(kah2852) huang HW11 radin (54915)
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001
assuming that the pattern of the rst few
terms continues.
4
3
n
1. an =
3
4
n
2. an =
5
4
n
3. an =
5
4
n1
4. an =
3
4
n1
5. an =
4
3
n1
6. an =
10.0 points
Find a formula for the general term an of
the sequence
{a n }
n=1
=
1
2, 6, 10, 14, . . .
,
assuming that the pattern of the rst few
terms continues.
1. an = 5n 3
2. an = 4n 2 correct
3. an = 3n 1
correct
Explanation:
By inspection, consecutive terms an1 and
an in the sequence
4. an = n + 4
5. an = n + 3
Explanation:
By inspection, consecutive terms an1 and
an in the sequence
{a n } =
n=1
64
4 16
, , ...
1, ,
39
27
{a n } =
n=1
have the property that
2, 6, 10, 14, . . .
have the property that
an = ran1 =
an an1 = d = 4 .
an = an1 + d = an2 + 2d = . . .
an = ran1 = r 2 an2 = . . .
= a1 + (n 1)d = 2 + 4(n 1) .
Consequently,
= r n1 a1 =
an =
keywords:
4
3
n1
4
3
n1
since a1 = 1.
10.0 points
Find a formula for the general term an of
the sequence
64
4 16
, , ...
1, ,
39
27
Consequently,
a n = 4n 2 .
{a n } =
n=1
4
an1 .
3
Thus
Thus
002
,
keywords: sequence, common ratio
003
10.0 points
a1 .
huang (kah2852) HW11 radin (54915)
Determine if the sequence {an } converges
when
1
4
an =
,
ln
n
5n + 1
and if it does, nd its limit.
1. limit = ln 5
3. limit =
5
4
Explanation:
After division by n4 we see that
2
3
4
4. limit = ln
5
3. limit = ln
an
5. limit = 0 correct
Explanation:
After division by n we see that
4
=
5n + 1
4
n
5+
1
n
Now
as n ; in particular, the denominator
converges and has limit 7 = 0. Thus by
properties of limits
,
1
1
14
ln ln 5 +
nnn
n
{an } diverges
.
But by known limits (or use LHospital),
14
ln ,
nn
1
1
ln 5 +
n
n
5
5
+4
n n.
=
4
2
7+ 2 + 4
n
n
6n
5
5
2
4
, 4 , 2 , 4 0
nn
n
n
so by properties of logs,
an =
5
2
4. limit = 0
5. limit =
2. the sequence diverges
2
0
as n . Consequently, the sequence {an }
converges and has
since the sequence {6n} diverges.
005
10.0 points
Determine if the sequence {an }n converges
when
5n
an =
6n 5
and if it does, nd its limit when
limit = 0
.
1. converges with limit =
004
10.0 points
Determine if the sequence {an } converges,
and if it does, nd its limit when
an =
1. limit =
6 n5 5 n3 + 5
.
7 n4 + 2 n2 + 4
6
7
2. the sequence diverges correct
5
correct
6
2. converges with limit = 5
3. converges with limit =
6
5
4. the sequence diverges
5. converges with limit = 0
Explanation:
huang (kah2852) HW11 radin (54915)
After division by n we see that
an =
5
5
6
n
as n . Thus {an } converges and has
5
6
limit =
007
as n . Thus {an }n converges and has
limit =
006
5
.
6
10
.
7
10.0 points
Determine if the sequence {an } converges,
and if it does, nd its limit when
an =
10.0 points
Determine whether the sequence {an } converges or diverges when
3
6n + (1)n
.
4n + 3
1. converges with limit =
3
correct
2
2. converges with limit =
5
4
and if it does, nd its limit
3. converges with limit =
1. limit =
10
correct
7
6
7
4. converges with limit =
2. limit =
10
21
7
4
5. sequence does not converge
an =
14 n2
2n2 + 3
,
7n + 2
n+1
3. limit = 0
Explanation:
After division by n we see that
4. the sequence diverges
5. limit =
an =
5
7
Explanation:
After bringing the two terms to a common
denominator we see that
an
14 n3 + 14 n2 (7n + 2) 2n2 + 3
=
(7n + 2) (n + 1)
(1)n
n.
3
4+
n
(1)n
,
n
3
0
n
6+
But
3
as n , so an 2 as n . Consequently, the sequence converges and has
10n2 21n 6
.
=
7n 2 + 9n + 2
Thus
an =
But
21
6
2
n
n.
9
2
7+ + 2
nn
10
9
2
21 6
, 2,
, 2 0
nn
nn
limit =
008
3
.
