∞
s
k
=1
4
k
2 + 3
k
B.
∞
s
n
= 2
3 + (

1)
n
2 +
n
√
n
are convergent.
1.
A only
2.
both of them
3.
B only
correct
4.
neither of them
Explanation:
A. Divergent: use Limit Comparison Test
and Geometric Series.
B. Convergent: use Limit Comparison Test
and
p
series Test with
p
=
3
2
.
016
10.0 points
Determine whether the series
∞
s
k
=1
sin
p
2
k
P
is absolutely convergent, conditionally con
vergent or divergent.
1.
conditionally convergent
2.
absolutely convergent
3.
divergent
correct
Explanation:
All the terms in the series
∞
s
k
=1
sin
p
2
k
P
are positive for
k
≥
2, so the given series ei
ther converges absolutely or it is divergent.
choice (hac762) – HW Quest Week 8 – cepparo – (53850)
8
To determine which we use the Limit Com
parison Test with
a
k
= sin
p
2
k
P
,
b
k
=
2
k
.
For then
lim
k
→∞
a
k
b
k
= lim
k
→∞
sin
p
2
k
P
2
k
= lim
θ
→
0
sin(
θ
)
θ
= 1
>
0
.
Thus the given series
∞
s
k
=1
sin
p
2
k
P
converges if and only if the series
∞
s
k
= 1
2
k
converges. But by the
p
series test with
p
= 1
(or because the harmonic series is divergent),
this last series is divergent. Consequently, the
given series is
divergent
.
017
10.0 points
Compare the radius of convergence,
R
1
, of
the series
∞
s
n
=0
c
n
x
n
with the radius of convergence,
R
2
, of the
series
∞
s
n
= 1
n c
n
x
n

1
when
lim
n
→∞
v
v
v
c
n
+1
c
n
v
v
v
= 3
.
1.
2
R
1
=
R
2
= 3
2.
R
1
= 2
R
2
= 3
3.
2
R
1
=
R
2
=
1
3
4.
R
1
= 2
R
2
=
1
3
5.
R
1
=
R
2
= 3
6.
R
1
=
R
2
=
1
3
correct
Explanation:
When
lim
n
→∞
v
v
v
c
n
+1
c
n
v
v
v
= 3
,
the Ratio Test ensures that the series
∞
s
n
=0
c
n
x
n
is
(i) convergent when

x

<
1
3
, and
(ii) divergent when

x

>
1
3
.
On the other hand, since
lim
n
→∞
v
v
v
(
n
+ 1)
c
n
+1
nc
n
v
v
v
= lim
n
→∞
v
v
v
c
n
+1
c
n
v
v
v
,
the Ratio Test ensures also that the series
∞
s
n
=1
n c
n
x
n

1
is
(i) convergent when

x

<
1
3
, and
(ii) divergent when

x

>
1
3
.
Consequently,
R
1
=
R
2
=
1
3
.
018
10.0 points
choice (hac762) – HW Quest Week 8 – cepparo – (53850)
9
Find a power series representation for the
function
f
(
x
) =
1
x

2
.
1.
f
(
x
) =

∞
s
n
=0
1
2
n
+1
x
n
correct
2.
f
(
x
) =
∞
s
n
=0
(

1)
n
2
n
x
n
3.
f
(
x
) =
∞
s
n
=0
(

1)
n

1
2
n
+1
x
n
4.
f
(
x
) =

∞
s
n
=0
2
n
x
n
5.
f
(
x
) =
∞
s
n
=0
1
2
n
+1
x
n
Explanation:
We know that
1
1

x
= 1 +
x
+
x
2
+
. . .
=
∞
s
n
= 0
x
n
.
On the other hand,
1
x

2
=

1
2
p
1
1

(
x/
2)
P
.
Thus
f
(
x
) =

1
2
∞
s
n
= 0
p
x
2
P
n
=

1
2
∞
s
n
= 0
1
2
n
x
n
.
Consequently,
f
(
x
) =

∞
s
n
= 0
1
2
n
+1
x
n
with

x

<
2.
019
10.0 points
Find a power series representation for the
function
f
(
z
) =
1
4

z
3
.
1.
f
(
z
) =
∞
s
n
=0
z
3
n
4
3
n
2.
f
(
z
) =

∞
s
n
=0
4
n
z
3
n
3.
f
(
z
) =

∞
s
n
=0
z
n
4
n
+1
4.
f
(
z
) =
∞
s
n
=0
4
n
z
3
n
5.
f
(
z
) =
∞
s
n
=0
z
3
n
4
n
+1
correct
6.
f
(
z
) =

∞
s
n
=0
z
3
n
4
3
n
Explanation:
After simpli±cation,
f
(
z
) =
1
4

z
3
=
1
4
p
1
1

(
z
3
/
4)
P
.
On the other hand, we know that
1
1

t
=
∞
s
n
=0
t
n
.
Replacing
t
with
z
3
/
4, we thus obtain
f
(
z
) =
1
4
∞
s
n
=0
z
3
n
4
n
=
∞
s
n
=0
z
3
n
4
n
+1
.