Thomas Calculus 13th [Solutions]

744 Chapter 10 Infinite Sequences and Series
Root: lim n a 1 1
n lim n
; let
p
p
n
n (ln n)
1/ n
lim (ln n)
n
1
n ln n
1/ n
f ( n) (ln n ) , then ln f ( n)
ln(ln n)
n
ln(ln ) 1
1/ ln ( ) 0
lim ln f ( n n n f n
) lim lim lim 0 lim (ln ) lim 1;
n n
n
n
1
n
nln
n
n e e
n n
therefore lim n a 1 1
n
1 no conclusion
p p
n
1/ n (1)
lim (ln n)
n
65.
a n
n for every n and the series
2 n
n
n
2
n 1
converges by the Ratio Test since lim ( n 1) n
2 1
1
2 2
1
n
n
n
n
a n converges by the Direct Comparison Test
1
66.
n
2
2
n!
0
for all n 1;
( 1)
2
2
n
2
( n 1)!
n 2 n 1
n!
2 n 1
n n
2
2
n
( n 1) n! n
2
2
n 1 n 1 1
n!
lim lim 2 lim 2 lim 2 4 lim 2 4 ln 4
n n n n n
1
n
n
2
2
n!
1
diverges
10.6 ALTERNATING SERIES AND CONDITIONAL CONVERGENCE
1. converges by the Alternating Convergence Test since: u 1
n 0 for all n 1;
n
n 1 n 1 n n 1 n 1 1 u 1
n 1 u n;
lim u lim 1 0.
n n
n
n n n
2. converges absolutely converges by the Alternating Convergence Test since | a 1
n|
which is a
3/2
n
n 1 n 1
convergent p-series.
3. converges converges by Alternating Series Test since: u 1
n 0 for all n 1;
n
n3
n 1 n n 1 n
n 1 n 1 n 3 3 ( n 1)3 n3 1 1 u
1
1 ;
n n n un
lim u 1
n lim 0.
n
( n 1)3 n3
n n n3
4. converges converges by Alternating Series Test since: u 4
n 0 for all n 2;
2
(ln n)
2 2
n 2 n 1 n ln( n 1) ln n ln( n 1) (ln n ) 1 1 4 4 u
2 2 2 2 n 1 un
;
ln( n 1) (ln n) ln( n 1) (ln n)
lim u
4
n lim 0.
2
n n (ln n)
5. converges converges by Alternating Series Test since: u n
n 0 for all n 1;
n
2 1
2 2 3 2 3 2 2 3 2
n 1 2n 2n n n 1 n 2n 2n n n n 1 n n 2n 2 n n n 1
2 2 n n 1
lim u lim n 0.
n n
n ( n 1) 1 n 1 ( n 1) u 2 2 n 1 u n;
n
n 1 ( n 1) 1
n
2 1
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