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Version 013 – Exam 3 – cheng – (58520)
1
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beFore answering.
001
10.0 points
Determine iF the sequence
{
a
n
}
converges
when
a
n
=
(2
n
−
1)!
(2
n
+ 1)!
,
and iF it converges, fnd the limit.
1.
converges with limit = 4
2.
converges with limit =
1
4
3.
converges with limit = 0
correct
4.
converges with limit = 1
5.
does not converge
Explanation:
By defnition,
m
! is the product
m
! = 1
.
2
.
3
. . . . .
m
oF the frst
m
positive integers. When
m
=
2
n
−
1, thereFore,
(2
n
−
1)! = 1
.
2
.
3
. . . .
(2
n
−
1)
,
while
(2
n
+ 1)! = 1
.
2
.
3
. . . . .
(2
n
−
1)2
n
(2
n
+ 1)
.
when
m
= 2
n
+ 1. But then,
(2
n
−
1)!
(2
n
+ 1)!
=
1
2
n
(2
n
+ 1)
.
Consequently, the given sequence
converges with limit = 0
.
002
10.0 points
Determine iF the sequence
{
a
n
}
converges,
and iF it does, fnd its limit when
a
n
=
4
n
+ (
−
1)
n
2
n
+ 4
.
1.
converges with limit =
5
2
2.
converges with limit =
2
3
3.
converges with limit = 2
correct
4.
converges with limit =
3
2
5.
sequence does not converge
Explanation:
AFter division by
n
we see that
a
n
=
4 +
(
−
1)
n
n
2 +
4
n
.
But
(
−
1)
n
n
,
4
n
−→
0
as
n
→ ∞
,
so
a
n
→
2 as
n
→ ∞
. Conse
quently, the sequence converges and has
limit = 2
.
003
10.0 points
±ind all values oF
x
For which the series
∞
s
n
= 0
(3
x
−
5)
n
4
n
converges.
1.
converges only on
p
1
3
,
−
1
3
P
2.
converges only on
p
−
3
,
3
P
3.
converges only on
p
−
3
,
−
1
3
P
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View Full DocumentVersion 013 – Exam 3 – cheng – (58520)
2
4.
converges only on
p
1
3
,
3
P
correct
5.
converges only on
p
−
5
3
,
5
3
P
Explanation:
A geometric series
∑
∞
n
= 0
ar
n
(i) converges when

r

<
1, and
(ii) diverges when

r
 ≥
1,
Now for the given series,
a
= 1
,
r
=
3
x
−
5
4
,
so it converges exactly when
−
1
<
3
x
−
5
4
<
1
.
By solving these inequalities we thus see that
the given series converges only on the interval
p
1
3
,
3
P
.
004
10.0 points
First ±nd
a
n
so that
∞
s
n
=1
a
n
= 6 +
3
√
2
+
2
√
3
+
3
4
+
6
5
√
5
+
. . .
and then determine whether the series con
verges or diverges.
1.
a
n
=
3
2
n
3
/
2
,
series converges
2.
a
n
=
6
n
3
/
2
,
series converges
correct
3.
a
n
=
3
2
n
3
/
2
,
series diverges
4.
a
n
=
6
n
1
/
2
,
series diverges
5.
a
n
=
6
n
3
/
2
,
series diverges
6.
a
n
=
6
n
1
/
2
,
series converges
Explanation:
By inspection
a
n
=
6
n
3
/
2
.
To test for convergence we use the Integral
test with
f
(
x
) =
6
x
3
/
2
.
This is a positive, continuous, decreasing
function on [1
,
∞
). Furthermore,
i
∞
1
f
(
x
)
dx
=
lim
n
→∞
i
n
1
6
x
3
/
2
dx
= lim
n
→∞
b
−
12
x
1
/
2
B
n
1
= 12
,
so the improper integral
i
∞
1
f
(
x
)
dx
is convergent, which by the Integral test
means that the
series converges
.
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 Fall '08
 RAdin
 Calculus

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