2
10.0 points
Determine if the sequence {an } converges
when
n2 n
,
an =
(n 8)2 n
and if it does, nd its limit
huang (kah2852) HW11 radin (54915)
16
1. limit = e
Explanation:
After simplication
correct
5n
=4
4n1
2. limit = e4
3. sequence diverges
5
4
n
.
But the sequence {r n }
16
4. limit = e
(a) converges with limit 0 when |r | < 1,
(b) and diverges for all |r | 1.
5. limit = 1
Consequently, the given sequence {an }
6. limit = e4
diverges .
Explanation:
By the Laws of Exponents,
an =
4
n8
n
2 n
=
8
1
n
=
1
8
n
2 n
010
Determine whether the sequence {an } converges or diverges when
n 2
.
But
xn
1+
ex
n
as n . Consequently, {an } converges
and has
an = (1)n
009
e8
2
= e16
1. limit =
8
5
10.0 points
an =
5n
4n1
,
and if it converges, nd the limit.
1. converges with limit = 5
3. limit =
5. limit = 0
Explanation:
After division,
8
3+
3n+8
n.
=
5
4n+5
4+
n
3. converges with limit = 1
5. diverges correct
3
4
4. sequence diverges correct
2. converges with limit = 4
4. converges with limit = 0
,
3
4
.
Determine whether the sequence {an } converges or diverges when
3n+ 8
4n+ 5
and if it does, nd its limit.
2. limit =
limit =
10.0 points
Now
85
,
0 as n , so
nn
lim
n
3
3n+8
=
= 0.
4n+5
4
huang (kah2852) HW11 radin (54915)
Thus as n , the values of an oscillate be3
tween values ever closer to 4 . Consequently,
the sequence diverges .
5
Determine if the sequence {an } converges
when
(2n + 1)!
an =
,
3n2 (2n 1)!
and if it converges, nd the limit.
011
10.0 points
Determine if the sequence {an } converges,
when
an = 9 cos
4n + 5
12 n + 6
2. does not converge
5
6
3
4
2
3
5. converges with limit =
2. limit = 3
3. converges with limit =
4. converges with limit =
5
6
3. limit = 9 cos
3
2
Explanation:
By m! denition, is the product
4. sequence does not converge
9
correct
2
9
3
6. limit =
2
(2n 1)! = 1 2 3 . . . (2n 1) ,
while
Explanation:
After division,
5
4 +
4n +5
n.
=
6
12 n + 6
12 +
n
56
,
0 as n , so
nn
4n +5
lim
=.
n 12 n + 6
3
Consequently, since cos x is continuous as a
function of x, the sequence {an } converges
and has
limit = 9 cos
012
m! = 1 2 3 . . . m
of the rst m positive integers. When m =
2n 1, therefore,
5. limit =
But
4
correct
3
,
and if it does, nd its limit.
1. limit = cos
1. converges with limit =
9
.
=
3
2
10.0 points
(2n + 1)! = 1 2 3 . . . (2n 1)2n(2n + 1) .
when m = 2n + 1. But then,
(2n + 1)!
2n(2n + 1)
4
=
2 (2n 1)!
2
3n
3n
3
as n . Consequently, the given sequence
converges with limit =
013
4
.
3
10.0 points
Determine whether the sequence {an } converges or diverges when
an =
ln(3n4)
,
ln(4n3)
huang (kah2852) HW11 radin (54915)
6
and if it converges, nd the limit.
3. converges with limit = 0 correct
1. converges with limit = 0
3
2. converges with limit =
4
3. converges with limit =
ln 3
ln 4
4. converges with limit =
4
correct
3
4. diverges
5. converges with limit = 4
Explanation:
Since
0 cos2 n 1,
we see that
4
5
an
.
5 + 2n
5 + 2n
5. diverges
Explanation:
By properties of logs,
But
lim
4
ln(3n ) = ln 3 + 4 ln n ,
ln(4n3 ) = ln 4 + 3 ln n .
Thus
n
5
4
= 0 = lim
,
n 5 + 2n
5 + 2n
so the Squeeze Theorem applies and ensures
the sequence {an }
converges with limit = 0 .
an
ln 3
4+
ln 3 + 4 ln n
ln n .
=
=
ln 4
ln 4 + 3 ln n
3+
ln n
On the other hand,
lim
n
ln 3
ln 4
= lim
= 0.
n ln n
ln n
Properties of limits thus ensure that the given
sequence
4
converges with limit =
.
3
014
015
10.0 points
Determine if the limit
lim
n ( n + 7 n 5)
n
exists, and nd its value when it does.
