Thomas Calculus 13th [Solutions]

760 Chapter 10 Infinite Sequences and Series
41.
un
1
1 1
3 n
x
1 1 1
n
un
n
n n
3 x
n
3 3 3
3 n
n
1
3
n
, which diverges; at
1
n
n
x we have
1
3
3
n 0 n 0
n 0 n 0
lim 1 lim 1 x lim 3 1 x x ; at x 1
we have
3
3 1, which diverges. The series
n
n n n
3 x (3 x ) , is a convergent geometric series when
1 x 1
and the sum is
1
0 n 0
3 3
1 3 x
.
42.
x
n 1
u e 4
n 1
x x x
lim 1 lim 1 e 4 lim 1 1 e 4 1 3 e 5 ln 3 x ln 5;
u
x
n
n
n n e 4
n
at x ln 3 we have
which diverges. The series
1 1
1 e 4 5
x
e x
.
ln 3
n
n
e 4 ( 1) , which diverges; at x ln 5 we have
n 0 n 0
x
4 n
n 0
n
4 1,
ln 5
e
n 0 n 0
e is a convergent geometric series when ln 3 x ln 5 and the sum is
43.
lim 2 2 2
1
1 lim n
u ( 1) n
n
x 4
1 ( x 1) 2
1 2
4
lim |1| 1 ( x 1) 4 x 1 2 2 x
u
1 2
n
n
n
n n 4 ( x 1)
n
n
n
1 x 3; at x 1 we have
n
4
1,
n
4
0 n 0
n
x 1
2
2
0
2n
n
( 2) n
4
n
4 4
n 0 n 0 n 0
1,
which diverges; at x 3 we have
a divergent series; the interval of convergence is 1 x 3; the series
2n
n
2
4
n 0
2n
( x 1)
n
4
n 0
1 1 4
1
x x 4 x 2x
1
is a convergent geometric series when 1 x 3 and sum is
2 2
1 4 ( 1) 2
2 4
4
3 2x x
2
44.
lim 2 2 2
1
1 lim n
u ( 1) n
n
x 9
1 ( x 1) 2
1 2
9
lim |1| 1 ( x 1) 9 x 1 3 3 x
u
1 3
n
n
n
n n 9 ( x 1)
n
4 x 2; when x 4 we have
2n
n
( 3)
n 0
9
n 0
1
which diverges; at x 2 we have
which also diverges; the interval of convergence is 4 x 2; the series
2n
n
n
3
9
0 n 0
( x
2n
1)
n
x 1
2
n 0
n
9
n 0
3
1 1
9 9
1
x 1 9 ( x 1) 2 9 x 2x 1 8 2x x
convergent geometric series when 4 x 2 and the sum is
2 2 2
3 9
1
is a
45.
n 1
u
x 2
n
n 1
lim 1 lim
2
1 x 2 2 2 x 2 2 0 x 4 0 x 16;
u
n 1
n
n
n n 2 x 2
when x 0 we have
n
n
( 1) ,
0
a divergent series; when x 16 we have
n
n
(1) ,
0
a divergent series; the
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