HOMEWORK 6: SOLUTIONS/HINTS
14.2
(a) As
N
X
n
=1
n
±
1
n
2
=
N
X
n
=1
1
n
±
N
X
n
=1
1
n
2
;
we see that the series diverges.
(d) We have lim
±
±
±
a
n
+1
a
n
±
±
±
=
1
3
, hence it converges by the ratio test.
(e) As lim
±
±
±
a
n
+1
a
n
±
±
±
= 0 the series converges.
14.3 The series in (e) does not converge, as sin(
n±
9
) is not a zero sequence. All
the other series converge: (a) and (f) by the ratio test, (b)(d) by the
comparison test.
14.4 The series in (a) and (c) converge by the comparison test respectively the
ratio test. The series in (b) diverges by the comparison test.
14.6 As (
b
n
) is bounded, there is an
M >
0 such that
j
b
n
j
< M
for all
n:
Then
P
M
j
a
n
j
converges (by exc. 14.5 (b)) and
j
a
n
b
n
j
< M
j
a
n
j
, therefore
P
a
n
b
n
converges by the comparison test. For part (b), just take
b
n
= 1 in
(a) to get the statement of Corollary 14.7.
14.12 Since lim inf
j
a
n
j
= 0, there is a subsequence (
a
n
i
) of (
a
n
) such that lim
a
n
i
=
0. Set
b
i
:=
a
n
i
, then as lim
b
i
= 0 there exists an
i
1
2
N
such that
j
b
i
j
<
1
1
2
for all
i
²
i
1
. Now by induction we ±nd an increasing sequence of integers
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 Spring '11
 Peter

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