cruz (fmc326) – HW02 – kalahurka – (55295)
1
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001
10.0points
Which, if any, of the following series converge?
(
A
)
∞
summationdisplay
m
=2
5 ln
m
m
2
(
B
)
∞
summationdisplay
n
=2
1
n
(ln
n
)
2
1.
neither
A
nor
B
2.
A
and
B
correct
3.
A
only
4.
B
only
Explanation:
(
A
) The function
f
(
x
) =
5 ln
x
x
2
is continous and positive on [2
,
∞
); in addi-
tion, since
f
′
(
x
) = 5
parenleftbigg
1
−
2 ln
x
x
3
parenrightbigg
<
0
on [2
,
∞
),
f
is also decreasing on this inter-
val. This suggests applying the Integral Test.
Now, after Integration by Parts, we see that
integraldisplay
t
2
f
(
x
)
dx
= 5
bracketleftBig
−
ln
x
x
−
1
x
bracketrightBig
t
2
,
and so
integraldisplay
∞
2
f
(
x
)
dx
= 5(1 + ln 2)
.
The Integral Test thus ensures that the given
series
converges
.
(
B
) The given series is of the form
∞
summationdisplay
n
=2
f
(
x
)
with
f
(
x
) =
1
x
(ln
x
)
2
,
x
≥
2
.
Since
f
is positive and decreasing on [2
,
∞
),
Integral Test ensures that the given series is
convergent if the improper integral
I
=
integraldisplay
∞
2
f
(
x
)
dx
converges.
But, after the substitution
u
=
ln
x
we see that
I
=
lim
t
→ ∞
bracketleftBig
−
1
u
bracketrightBig
t
ln2
=
1
ln 2
.
Consequently, the series
converges
.
keywords:
002
10.0points
What is the smallest number of terms of
the series
∞
summationdisplay
m
=1
2
(
m
+ 1)(ln(
m
+ 1))
2
you would need to add to find its sum to
within
1
10
?
1.
e
20
terms
correct
2.
e
40
terms
3.
e
35
terms
4.
e
30
terms

cruz (fmc326) – HW02 – kalahurka – (55295)
2
5.
e
25
terms
Explanation:
If
f
is a function which is continuous, pos-
itive and decreasing on [1
,
∞
), then the in-
equalities
∞
summationdisplay
m
=
M
+1
f
(
m
)
<
integraldisplay
∞
M
f
(
x
)
dx <
∞
summationdisplay
m
=
M
f
(
m
)
hold for any
M
≥
1. Thus, if
s
=
∞
summationdisplay
m
=1
f
(
m
)
,
S
M
=
M
summationdisplay
m
=1
f
(
m
)
,
then
s
−
S
M
<
integraldisplay
∞
M
f
(
m
)
dx < s
−
S
M
−
1
.
Consequently, if we want the smallest value of
M
so that
S
M
is within
1
10
of the sum
s
of the
series, we need to find the smallest value of
M
so that
integraldisplay
∞
M
f
(
x
)
dx
≤
1
10
.
But for the given series
f
(
x
) =
2
(
x
+ 1)(ln(
x
+ 1))
2
,
a change of variable
u
= ln(
x
+1) ensures that
integraldisplay
∞
M
f
(
x
)
dx
=
integraldisplay
∞
M
,
2
(
x
+ 1)(ln(
x
+ 1))
2
dx
=
lim
t
→ ∞
bracketleftBig
−
2
u
bracketrightBig
t
ln(
M
+1)
=
2
ln(
M
+ 1)
.
Thus
integraldisplay
∞
M
f
(
x
)
dx
≤
1
10
=
⇒
2
ln(
M
+ 1)
≤
1
10
,
and so the smallest number of terms you
would need to add is
e
20
terms
.
003
10.0points
Use the Remainder Estimate
integraldisplay
∞
n
+1
f
(
x
)
dx
≤
R
n
≤
integraldisplay
∞
n
f
(
x
)
dx
for the Integral Test applied with
f
(
x
) =
4
x
3
to determine the smallest value of
n
so that
the error
R
n
=
s
−
S
n
in using the
n
th
partial
sum
S
n
to estimate the sum
s
of the series
∞
summationdisplay
k
=1
4
k
3
is less than 1
/
10
2
.
1.
n
= 13
2.
n
= 14
3.
n
= 15
correct
4.
n
= 11
5.
n
= 12
Explanation:
We have to find the smallest value of
n
so
that
s
−
S
n
≤
integraldisplay
∞
n
4
x
3
dx <
1
10
2
.
Now
integraldisplay
∞
n
4
x
3
dx
=
lim
t
→ ∞
integraldisplay
t
n
4
x
3
dx ,
while
integraldisplay
t
n
4
x
3
dx
=
bracketleftBig
−
2
x
2
bracketrightBig
t
n
=
2
n
2
−
2
t
2
−→
2
n
2
as
t
→ ∞
. Thus we have to solve the inequal-
ity
2
n
2
<
1
10
2
.

cruz (fmc326) – HW02 – kalahurka – (55295)
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