pokharel (yp624) – HW11 – Radin – (57410)
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001
10.0 points
Determine if the sequence
{
a
n
}
converges
when
a
n
=
1
n
ln
parenleftbigg
4
2
n
+ 4
parenrightbigg
,
and if it does, find its limit.
1.
the sequence diverges
2.
limit =
−
ln 2
3.
limit = ln
2
3
4.
limit = ln 2
5.
limit = 0
correct
Explanation:
After division by
n
we see that
4
2
n
+ 4
=
4
n
2 +
4
n
,
so by properties of logs,
a
n
=
1
n
ln
4
n
−
1
n
ln
parenleftbigg
2 +
4
n
parenrightbigg
.
But by known limits (or use L’Hospital),
1
n
ln
4
n
,
1
n
ln
parenleftbigg
2 +
4
n
parenrightbigg
−→
0
as
n
→ ∞
. Consequently, the sequence
{
a
n
}
converges and has
limit = 0
.
002
10.0 points
Find a formula for the general term
a
n
of
the sequence
{
a
n
}
∞
n
=1
=
braceleftBig
2
,
6
,
10
,
14
, . . .
bracerightBig
,
assuming that the pattern of the first few
terms continues.
1.
a
n
=
n
+ 4
2.
a
n
=
n
+ 3
3.
a
n
= 4
n
−
2
correct
4.
a
n
= 5
n
−
3
5.
a
n
= 3
n
−
1
Explanation:
By inspection, consecutive terms
a
n
−
1
and
a
n
in the sequence
{
a
n
}
∞
n
=1
=
braceleftBig
2
,
6
,
10
,
14
, . . .
bracerightBig
have the property that
a
n
−
a
n
−
1
=
d
= 4
.
Thus
a
n
=
a
n
−
1
+
d
=
a
n
−
2
+ 2
d
=
. . .
=
a
1
+ (
n
−
1)
d
= 2 + 4(
n
−
1)
.
Consequently,
a
n
= 4
n
−
2
.
keywords:
003
10.0 points
Find a formula for the general term
a
n
of
the sequence
{
a
n
}
∞
n
=1
=
braceleftBig
1
,
−
5
3
,
25
9
,
−
125
27
, . . .
bracerightBig
,
assuming that the pattern of the first few
terms continues.
1.
a
n
=
−
parenleftBig
5
3
parenrightBig
n
2.
a
n
=
parenleftBig
−
3
5
parenrightBig
n
−
1
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pokharel (yp624) – HW11 – Radin – (57410)
2
3.
a
n
=
parenleftBig
−
3
2
parenrightBig
n
−
1
4.
a
n
=
−
parenleftBig
3
5
parenrightBig
n
5.
a
n
=
−
parenleftBig
3
2
parenrightBig
n
6.
a
n
=
parenleftBig
−
5
3
parenrightBig
n
−
1
correct
Explanation:
By inspection, consecutive terms
a
n
−
1
and
a
n
in the sequence
{
a
n
}
∞
n
=1
=
braceleftBig
1
,
−
5
3
,
25
9
,
−
125
27
, . . .
bracerightBig
have the property that
a
n
=
ra
n
−
1
=
parenleftBig
−
5
3
parenrightBig
a
n
−
1
.
Thus
a
n
=
ra
n
−
1
=
r
2
a
n
−
2
=
. . .
=
r
n
−
1
a
1
=
parenleftBig
−
5
3
parenrightBig
n
−
1
a
1
.
Consequently,
a
n
=
parenleftBig
−
5
3
parenrightBig
n
−
1
since
a
1
= 1.
keywords: sequence, common ratio
004
10.0 points
Determine if the sequence
{
a
n
}
converges,
and if it does, find its limit when
a
n
=
3
n
5
−
n
3
+ 4
7
n
4
+ 4
n
2
+ 2
.
1.
limit =
−
1
4
2.
limit = 2
3.
the sequence diverges
correct
4.
limit = 0
5.
limit =
3
7
Explanation:
After division by
n
4
we see that
a
n
=
3
n
−
1
n
+
4
n
4
7 +
4
n
2
+
2
n
4
.
Now
1
n
,
4
n
4
,
4
n
2
,
2
n
4
−→
0
as
n
→ ∞
; in particular, the denominator
converges and has limit 7
negationslash
=
0.
Thus by
properties of limits
{
a
n
}
diverges
since the sequence
{
3
n
}
diverges.
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 Fall '08
 RAdin
 Calculus, Squeeze Theorem, Limit, lim

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