Thomas Calculus 13th [Solutions]

736 Chapter 10 Infinite Sequences and Series
q 2
q( q 1)(ln n) q( q 1)
lim lim 0; otherwise, we apply L'Hopital's Rule again. Since q is finite,
2 p r 2 p r 2 q
n ( p r) n n ( p r) n (ln n)
there is a positive integer k such that q k 0 k q 0. Thus, after k applications of L'Hopital's Rule we
q k
q( q 1) ( q k 1)(ln n) q( q 1) ( q k 1)
obtain lim lim 0. Since the limit is 0 in every case, by Limit
k p r k p r k q
n ( p r) n n ( p r) n (ln n)
(ln n)
Comparison Test, the series
p
n
n 1
q
converges.
62. Let q and p 1. If q 0, then
with 1 , which is a divergent
p
n
n 2
p -series. Then
q
(ln n) 1 ,
p
p
n n
n 2 n 2
(ln n)
q
n
p
p
with 1
n
p (ln n)
, where 0 p r 1. lim lim lim n
r
n 2 n
n 1/ n n n n (ln n)
1
( r p) n ( )
obtain lim r p r p n
lim r p
.
q 1 1
q 1
n ( q)(ln n)
n ( q)(ln n)
n
which is a divergent p-series. If q 0, compare
q
lim lim (ln n ) . If q 0 q 0, compare
n 1/ n n
(ln n)
q
q r p
r p r q
If q 1 0 q 1 0 and
since r p 0. Apply L'Hopital's to
r p q 1
( r p) n (ln n)
lim ,
n
( q)
2 r p 1 2 r p
( r p) n ( r p)
n
otherwise, we apply L'Hopital's Rule again to obtain lim lim . If
q 2 1
q 2
n ( q)( q 1)(ln n)
n ( q)( q 1)(ln n)
2 r p 2 r p q 2
q 2
( ) ( ) (ln )
q 2 0 q 2 0 and lim r p n lim r p n n
( )( 1)(ln )
( q)( q 1)
, otherwise, we apply
n q q n n
L'Hopital's Rule again. Since q is finite, there is a positive integer k such that q k 0 q k 0. Thus,
k r p
( r p)
n
after k applications of L'Hopital's Rule we obtain lim
q k
n ( q)( q 1) ( q k 1)(ln n)
( r p) (ln )
lim k n r p n
q k
. Since the limit is if q 0 or if q 0 and p 1, by Limit comparison test,
n
( q)( q 1) ( q k 1)
the series
q
(ln n)
n 1
p r
n
which is a divergent
diverges. Finally if q 0 and p 1 then
p -series. For n 3,
Comparison Test. Thus, if q and p 1, the series
q
(ln n) (ln n)
. Compare with 1,
p
n
n
n
n 2 n 2
n 2
(ln )
ln n 1 (ln n q n
) 1 q
1 .
n n
Thus
q
(ln n)
diverges by
n
n 2
q
(ln n)
n 1
p r
n
n
diverges.
q
63. Since 0 dn
9 for all n and the geometric series
n
9
110 n
converges to 1,
n
dn
1 10 n
converges.
64. Since
n
a n converges, a n 0 as n . Thus for all n greater than some N we have 0
1
a n and
2
thus 0 sin an
a n.
Thus
n
sin a n converges by Theorem 10.
1
65. Converges by Exercise 61 with q 3 and p 4.
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