Thomas Calculus 13th [Solutions]

Section 10.7 Power Series 753
6.
n 1
lim un
1 1 lim (2 x) 1 lim 2 x 1 2 x 1 1 x 1
u
(2 )
2 2
; when x 1 we have
n
n n n x
n
2
n
n
( 1) ,
1
a divergent series; when x 1 we have
2
1,
1
a divergent series
n
(a) the radius is 1 2 ; the interval of convergence is 1 x 1
2 2
(b) the interval of absolute convergence is 1 x 1
2 2
(c) there are no values for which the series converges conditionally
7.
n 1
un
1 ( n 1) x ( n 2) ( n 1)( n 2)
lim 1 lim 1 x lim 1 x 1 1 x 1; when x 1 we have
u ( 3) n
n n n
n nx
n
( n 3)( n)
( 1) n n
2
, a divergent series by the nth-term; Test; when x 1 we have
n
n 1
n
(a) the radius is 1; the interval of convergence is 1 x 1
(b) the interval of absolute convergence is 1 x 1
(c) there are no values for which the series converges conditionally
n ,
n 2
1
a divergent series
8.
n 1
lim un
1 1 lim ( x 2) n
1 1 x 2 lim n
( 2)
1
1 x 2 1 1 x
u 2 1 3 x 1;
n
n n n
n x
n
n
when x 3 we have 1,
a divergent series; when x 1 we have
n
n 1
n
(a) the radius is 1; the interval of convergence is 3 x 1
(b) the interval of absolute convergence is 3 x 1
(c) the series converges conditionally at x 1
1
n
( 1)
,
n
a convergent series
9.
lim u
1
1
1 lim n
n
x n n 3
n
1 x
1
3 lim n
1 lim n
1 1 x
3
(1)(1) 1 x
u 3
n n
n
n n
n n ( n 1) n 13 x
n n
( 1)
3 x 3; when x 3 we have
3/2
n
n 1
n
,
an absolutely convergent series; when x 3 we have
1 , a convergent p-series
3/2
n
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
10.
n 1
lim un
1 1 lim ( x 1) n 1 x 1 lim n
1 ( 1)
1
1 x 1 1 1 x
u
1 1 0 x 2; when
n
n n n n x
n
n
n
( 1)
x 0 we have , a conditionally convergent series; when x 2 we have 1 ,
1/2
1/2
n
n
n 1
n 1
(a) the radius is 1; the interval of convergence is 0 x 2
(b) the interval of absolute convergence is 0 x 2
(c) the series converges conditionally at x 0
a divergent series
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