rabinovich (ahr454) – HW12 – um – (55915)
1
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001
10.0 points
Which one of the following series is conver
gent?
1.
∞
n
= 1
(

1)
n

1
1 +
√
n
correct
2.
∞
n
= 1
(

1)
2
n
5
4 +
√
n
3.
∞
n
= 1
4
5 +
√
n
4.
∞
n
= 1
(

1)
3
5
4 +
√
n
5.
∞
n
= 1
(

1)
n

1
1 +
√
n
5 +
√
n
Explanation:
Since
∞
n
= 1
(

1)
3
5
4 +
√
n
=

∞
n
= 1
5
4 +
√
n
,
use of the Limit Comparison and
p
series
Tests with
p
=
1
2
shows that this series is
divergent. Similarly, since
∞
n
= 1
(

1)
2
n
5
4 +
√
n
=
∞
n
= 1
5
4 +
√
n
,
the same argument shows that this series as
well as
∞
n
= 1
4
5 +
√
n
is divergent.
On the other hand, by the Divergence Test,
the series
∞
n
= 1
(

1)
n

1
1 +
√
n
5 +
√
n
is divergent because
lim
n
→ ∞
(

1)
n

1
1 +
√
n
5 +
√
n
= 0
.
This leaves only the series
∞
n
= 1
(

1)
n

1
1 +
√
n
.
To see that this series is convergent, set
b
n
=
1
1 +
√
n
.
Then
(i)
b
n
+1
≤
b
n
,
(ii)
lim
n
→ ∞
b
n
= 0
.
Consequently, by the Alternating Series Test,
the series
∞
n
= 1
(

1)
n

1
1 +
√
n
is convergent.
002 (part 1 of 3) 10.0 points
Decide whether the series
∞
n
= 1
(

1)
n

1
3
n

1
converges or diverges.
1.
diverges
2.
converges
correct
Explanation:
The given series has the form
∞
n
= 1
(

1)
n

1
f
(
n
)
with
f
defined by
f
(
x
) =
1
3
x

1
.
rabinovich (ahr454) – HW12 – um – (55915)
2
Now
f
(
n
) =
1
3
n

1
>
1
3
n
+ 2
=
f
(
n
+ 1) ;
on the other hand,
lim
n
→ ∞
1
3
n

1
= 0
.
Consequently, by the Alternating Series Test,
the given series
converges
.
003 (part 2 of 3) 10.0 points
Decide whether the series
∞
n
= 1
(

1)
n

1
√
3
n
2 +
√
n
converges or diverges.
1.
converges
2.
diverges
correct
Explanation:
The given series has the form
∞
n
= 1
(

1)
n

1
f
(
n
)
with
f
defined by
f
(
x
) =
√
3
x
2 +
√
x
.
Now
lim
x
→ ∞
√
3
x
2 +
√
x
=
√
3
.
Consequently, by the Divergence Test, the
given series
diverges
.
004 (part 3 of 3) 10.0 points
Decide whether the series
∞
n
= 1
(

1)
n

1
sin
1
n
converges or diverges.
1.
converges
correct
2.
diverges
Explanation:
The given series has the form
∞
n
= 1
(

1)
n

1
f
(
n
)
with
f
defined by
f
(
x
) = sin
1
x
.
Now by the Chain Rule
f
(
x
) =

1
x
2
cos
1
x
,
so
f
(
x
)
<
0 for large, positive
x
. Thus
f
(
n
) = sin
1
n
>
sin
1
n
+ 1
=
f
(
n
+ 1)
,
while
lim
x
→ ∞
f
(
x
) = sin(0) = 0
.
Consequently, by the Alternating Series Test,
the given series
converges
.
005
10.0 points
Determine whether the series
∞
n
= 0
n
n
2
+ 5
cos(2
n
π
)
converges or diverges.
rabinovich (ahr454) – HW12 – um – (55915)
3
1.
series converges
2.
series diverges
correct
Explanation:
Since cos(2
n
π
) = 1, the given series can be
rewritten as the series
∞
n
= 0
n
n
2
+ 5
=
∞
n
= 0
f
(
n
)
of positive, nonalternating terms where
f
(
x
) =
x
x
2
+ 5
.
Now
lim
x
→ ∞
xf
(
x
) =
lim
x
→ ∞
x
2
x
2
+ 5
= 1
.
Thus by the Limit Comparison Test, the given
series converges if and only if the series
∞
n
= 1
1
n
converges. But, by applying the
p
series Test
with
p
= 1, we see that this last series di
verges. Consequently, the given series
diverges
.
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 Gramlich
 Accounting, Mathematical Series, lim