Thomas Calculus 13th [Solutions]

706 Chapter 10 Infinite Sequences and Series
81.
lim tan
n
1
n converges 82.
2
1 1
lim tan n 0 0 converges
n n
2
83. lim n
1 1 1 1
3 lim n n
2
3 0
n
n
n
2
converges (Theorem 5, #4)
84.
2
n 2
ln
2 1 0
lim n n n n
lim exp lim exp n 1
n e converges
2
n n n n n
85.
200 199 198
(ln n) 200 (ln n) 200199 (ln n) lim lim lim lim 200! 0
n
n
n
n
n
n
n
n
converges
86.
5 ln n
4
5 n
4 3
ln n 10 ln n 80 ln n
lim lim lim lim lim 3840 0
n n n n n n n n n
1
2 n
converges
87.
2
2 2 n n n
lim n n n lim n n n lim n lim 1 1 converges
n n n n n n n n n n
2
2 2
1 1
1
n
88.
89.
1 1
2 2 2 2
1 1
2
1 1
n 1 n n n 1 n n
n
n
2 2 2 2 2 2
1 1
1 1 1
n
1
n
lim lim lim lim 2
n n n n n n n n n n n n n
converges
1 1 ln
lim n
lim lim 1
n 1 x n n
n
n
n
0 converges (Theorem 5, #1)
n
n
90. lim 1 dx lim 1 1 lim 1 1 1 1 if p 1 converges
1
p 1 p p 1 1 1 p p 1
n x n x n n p 1
91. Since a n converges lim a 72 72
n L lim an 1 lim L L(1 L) 72
n n n
1 an
1 L
2
L L 72 0 L 9 or L 8; since a n 0 for n 1 L 8
an
6
92. Since a n converges lim a lim 6
1 lim L
n L an L L( L 2) L 6
n n n
an
2 L 2
2
L L 6 0 L 3 or L 2; since a n 0 for n 2 L 2
93. Since a n converges
L 2 or 4;
94. Since a n converges
L 2 or 4;
2
lim an L lim an 1 lim 8 2an
L 8 2L L 2L
8 0
n n n
L since a n 0 for n 3 L 4
2
lim an L lim an 1 lim 8 2an
L 8 2L L 2L
8 0
n n n
L since a n 0 for n 2 L 4
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