Version 060 – EXAM 3 – Radin – (57410)
1
This
print-out
should
have
20
questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
Determine whether the sequence
{
a
n
}
con-
verges or diverges when
a
n
=
(
−
1)
n
−
1
n
n
2
+ 2
,
and if it converges, find the limit.
1.
converges with limit = 0
correct
2.
converges with limit =
1
2
3.
converges with limit =
−
1
2
4.
converges with limit = 2
5.
converges with limit =
−
2
6.
sequence diverges
Explanation:
After division,
a
n
=
(
−
1)
n
−
1
n
n
2
+ 2
=
(
−
1)
n
−
1
n
+
2
n
.
Consequently,
0
≤ |
a
n
|
=
1
n
+
2
n
≤
1
n
.
But 1
/n
→
0 as
n
→ ∞
, so by the Squeeze
theorem,
lim
n
→ ∞
|
a
n
|
= 0
.
But
−|
a
n
| ≤
a
n
≤ |
a
n
|
,
so by the Squeeze theorem again the given
sequence
{
a
n
}
converges and has
limit = 0
.
keywords:
002
10.0 points
Which of the following sequences converge?
A
.
braceleftbigg
5
e
n
4 +
e
n
bracerightbigg
B
.
braceleftbigg
e
n
+ 5
3
n
+ 2
bracerightbigg
1.
A
only
correct
2.
neither of them
3.
B
only
4.
both
A
and
B
Explanation:
A
. After simplification,
5
e
n
4 +
e
n
=
5
4
e
−
n
+ 1
.
Thus
5
e
n
4 +
e
n
−→
5
as
n
→ ∞
since
e
−
x
→
0 as
x
→ ∞
.
B
. To test for convergence of the sequence
braceleftbigg
e
n
+ 5
3
n
+ 2
bracerightbigg
we use L’Hospital’s Rule with
f
(
x
) =
e
x
+ 5
,
g
(
x
) = 3
x
+ 2
since
f
(
x
)
−→ ∞
,
g
(
x
)
−→ ∞
as
x
→ ∞
. Thus
lim
n
→ ∞
e
n
3
n
+ 2
=
lim
x
→ ∞
f
′
(
x
)
g
′
(
x
)
.
This
preview
has intentionally blurred sections.
Sign up to view the full version.
Version 060 – EXAM 3 – Radin – (57410)
2
But
f
′
(
x
)
g
′
(
x
)
=
e
x
3
−→ ∞
as
n
→ ∞
.
Consequently,
only
A
converges
.
003
10.0 points
Determine if the sequence
{
a
n
}
converges,
and if it does, find its limit when
a
n
=
parenleftbigg
1
−
3
4
n
parenrightbigg
5
n
.
1.
limit =
e
−
15
4
correct
2.
limit =
e
3
4
3.
limit =
e
15
4
4.
limit = 1
5.
the sequence diverges
6.
limit =
e
−
3
4
7.
limit =
e
−
12
5
Explanation:
We know that
lim
n
→ ∞
parenleftBig
1
−
x
n
parenrightBig
n
=
e
−
x
.
Now
a
n
=
bracketleftbiggparenleftbigg
1
−
3
4
n
parenrightbigg
n
bracketrightbigg
5
.
By the properties of limits, therefore,
lim
n
→ ∞
a
n
=
bracketleftbigg
lim
n
→ ∞
parenleftbigg
1
−
3
4
n
parenrightbigg
n
bracketrightbigg
5
,
so the sequence converges and has
limit =
parenleftBig
e
−
3
4
parenrightBig
5
=
e
−
15
4
.
004
10.0 points
Determine whether the infinite series
∞
summationdisplay
n
=1
ln
parenleftbigg
4
n
5
n
+ 4
parenrightbigg
is convergent or divergent, and if convergent,
find its sum.
1.
converges with sum = ln
parenleftbigg
4
5
parenrightbigg
2.
converges with sum = ln
parenleftbigg
4
9
parenrightbigg
3.
converges with sum = ln
parenleftbigg
5
4
parenrightbigg
4.
diverges
correct
5.
converges with sum = ln
parenleftbigg
9
4
parenrightbigg
Explanation:
An infinite series
summationdisplay
a
n
diverges when
lim
n
→∞
a
n
negationslash
= 0
.
For the given series
a
n
= ln
parenleftbigg
4
n
5
n
+ 4
parenrightbigg
.

This is the end of the preview.
Sign up
to
access the rest of the document.
- Fall '08
- RAdin
- Mathematical Series, lim
-
Click to edit the document details