Version 101 – Exam01 – gilbert – (55485)
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001
10.0points
To apply the ratio test to the infinite series
summationdisplay
n
a
n
, the value
λ
=
lim
n
→ ∞
a
n
+1
a
n
has to be determined.
Compute
λ
for the series
∞
summationdisplay
n
=1
4
n
3
n
4
+ 8
.
1.
λ
= 4
correct
2.
λ
= 0
3.
λ
=
4
3
4.
λ
=
1
2
5.
λ
=
4
11
Explanation:
By algebra,
a
n
+1
a
n
=
4
n
+1
4
n
bracketleftbigg
3
n
4
+ 8
3(
n
+ 1)
4
+ 8
bracketrightbigg
.
But
3
n
4
+ 8
3(
n
+ 1)
4
+ 8
=
3 +
8
n
4
3
parenleftbigg
n
+ 1
n
parenrightbigg
4
+
8
n
4
.
Since
parenleftbigg
n
+ 1
n
parenrightbigg
4
=
parenleftbigg
1 +
1
n
parenrightbigg
4
→
1
as
n
→ ∞
, we see that
lim
n
→ ∞
a
n
+1
a
n
=
lim
n
→ ∞
4
parenleftBig
3 +
8
n
4
parenrightBig
3
parenleftBig
n
+ 1
n
parenrightBig
4
+
8
n
4
= 4
.
Consequently,
λ
= 4
.
002
10.0points
Compute the degree 2 Taylor polynomial
for
f
centered at
x
= 4 when
f
(
x
) =
√
x .
1.
T
2
(
x
) = 4 +
1
8
(
x

4) +
1
32
(
x

4)
2
2.
T
2
(
x
)
=
2 +
1
4
(
x

4)

1
64
(
x

4)
2
correct
3.
T
2
(
x
) = 2

1
8
(
x

4) +
1
64
(
x

4)
2
4.
T
2
(
x
) = 2 +
1
4
(
x

4)

1
32
(
x

4)
2
5.
T
2
(
x
) = 2

1
4
(
x

4) +
1
64
(
x

4)
2
6.
T
2
(
x
) = 4

1
8
(
x

4)

1
32
(
x

4)
2
Explanation:
The degree 2 Taylor polynomial centered at
x
= 4 for a general
f
is given by
T
2
(
x
) =
f
(4)+
f
′
(4) (
x

4)+
f
′′
(4)
2!
(
x

4)
2
.
Now when
f
(
x
) =
√
x
,
f
′
(
x
) =
1
2
√
x
,
f
′′
(
x
) =

1
4
x
√
x
,
in which case,
f
(4) = 2
,
f
′
(4) =
1
4
,
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Version 101 – Exam01 – gilbert – (55485)
2
while
f
′′
(4)
2!
=

1
64
.
Consequently,
T
2
(
x
) = 2 +
1
4
(
x

4)

1
64
(
x

4)
2
.
003
10.0points
Which one of the following properties does
the series
∞
summationdisplay
k
=1
(

1)
k
−
1
3
√
1 +
k
2
have?
1.
divergent
2.
conditionally convergent
correct
3.
absolutely convergent
Explanation:
The given series has the form of an alter
nating series
∞
summationdisplay
k
=1
(

1)
k
−
1
a
k
,
a
k
=
3
√
1 +
k
2
.
To check for absolute convergence we apply
the Limit Comparison Test with
a
k
=
3
√
1 +
k
2
,
b
k
=
1
k
.
For then
lim
k
→ ∞
a
k
b
k
=
lim
k
→ ∞
3
k
√
1 +
k
2
= 3
>
0
.
Thus the series
∑
a
k
converges if and only if
the series
∞
summationdisplay
k
=1
1
k
converges.
But, by the
p
series test with
p
= 1, this last series diverges; in particular,
the given series does not converge absolutely.
To check if the given series converges con
ditionally, consider first the function
f
(
x
) =
3
√
1 +
x
2
.
Then, by the Chain Rule,
f
′
(
x
) =

3
x
(1 +
x
2
)
3
/
2
<
0
for all
x >
0, while
lim
x
→ ∞
f
(
x
) = 0
.
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 Spring '07
 TextbookAnswers
 Mathematical Series, lim g

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