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schuem – Exam3 Practice Problems – Schultz – (58385)
1
This printout should have 23 questions.
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beFore answering.
001
10.0 points
Determine whether the sequence
{
a
n
}
con
verges or diverges when
a
n
= (
−
1)
n
p
2
n
+ 1
5
n
+ 1
P
,
and iF it does, fnd its limit.
1.
limit =
2
5
2.
limit =
±
2
5
3.
limit = 1
4.
sequence diverges
correct
5.
limit = 0
Explanation:
AFter division,
2
n
+ 1
5
n
+ 1
=
2 +
1
n
5 +
1
n
.
Now
1
n
,
1
n
→
0 as
n
→ ∞
,
so
lim
n
→∞
2
n
+ 1
5
n
+ 1
=
2
5
n
= 0
.
Thus as
n
→ ∞
, the values oF
a
n
oscillate be
tween values ever closer to
±
2
5
. Consequently,
the sequence diverges
.
002
10.0 points
Determine whether the series
3 + 2 +
4
3
+
8
9
+
···
is convergent or divergent, and iF convergent,
fnd its sum.
1.
convergent with sum =
1
4
2.
convergent with sum = 9
correct
3.
convergent with sum = 4
4.
convergent with sum =
1
9
5.
divergent
Explanation:
The series
3 + 2 +
4
3
+
8
9
+
···
=
∞
s
n
=1
a r
n
−
1
is an infnite geometric series in which
a
= 3
and
r
=
2
3
. But such a series is
(i) convergent with sum
a
1
−
r
when

r

<
1,
(ii) divergent when

r
 ≥
1
.
Thus the given series is
convergent with sum = 9
.
003
10.0 points
Determine whether the infnite series
∞
s
n
= 1
(
n
+ 1)
2
n
(
n
+ 2)
converges or diverges, and iF converges, fnd
its sum.
1.
converges with sum =
1
2
2.
converges with sum =
1
8
3.
converges with sum =
1
4
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View Full Document schuem – Exam3 Practice Problems – Schultz – (58385)
2
4.
diverges
correct
5.
converges with sum = 1
Explanation:
By the Divergence Test, an infnite series
∑
n
a
n
diverges when
lim
n
→∞
a
n
n
= 0
.
Now, For the given series,
a
n
=
(
n
+ 1)
2
n
(
n
+ 2)
=
n
2
+ 2
n
+ 1
n
2
+ 2
n
.
But then,
lim
n
→∞
a
n
= 1
n
= 0
.
Consequently, the Divergence Test says that
the given series
diverges
.
keywords: infnite series, Divergence Test, ra
tional Function
004
10.0 points
To apply the root test to an infnite series
∑
n
a
n
the value oF
ρ
=
lim
n
→∞

a
n

1
/n
has to be determined.
Compute the value oF
ρ
For the series
∞
s
n
= 1
p
3
n
+ 5
4
n
P
2
n
.
1.
ρ
=
3
4
2.
ρ
=
25
9
3.
ρ
=
9
16
correct
4.
ρ
=
25
16
5.
ρ
=
5
4
Explanation:
AFter division,
3
n
+ 5
4
n
=
3 + 5
/n
4
,
so

a
n

1
/n
=
±
3 + 5
/n
4
²
2
.
On the other hand,
lim
n
→∞
3 + 5
/n
4
=
3
4
.
Consequently,
ρ
=
9
16
.
keywords:
/* IF you use any oF these, fx the comment
symbols.
005
10.0 points
Determine whether the series
∞
s
k
= 1
(
−
1)
k
−
1
cos
±
1
3
k
²
is absolutely convergent, conditionally con
vergent or divergent.
1.
absolutely convergent
2.
conditionally convergent
3.
series is divergent
correct
Explanation:
The given series can be written as
∞
s
k
= 1
(
−
1)
k
−
1
a
k
schuem – Exam3 Practice Problems – Schultz – (58385)
3
with
a
k
= cos
p
1
3
k
P
>
0
,
and so is an alternating series. But
lim
k
→∞
cos
p
1
3
k
P
= 1
,
in which case
lim
k
→∞
(
−
1)
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This note was uploaded on 09/14/2009 for the course M 408 L taught by Professor Cepparo during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Cepparo

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