Su, Yung – Homework 12 – Due: Nov 20 2007, 3:00 am – Inst: Shinko Harper
1
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The due time is Central
time.
001
(part 1 of 1) 10 points
Determine whether the series
7
10

8
11
+
9
12

10
13
+
11
14

. . .
is absolutely convergent, conditionally con
vergent or divergent.
1.
absolutely convergent
2.
conditionally convergent
3.
divergent
correct
Explanation:
In summation notation,
7
10

8
11
+
9
12

10
13
+
11
14

. . .
=
∞
X
n
= 7
a
n
,
with
a
n
given by
a
n
= (

1)
n
n
n
+ 3
.
However,
lim
n
→∞
n
n
+ 3
= 1
,
so that as
n
→ ∞
,
a
n
oscillates between val
ues approaching 1 and

1.
In particular,
therefore,
lim
n
→ ∞
a
n
6
= 0
.
Consequently, by the Divergence Test, the
series is divergent
.
keywords:
002
(part 1 of 1) 10 points
Which one of the following series is conver
gent?
1.
∞
X
n
= 1
(

1)
n

1
6 +
√
n
correct
2.
∞
X
n
= 1
(

1)
3
2
5 +
√
n
3.
∞
X
n
= 1
(

1)
2
n
2
5 +
√
n
4.
∞
X
n
= 1
(

1)
n

1
6 +
√
n
2 +
√
n
5.
∞
X
n
= 1
5
2 +
√
n
Explanation:
Since
∞
X
n
= 1
(

1)
3
2
5 +
√
n
=

∞
X
n
= 1
2
5 +
√
n
,
use of the Limit Comparison and
p
series
Tests with
p
=
1
2
shows that this series is
divergent. Similarly, since
∞
X
n
= 1
(

1)
2
n
2
5 +
√
n
=
∞
X
n
= 1
2
5 +
√
n
,
the same argument shows that this series as
well as
∞
X
n
= 1
5
2 +
√
n
is divergent.
On the other hand, by the Divergence Test,
the series
∞
X
n
= 1
(

1)
n

1
6 +
√
n
2 +
√
n
is divergent because
lim
n
→ ∞
(

1)
n

1
6 +
√
n
2 +
√
n
6
= 0
.
This leaves only the series
∞
X
n
= 1
(

1)
n

1
6 +
√
n
.
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Su, Yung – Homework 12 – Due: Nov 20 2007, 3:00 am – Inst: Shinko Harper
2
To see that this series is convergent, set
b
n
=
1
6 +
√
n
.
Then
(i)
b
n
+1
≤
b
n
,
(ii)
lim
n
→ ∞
b
n
= 0
.
Consequently, by the Alternating Series Test,
the series
∞
X
n
= 1
(

1)
n

1
6 +
√
n
is convergent.
keywords:
003
(part 1 of 1) 10 points
Determine whether the series
∞
X
n
= 1
(

1)
n

1
e
1
/n
6
n
is absolutely convergent, conditionally con
vergent or divergent.
1.
divergent
2.
absolutely convergent
3.
conditionally convergent
correct
Explanation:
Since
∞
X
n
= 1
(

1)
n

1
e
1
/n
6
n
=

1
6
∞
X
n
= 1
(

1)
n
e
1
/n
n
,
we have to decide if the series
∞
X
n
= 1
(

1)
n
e
1
/n
n
is absolutely convergent, conditionally con
vergent or divergent.
First we check for absolute convergence.
Now, since
e
1
/n
≥
1 for all
n
≥
1,
e
1
/n
6
n
≥
1
6
n
>
0
.
But by the
p
series test with
p
= 1, the series
∞
X
n
= 1
1
6
n
diverges, and so by the Comparison Test, the
series
∞
X
n
= 1
e
1
/n
6
n
too diverges; in other words, the given series
is not absolutely convergent.
To check for conditional convergence, con
sider the series
∞
X
n
= 1
(

1)
n
f
(
n
)
where
f
(
x
) =
e
1
/x
x
.
Then
f
(
x
)
>
0 on (0
,
∞
). On the other hand,
f
0
(
x
) =

1
x
2
e
1
/x
x

e
1
/x
x
2
=

e
1
/x
‡
1
x
+
x
·
x
2
=

e
1
/x
n
1 +
x
2
x
3
o
.
Thus
f
0
(
x
)
<
0 on (0
,
∞
), so
f
(
n
)
> f
(
n
+ 1)
for all
n
. Finally, since
lim
x
→ ∞
e
1
/x
= 1
,
we see that
f
(
n
)
→
0 as
n
→ ∞
.
By the
Alternating Series Test, therefore, the series
∞
X
n
= 1
(

1)
n
f
(
n
)
is convergent.
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 Spring '09
 Harper
 Calculus, Mathematical Series, lim

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