1. limit doesnt exist
2. limit = 1
3. limit = 6 correct
10.0 points
4. limit = 2
Determine whether the sequence {an } converges or diverges when
an =
4 + cos2 n
.
5 + 2n
1. converges with limit =
4
5
2. converges with limit =
2
3
5. limit = 12
Explanation:
By rationalization,
n+7
(n + 7) (n 5)
n5 =
n+7+ n5
=
12
.
n+7+ n5
huang (kah2852) HW11 radin (54915)
On the other hand,
n
=
n+7+ n5
for all n. But
n
1+
1+
7
= lim
n
n
7
+
n
1
5
.
n
1
7
1+ +
n
lim
as n . By the Pinching theorem, therefore, the sequence converges and
limit = 0 .
5
= 1,
n
it thus follows by properties of limits that
n
20
0
n
1
Since
lim
7
5
1
n
exists and has value 2. Consequently, again
by properties of limits, the limit
lim
n ( n + 7 n 5)
017
10.0 points
Determine if the sequence
{6, 0, 6, 0, 0, 6, 0, 0, 0, 6, . . . }
converges and, if it does, nd its limit.
1. limit = 0
n
2. sequence diverges correct
exists and
limit = 6 .
016
10.0 points
Determine if the sequence
22552255
, , , , , , , , ...
13245768
converges and, if it does, nd its limit.
3. limit = 6
4. limit = 5
5. limit = 12
Explanation:
Since a sequence {an }n converges and has
limit L precisely when all the values of an are
as close as we please to L for it all suciently
large n, the
1. limit = 2
sequence diverges
2. limit = 0 correct
because the values of an oscillate from 0 to 6
and back as n .
3. sequence diverges
4. limit =
018
7
2
10.0 points
Determine whether the sequence an converges or diverges when
5. limit = 5
Explanation:
Each term an in the sequence satises the
inequalities
0 < an <
20
n
an = tan1
4n
4n + 3
and if it converges, nd its limit.
1. diverges
,
huang (kah2852) HW11 radin (54915)
2. converges with limit =
3. converges with limit =
4. converges with limit =
5. converges with limit =
6
3
2
correct
4
Explanation:
Since
4 9 14 (5n 1)
(5n)n
=
9 14
5n 1
4
4
5n 5n 5n
5n
5n
for all n 1, we see that the inequalities
Explanation:
Now
0 an
4n
= lim
n
n 4n + 3
4
lim
4+
3
n
=1
On the other hand,
lim f
n
4n
4n + 3
=f
4n
lim
n 4n + 3
limit = tan1 (1) =
019
.
4
10.0 points
Determine whether the sequence {an } converges or diverges when
4 9 14 (5n 1)
,
(5n)n
and if it converges, nd its limit.
1. converges with limit =
1
6
2. converges with limit = 6
3. converges with limit = 5
4. converges with limit = 0 correct
lim
n
limit = 0 .
keywords:
020
10.0 points
Determine if the sequence {an }n converges
and, if it does, nd its limit when
an = n f
n+1
n
and f (x) = 2x2 + 4x + 2.
1. limit = 9
2. limit doesnt exist
3. limit = 10
4. limit = 8 correct
5. limit = 7
1
5
4
= 0.
5n
The Squeeze Theorem thus ensures that the
given sequence converges and has
5. sequence diverges
6. converges with limit =
4
5n
hold for all n 1. But
whenever f is a continous function. Consequently, the sequence an converges and has
an =
8
Explanation:
f (1)
huang (kah2852) HW11 radin (54915)
By the Newtonian quotient,
f (x) = lim
h0
f (x + h) f (x)
.
h
1
Setting h = n , we thus see that
lim n f
1
n
x+
n
For the given limit therefore,
lim an = f (1)
n
where f (x) = 4x + 4. Consequently, the
sequence {an }n converges and
limit = 8 .
021
10.0 points
Determine if the sequence {an } converges,
and if it does, nd its limit when
6n
an =
n
1
dx .
3x+4
1. limit = ln 6
2. the sequence diverges
3. limit =
1
ln 6 correct
3
4. limit =
1
ln 2
3
5. limit =
1
1
ln
3
6
Explanation:
After integration,
an =
1
ln(3 x + 4)
3
=
Now
as n . But ln x is continuous as a function of x. We thus see that {an } converges as
n and has
limit =
= f ( x) .
f ( x)
6n
n
18 n + 4
1
ln
3
3n+ 4
4
18 +
18 n + 4
n 6
=
4
3n+ 4
3+
n
.
9
1
ln 6 .
3