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Bryson, Spencer – Exam 3 – Due: May 1 2007, 11:00 pm – Inst: Mar Gonzalez
1
This
printout
should
have
18
questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
001
(part 1 oF 1) 10 points
Which oF the Following sequences converge?
A.
‰
4
n
2
5
n
2
+ 3
¾
,
B.
‰
4
n
2
n
+ 1
¾
.
1.
both oF them
2.
neither oF them
3.
A only
correct
4.
B only
Explanation:
A. AFter simplifcation,
4
n
2
5
n
2
+ 3
=
4
5 +
3
n
2
.
Consequently,
4
n
2
5
n
2
+ 3
→
4
5
as
n
→ ∞
, so the sequence converges.
B. AFter simplifcation,
4
n
2
n
+ 3
=
2
n
1 +
3
2
n
.
Consequently,
4
n
2
n
+ 3
→ ∞
as
n
→ ∞
, so the sequence does not converge.
keywords:
002
(part 1 oF 1) 10 points
IF the
n
th
partial sum oF an infnite series is
S
n
=
5
n
2

2
2
n
2
+ 3
,
what is the sum oF the series?
1.
sum =
19
8
2.
sum =
5
2
correct
3.
sum =
21
8
4.
sum =
9
4
5.
sum =
17
8
Explanation:
By defnition
sum =
lim
n
→∞
S
n
=
lim
n
→∞
‡
5
n
2

2
2
n
2
+ 3
·
.
Thus
sum =
5
2
.
keywords: partial sum, defnition oF series
003
(part 1 oF 1) 10 points
IF
∑
n
a
n
converges, which, iF any, oF the
Following statements are true:
(A)
lim
n
→∞
a
n
= 0
,
(B)
X
n

a
n

is convergent
.
1.
A only
correct
2.
B only
3.
neither A nor B
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2
4.
both A and B
Explanation:
(A) TRUE: to say that
∑
n
a
n
converges is
to say that the limit
lim
n
→∞
S
n
of its partial sums
S
n
=
a
1
+
a
2
+
...
+
a
n
converges. But then
lim
n
→∞
a
n
=
lim
n
→∞
(
S
n

S
n

1
) = 0
.
(B) FALSE: set
a
n
=
(

1)
n
n
.
Then
X
n

a
n

=
X
n
1
n
,
so by the
p
series test with
p
= 1, the series
X

a
n

diverges. On the other hand,
X
n
a
n
=
X
n
(

1)
n
n
converges by the Alternating Series Test.
keywords:
004
(part 1 of 1) 10 points
Determine the convergence or divergence of
the series
(
A
)
1 +
1
4
+
1
9
+
1
16
+
1
25
+
...,
and
(
B
)
∞
X
k
= 1
k
2
e

k
3
.
1.
both series convergent
correct
2.
A
convergent,
B
divergent
3.
both series divergent
4.
A
divergent,
B
convergent
Explanation:
(
A
) The given series has the form
1 +
1
4
+
1
9
+
1
16
+
1
25
+
...
=
∞
X
n
=1
1
n
2
.
This is a
p
series with
p
= 2
>
1, so the series
converges.
(
B
) The given series has the form
∞
X
k
= 1
f
(
k
)
with
f
de±ned by
f
(
x
) =
x
2
e

x
3
.
Note ±rst that
f
is continuous and positive on
[1
,
∞
); in addition, since
f
0
(
x
) =
e

x
3
(2
x
1

3
x
4
)
<
0
for
x >
1,
f
is decreasing on [1
,
∞
). Thus we
can use the Integral Test. Now, by substitu
tion,
Z
t
1
x
2
e

x
3
dx
=
•

1
3
e

x
3
‚
t
1
,
and so
Z
∞
1
x
2
e

x
3
dx
=
1
3
e
.
Since the integral converges, the series con
verges. This could also be established using
the Ratio Test.
keywords:
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