Thomas Calculus 13th [Solutions]

Section 10.7 Power Series 759
1 2
1 2
1
35. For the series x , recall 1 2
2 2 2
can rewrite the series as
n
(2n
1)
(2n
3)
n
n
n
n( n 1)
2
n( n 1)(2n
1)
6
n 3
2n
1
n 1 n 1
n( n 1)
2
n and 1 2
x x ; then
n
n
2 2 2
x lim 1 x 1 1 x 1; when x 1 we have
3
n
convergent series; when x 1 we have
3
2n
1
n 1
,
a divergent series.
(a) the radius is 1; the interval of convergence is 1 x 1
(b) the interval of absolute convergence is 1 x 1
(c) the series converges conditionally at x 1
n( n 1)(2n
1)
6
n so that we
u
u
n 2( n 1) 1 3x
n 1
n 1
3x
(2n
1)
n
n
lim 1 lim 1
2n
1
n 1
n
( 1) , a
conditionally
36. For the series
n 1
rewrite the series as
n 1 n ( x 3) , note that
( x 3)
n 1
n 1
n 1 n
n 2 n 1
n
n
n
; then
n
n 1 n n 1 n 1
1 n 1 n n 1 n
n 1 n so that we can
n 1
un
1 ( x 3) n 1 n
u n
n n n 2 n 1 ( x 3)
lim 1 lim 1
x 3 lim 1 x 3 1 2 x 4; when x 2 we have
n
convergent series; when x 4 we have
1
n 1
n 1
(a) the radius is 1; the interval of convergence is 2 x 4
(b) the interval of absolute convergence is 2 x 4
(c) the series converges conditionally at x 2
n
, a
divergent series;
( 1)
n 1
n 1
n
n
, a
conditionally
37.
n 1
un
1 ( n 1)! x 3 6 9 (3 n) ( n 1) x
x x R
3 6 9 (3 ) 3( 1) n
n
un
n n n n!
x
n
3( n 1) 3
lim 1 lim 1 lim 1 1 3 3
38.
2 1 2
2
u
2 4 6 (2 n) 2( n 1) x n
n 1
2 5 8 (3n
1) (2n
2) 4 x
lim 1 lim 1 x lim 1 1
u
2 2 n
2
n
9
n n 2 5 8 (3n
1) 3( n 1) 1 2 4 6 (2 n)
x
n
| x |
9 R 9
4 4
(3n
2)
39.
2 lim 1 2
1
1 lim ( 1)! n n
un
n x 2 (2 n)! 1 lim ( n 1)
1 2
2(2 2)(2 1) 1 x
x
8
1 x 8 R
u 8
n
n
n
n n
n n 2 2( n 1) ! n!
x
n
40.
2
n
n
1
lim n n n
u 1 lim
n
1 lim
n
n x x 1 x e 1 x e R e
n n
n 1
n
n 1
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