rabinovich (ahr454) – HW11 – um – (55915)
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001
10.0 points
Let
h
be a continuous, positive, decreasing
function on [3
,
∞
).
Compare the values of
the series
A
=
11
n
= 4
h
(
n
)
and the integral
B
=
11
3
h
(
z
)
dz .
1.
A
=
B
2.
A < B
correct
3.
A > B
Explanation:
In the figure
3
4
5
6
7
. . .
a
4
a
5
a
6
a
7
the bold line is the graph of
h
on [3
,
∞
) and
the areas of the rectangles the terms in the
series
∞
n
= 4
a
n
,
a
n
=
h
(
n
)
.
Clearly from this figure we see that
a
4
=
h
(4)
<
4
3
h
(
z
)
dz,
a
5
=
h
(5)
<
5
4
h
(
z
)
dz ,
while
a
6
=
h
(6)
<
6
5
h
(
z
)
dz,
a
7
=
h
(7)
<
7
6
h
(
z
)
dz ,
and so on. Consequently,
A < B
.
keywords: Szyszko
002
10.0 points
Which, if any, of the following series con
verge?
A
.
1 +
1
4
+
1
9
+
1
16
+
. . .
B
.
∞
n
= 1
n
1 +
n
2
1.
neither of them
2.
B only
3.
A only
correct
4.
both of them
Explanation:
A. Series is
∞
n
=1
1
n
2
. Use
f
(
x
) =
1
x
2
. Then
∞
1
f
(
x
)
dx
is convergent, so series converges.
rabinovich (ahr454) – HW11 – um – (55915)
2
B. Use
f
(
x
) =
x
1 +
x
2
. Then
∞
1
f
(
x
)
dx
is divergent (ln integral), so series diverges.
003
10.0 points
Determine whether the series
∞
n
= 1
6

4
√
n
n
3
converges or diverges.
1.
series is convergent
correct
2.
series is divergent
Explanation:
We check separately the convergence or di
vergence of
∞
n
= 1
6
n
3
,
∞
n
= 1
4
√
n
n
3
=
∞
n
= 1
4
n
5
/
2
.
Now both are pseries: the first being of the
form
∞
n
= 1
6
n
3
=
∞
n
= 1
6
n
p
,
p
= 3
,
while the second one has the form
∞
n
= 1
4
n
5
/
2
=
∞
n
= 1
4
n
p
,
p
=
5
2
.
Since
p >
1 in both cases, each series con
verges. As the di
ff
erence of convergent series,
therefore, the given series
is convergent
.
004
10.0 points
Determine whether the series
∞
n
= 2
n
4(ln(
n
))
2
is convergent or divergent.
1.
divergent
correct
2.
convergent
Explanation:
By the Divergence Test, a series
∞
n
=
N
a
n
will be divergent for each fixed choice of
N
if
lim
n
→ ∞
a
n
= 0
since it is only the behaviour of
a
n
as
n
→ ∞
that’s important.
Now, for the given series,
N
= 2 and
a
n
=
n
4(ln(
n
))
2
.
But by L’Hospital’s Rule applied twice,
lim
x
→ ∞
x
(ln(
x
))
2
=
lim
x
→ ∞
1
(2 ln(
x
))
/x
=
lim
x
→ ∞
x
2 ln(
x
)
=
lim
x
→ ∞
1
2
/x
=
∞
.
Consequently, by the Divergence Test, the
given series is
divergent
.
005
10.0 points
If the improper integral
∞
1
1
x
p
dx
converges, which of the following statements
is (are) always true?
(
A
)
n
1
n
p
converges;
(
B
)
n
1
n
p
+1
diverges;
rabinovich (ahr454) – HW11 – um – (55915)
3
(
C
)
n
1
n
p

1
converges;
(
D
)
n
1
n
p

1
diverges;
(
E
)
n
1
n
p
+1
converges
.
1.
A, D
and
E
2.
B
and
D
only
3.
A
and
E
only
correct
4.
A, C
and
E
only
5.
A
only
Explanation:
To apply the Integral test we need to start
with a function
f
which is positive, continuous
and decreasing on [1
,
∞
). Then the integral
test says that the improper integral
∞
1
f
(
x
)
dx
converges if and only if the infinite series
∞
n
= 1
f
(
n
)
converges.
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 Fall '11
 Gramlich
 Accounting, Calculus, Mathematical Series, lim