tgo72 – Homework 12 – Gompf – (58370)
1
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printout
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have
20
questions.
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before answering.
001
10.0 points
If
a
m
, b
m
,
and
c
m
satisfy the inequalities
0
< a
m
≤
c
m
≤
b
m
,
for all
m
, what can we say about the series
(
A
) :
∞
summationdisplay
m
=1
a
m
,
(
B
) :
∞
summationdisplay
m
=1
b
m
if we know that the series
(
C
) :
∞
summationdisplay
m
=1
c
m
is convergent but know nothing else about
a
m
and
b
m
?
1.
(
A
) converges
,
(
B
) need not converge
correct
2.
(
A
) diverges
,
(
B
) diverges
3.
(
A
) converges
,
(
B
) diverges
4.
(
A
) need not converge
,
(
B
) converges
5.
(
A
) converges
,
(
B
) converges
6.
(
A
) diverges
,
(
B
) converges
Explanation:
Let’s try applying the Comparison Test:
(i) if
0
< a
m
≤
c
m
,
summationdisplay
m
c
m
converges
,
then the Comparison Test applies and says
that
summationdisplay
a
m
converges;
(ii) while if
0
< c
m
≤
b
m
,
summationdisplay
m
c
m
converges
,
then the Comparison Test is inconclusive be
cause
summationdisplay
b
m
could converge, but it could di
verge  we can’t say precisely without further
restrictions on
b
m
.
Consequently, what we can say is
(
A
) converges
,
(
B
) need not converge
.
002
10.0 points
Determine whether the series
∞
summationdisplay
k
=1
cos
2
k
k
2
+ 7
converges or diverges.
1.
series is convergent
correct
2.
series is divergent
Explanation:
Note first that the inequalities
0
<
cos
2
k
k
2
+ 7
≤
1
k
2
+ 7
≤
1
k
2
hold for all
n
≥
1. On the other hand, by the
p
series test the series
∞
summationdisplay
n
=1
1
k
2
is convergent since
p
= 2
>
1. Thus, by the
comparison test, the given
series is convergent
.
003
10.0 points
Which of the following infinite series con
verges?
1.
∞
summationdisplay
n
=1
parenleftbigg
3
2
parenrightbigg
n
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tgo72 – Homework 12 – Gompf – (58370)
2
2.
∞
summationdisplay
n
=1
3
n
n
(
n
+ 4)
n
3.
∞
summationdisplay
n
=1
parenleftbigg
4
n
5
n
+ 2
parenrightbigg
n
correct
4.
∞
summationdisplay
k
=2
2
5
k
ln
k
+ 3
k
5.
∞
summationdisplay
n
=1
5
3
n
−
2
Explanation:
We test the convergence of each of the five
series.
(i) For the series
∞
summationdisplay
k
=2
2
5
k
ln
k
+ 3
k
the limit comparison test can be used, com
paring it with
∞
summationdisplay
k
=2
1
k
ln
k
.
Now, after division,
k
ln
k
parenleftbigg
2
5
k
ln
k
+ 3
k
parenrightbigg
=
2
5 +
3
ln
k
.
Since
2
5 +
3
ln
k
−→
2
5
>
0
as
k
→ ∞
, the limit comparison test applies.
But by the Integral test, the series
∞
summationdisplay
k
=2
1
k
ln
k
does not converge. Thus
∞
summationdisplay
k
=2
2
5
k
ln
k
+ 3
k
does not converge.
(ii) If a series
∑
n
a
n
converges, then
a
n
−→
0
as
n
→ ∞
. But for the series
∞
summationdisplay
n
=1
3
n
n
(
n
+ 4)
n
we see that
a
n
=
3
n
n
(
n
+ 4)
n
= 3
parenleftbigg
n
n
+ 4
parenrightbigg
n
= 3
parenleftbigg
n
+ 4
n
parenrightbigg
−
n
= 3
braceleftbiggparenleftbigg
1 +
4
n
parenrightbigg
n
bracerightbigg
−
1
.
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 Spring '08
 RAdin
 Calculus, Mathematical Series, 1 K, 1 cm, 1 3K, 1 1 3k

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