lee (jsl2382) – convergence tests and taylor series – tran – (56320)
1
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This homework includes convergence tests
(oF all kinds), and Taylor series.
001
10.0 points
Determine the interval oF convergence oF
the series
∞
s
n
= 1
(
−
2)
n
n
+ 4
x
n
.
1.
p
−
1
2
,
1
2
b
correct
2.
(
−
2
,
2]
3.
(
−∞
,
∞
)
4.
[
−
4
,
4)
5.
converges only at
x
= 0
6.
B
−
1
4
,
1
4
P
7.
none oF the other answers
Explanation:
The given series has the Form
∞
s
n
= 1
a
n
x
n
with
a
n
=
(
−
2)
n
n
+ 4
.
Now For this series,
(i)
R
= 0
if it converges only at
x
= 0,
(ii)
R
=
∞
if it converges for all
x
,
while iF
R >
0,
(iii)
it converges when

x

< R
, and
(iv)
diverges when

x

> R
.
But
lim
n
→∞
v
v
v
a
n
+1
a
n
v
v
v
=
lim
n
2(
n
+ 4)
n
+ 5
= 2
.
By the Ratio Test, thereFore, the given series
converges when

x

<
1
/
2 and diverges when

x

>
1
/
2. On the other hand, at the points
x
=
−
1
/
2 and
x
= 1
/
2 the series reduces to
∞
s
n
= 1
1
n
+ 4
,
∞
s
n
= 1
(
−
1)
n
n
+ 4
respectively. Now, by the
p
series Test with
p
= 1, the frst oF these series diverges, while
by the Alternating Series Test, the second
conveges. Consequently, the given series has
interval oF convergence =
p
−
1
2
,
1
2
b
.
keywords:
002
10.0 points
Determine the convergence or divergence oF
the series
(
A
)
∞
s
k
= 1
3 ln(2
k
)
k
2
,
and
(
B
)
∞
s
k
= 1
sin
2
k
k
2
+ 4
.
1.
A
converges,
B
diverges
2.
both series diverge
3.
both series converge
correct
4.
A
diverges,
B
converges
Explanation:
(
A
) The Function
f
(
x
) =
3 ln 2
x
x
2
lee (jsl2382) – convergence tests and taylor series – tran – (56320)
2
is continuous and positive on [1
,
∞
); in addi
tion, since
f
′
(
x
) = 3
p
1
−
2 ln2
x
x
3
P
<
0
on [1
,
∞
),
f
is also decreasing on this inter
val. This suggests applying the Integral Test.
Now, after Integration by Parts, we see that
i
t
1
f
(
x
)
dx
= 3
b
−
ln(2
x
)
x
−
1
x
B
t
1
,
and so
i
∞
1
f
(
x
)
dx
= 3(1 + ln 2)
.
The Integral Test thus ensures that series (
A
)
converges
.
(
B
) Note Frst that the inequalities
0
<
sin
2
k
k
2
+ 4
≤
1
k
2
+ 4
≤
1
k
2
hold for all
n
≥
1. On the other hand, by the
p
series test the series
∞
s
k
= 1
1
k
2
is convergent since
p
= 2
>
1. Thus, by the
comparison test, series (
B
)
converges
.
003
10.0 points
To apply the root test to an inFnite series
s
k
a
k
,
the value of
ρ
=
lim
k
→∞

a
k

1
/k
has to be determined.
Compute the value of
ρ
for the series
∞
s
k
= 1
2
k
k
(ln
k
+ 5)
k
.
1.
ρ
=
∞
2.
ρ
= 0
correct
3.
ρ
= 2
4.
ρ
= 10
5.
ρ
= 5
Explanation:
±or the given series
(
a
k
)
1
/k
= 2
1
/k
p
ln
k
+ 5
k
P
= 2
1
/k
p
ln
k
k
+
5
k
P
.
But
2
1
/k
−→
1
,
ln
k
k
0
as
k
→ ∞
. Consequently,
ρ
= 0
.
004
10.0 points
Which of the following series
(
A
)
∞
s
n
= 3
5
n
+ 7
(
n
ln
n
)
2
+ 6
(
B
)
∞
s
n
= 1
√
n
−
8
√
n
+ 7
(
C
)
∞
s
n
= 1
p
8
n
+ 7
6
n
−
5
P
n
diverge(s)?
1.
B
and
C
correct
2.
C
only
3.
all of
A, B, C
4.
A
and
B
lee (jsl2382) – convergence tests and taylor series – tran – (56320)
3
5.
B
only
Explanation:
(
A
) After division,
5
n
+ 7
(
n
ln
n
)
2
+ 6
=
5 +
7
n
n
(ln
n
)
2
+
6
n
.