1
n
u
n
1(2 5) n
1
t
n
n x t
n
x x
u n
n n n
n
n(2x
5)
n lim n
1 2 x 5 1
n
n n n
1 2x 5 1 3 x 2; when x 3 we have
n
1
n
( 1) n , a divergent series since lim n 1;
n
n
when x 2 we have
n 1
n
n , a divergent series
(a) the radius is 1 ; the interval of convergence is 3 2
2 x
(b) the interval of absolute convergence is 3 x 2
(c) there are no values for which the series converges conditionally
21. First, rewrite the series as
For the series
For the series
n 1
n 1
n 1 1 1
x n x n n x
n
n 1 n 1 n 1
n 1
2( 1) :
n 1
x
n
( 1) ( x 1)
2 ( 1) ( 1) 2( 1) ( 1) ( 1) .
n
u
2( x 1)
x x x
u
n 1
n n n 2( x 1)
n
n 1
lim 1 lim 1 1 lim 1 1 1 2 0;
1
1
( 1) ( 1)
: lim 1 lim n n
un
x
1 x 1 lim 1 x 1 1
u
n n 1
n
n n ( 1) ( x 1)
n
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