Thomas Calculus 13th [Solutions]

Chapter 10 Practice Exercises 787
15. converges to ln 2, since
1/ n
1
n
2
1/ n
ln 2
1/ n
2 1
1/ n 0
lim an
lim n 2 1 lim lim lim 2 ln 2 2 ln 2 ln 2
n n n n n
n
2
1
n
2
16. converges to 1, since
2
2n
1
ln(2 1) 0
lim a n
n
n lim 2 n 1 lim exp lim exp 1
n n n
n
n
1
e
17. diverges, since
( n 1)!
lim an
lim lim n 1
n n
n!
n
18. converges to 0, since
n
( 4)
lim a n lim 0
n n
n!
by Theorem 5
19.
1 1 1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2 2 2
1
s
(2n 3)(2n 1) 2n 3 2n 1 n 3 5 5 7 2n 3 2n 1 3 2n
1
lim
n
sn
lim
n
1
2
1 1
6 2n
1 6
20. 2 2 2 s 2 2 2 2 2 2 2 2 2
( 1) 1 n
lim s lim 1 1
n n n n 2 3 3 4 n n 1 2 n 1 n
n n
n 1
21. 9 3 3 s 3 3 3 3 3 3 3 3 3 3
(3n 1)(3n 2) 3n 1 3n 2 n 2 5 5 8 8 11 3n 1 3n 2 2 3n
2
lim s 3 3 3
n lim
n n
2 3n
2 2
22. 8 2 2 s 2 2 2 2 2 2 2 2 2 2
(4n 3)(4n 1) 4n 3 4n 1 n 9 13 13 17 17 21 4n 3 4n 1 9 4n
1
lim s
2 2 2
n lim
n n
9 4n
1 9
n
23. e 1 , a convergent geometric series with r 1 and a 1 the sum is
n
e
e
n 0 n 0
1
1
1 e
e1
e
n
24. ( 1) 3 3 1
n
4
4 4
n 1 n 0
1
3
4
1
4
3
5
n
a convergent geometric series with r 1 and a 3 the sum is
4
4
25. diverges, a p-series with p 1
2
26. 5 5 1 , diverges since it is a nonzero multiple of the divergent harmonic series
n n
n 1 n 1
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