Thomas Calculus 13th [Solutions]

Section 10.6 Alternating Series and Conditional Convergence 745
6. diverges diverges by
exist
th
n Term Test for Divergence since:
2
1 n 2
2 2
lim n 5 1 lim ( 1) n 5
n n 4 n
n 4
does not
7. diverges diverges by
th
n Term Test for Divergence since: 2 n 1
lim lim ( 1) 2
2 2
n n n
n
n
n
does not exist
8. converges absolutely converges by the Absolute Convergence Test since | a 10
n | ( n 1)!
, which
n 1 n 1
an
1
converges by the Ratio Test, since lim lim 10
2
0 1
n
an
n
n
n
9. diverges by the nth-Term Test since for
n
n 1
n
n 10 n 1 lim n 0 ( 1) n diverges
10
n
10 10
n 1
10. converges by the Alternating Series Test because f ( x) ln x an increasing function of x 1 is decreasing
ln x
un
u n 1 for n 1; also u n 0 for n 1 and lim 1 0
n
ln n
11. converges by the Alternating Series Test since f ( x ) ln x ( ) 1 ln x 0
x
f x when x e f ( x ) is
2
x
decreasing un
u n 1 ; also u n 0 for n 1 and
1
n
lim u lim ln n
n
lim 0
n n
n
n
1
12. converges by the Alternating Series Test since
1
f ( x) ln 1 x f ( x ) 1 0 for x 0 f ( x ) is
x( x 1)
decreasing un
u n 1 ; also u n 0 for n 1 and lim u lim ln 1 1 ln lim 1 1
n ln1 0
n n
n
n
n
13. converges by the Alternating Series Test since f ( x) x 1 1 2
1 ( ) x x
x
f x 0 f ( x ) is decreasing
2
2 x( x 1)
un
u n 1 ; also u n 0 for n 1 and
n 1
lim un
lim 0
n n
n 1
14. diverges by the nth-Term Test since
1
n
3 n 1
3 1
lim lim 3 0
n n 1 n 1
1
n
n
15. converges absolutely since | a 1
n | a convergent geometric series
10
n 1 n 1
16. converges absolutely by the Direct Comparison Test since
of a convergent geometric series
n 1
n
( 1) (0.1) 1 1
n
n
(10) n 10
n
which is the nth term
17. converges conditionally since 1 1
n n 1
is a divergent p-series
0
and lim 1 0
n n
convergence; but | a 1
n|
1/2
n
n 1 n 1
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