Thomas Calculus 13th [Solutions]

710 Chapter 10 Infinite Sequences and Series
123.
n 1 n
4 3
4
3
n 1 1
n
4
4 so
3
n
3
n
3
n
3
n
3
n n 1 4 4 1
4 4 4 4 4
a a and 4
3
4;
sequence is nonincreasing and bounded below by 4 it converges
4
n
thus the
124. a1 1, a2 2 3,
2 2
a 3 2(2 3) 3 2 2 1 3,
2 2 3 3
a 4 2 2 2 1 3 3 2 2 1 3,
3 3 4 4
a 5 2 2 2 1 3 3 2 2 1 3, ,
a n
n 1 n 1 n 1 n 1
2 2 1 3 2 3 2 3
n 1
n n n 1 n n 1
n n
2 (1 3) 3 2 3; a a 1 2 3 2 3 2 2 1 2 so the sequence is
nonincreasing but not bounded below and therefore diverges
125. For a given , choose N to be any integer greater than 1/ . Then for n N ,
sin n sin n 1 1
0 .
n n n N
126. For a given , choose N to be any integer greater than 1/ e . Then for n N ,
1 1 1
1 1 .
2 2 2
n n N
127. Let 0 M 1 and let N be an integer greater than 1
.
M Then n N n n nM M
1
n M nM n M ( n 1) n M .
n 1
M
128. Since M 1 is a least upper bound and M 2 is an upper bound, M 1 M 2 . Since M 2 is a least upper bound and
M 1 is an upper bound, M 2 M 1 . We conclude that M1 M 2 so the least upper bound is unique.
( 1)
n
129. The sequence a n 1 is the sequence
1
,
3
,
1
,
3
, . This sequence is bounded above by 3 , but it
2
2 2 2 2
2
clearly does not converge, by definition of convergence.
M
M
130. Let L be the limit of the convergent sequence { a n}.
Then by definition of convergence, for 2
there
corresponds an N such that for all m and n, m N a L and n N a L Now
a a a L L a a L L a whenever m N and n N.
m n m n m n
2 2
131. Given an 0, by definition of convergence there corresponds an N such that for all n N,
L1 a n and
L2 a n . Now L2 L1 L2 an an L1 L2 an an
L1 2 . L2 L 1 2 says that the
difference between two fixed values is smaller than any positive number 2 . The only nonnegative number
smaller than every positive number is 0, so L1 L 2 0 or L 1 L 2 .
m
2
n
2 .
132. Let k( n ) and i( n ) be two order-preserving functions whose domains are the set of positive integers and whose
ranges are a subset of the positive integers. Consider the two subsequences a k( n)
and a i( n),
where
ak( n) L 1, i( n) 2
a L and L 1 L 2 . Thus a ( ) a ( ) L1 L 2 0. So there does not exist N such that
k n
i n
for all m, n N a a . So by Exercise 128, the sequence { a } is not convergent and hence diverges.
m
n
n
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