gilbert HW02 (55035)
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1
assuming that the pattern of the first few
terms continues.
2 n
2.
001
3
n
10.0 points
Find a formula for the general term an
of the sequence
o
=
a
{an} n=1 =
2, 6, 10, 14, . . .
3.
3n
=
a
2
n
3 . a n = 2
,
assuming that the pattern of the first few
terms continues.
3
4 . a n = 3
2
1. an = n + 3
6.
2. an = 4n 2 correct
3. an = 5n 3
7.
4. an = 3n 1
5. an = n + 4
Explanation:
By inspection, consecutive terms
an1 an in the sequence
3
=
n1
n1
correct
4
n
3n
=
a
4
n
Explanation:
By inspection, consecutive terms an1
and an in the sequence
and
o
{an} n=1 =
a
n1
{an} n=1 =
2
1,
3,
have the property that
2, 6, 10, 14, . . .
have the property that
a
= ra
n1
n
an an1 = d = 4 .
4
8
, 27 , . . . o
9
=
2
3
a
n1
Thus
Thus
2
an = ran1 = r an2 = . . .
an = an1 + d = an2 + 2d = . . .
= a1 + (n 1)d = 2 + 4(n 1) .
Consequently,
a n = 4n 2
=r
n 1
a1
2n1
a1 .
= 3
Consequently,
.
n1
an = 32
keywords:
002
since a1 = 1.
10.0 points
Find a formula for the general term a n of
the sequence
{an}n =1 =
2
1 , 3 ,
9
4
8
, 27 , . . .
o,
keywords: sequence, common ratio
003 10.0 points
.
HW02 gilbert (55035)
2
5
Compute the value of
lim
n
when
lim
lim
n
n
1. limit =
12
3. converges with limit =
57
4. converges with limit =
32
= 2.
b
n
5 correct
5. sequence does not converge
Explanation:
After division by n we see that
n
5 + 1)
5
2. limit = 2
3. limit doesnt exist
(
n
an =
4+n
12
4. limit = 5
But
5
5. limit = 2
n
(1) ,
.
3
3
0
n
n
5
as n , so an 4 as n . Consequently, the sequence converges and has
Explanation:
By properties of limits
limit = 5
4
lim 4anbn
n2
= 4 lim
a
n
correct
4
3an bn
a = 6,
n
2. converges with limit =
4a n b n
lim
n
= 48
b
n
.
n
005
10.0 points
while
(3a b )
lim
n
=3
Determine if the sequence {a n} converges
2
n (2n 1)! ,
when an =
n
n
5
n
lim
a lim
n
n
b
n
= 20 6= 0.
Thus, by properties of limits again,
lim
n
4anbn
3an bn
=
1. converges with limit =
12
5
(2n + 1)!
and if it converges, find the limit.
5
4
.
2. converges with limit =
004
25
3. converges with limit =
5
10.0 points
Determine if the sequence {an} converges,
and if it does, find its limit when
n
an = 5n + (1) .
4n + 3
1. converges with limit = 1
2
4. converges with limit = 4
5
5. does not converge
Explanation:
correct
HW02 gilbert (55035)
3
But
By definition, m! is the product
ln 2
m! = 1 2 3 . . . m
of the first m positive integers. When m
= 2n 1, therefore,
(2n 1)! = 1 2 3 . . . (2n 1) ,
ln(n)
+ 1,
ln 5
ln(n)
+1
1
as n . Consequently, the given
sequence converges and has
while
limit = 1
.
(2n + 1)! = 1 2 3 . . . (2n 1)2n(2n + 1) .
007
when m = 2n + 1. But then,
2
2
5n (2n 1)! =
5n
(2n + 1)!
Determine if the sequence
5
2n(2n + 1)
{2, 0, 2, 0, 0, 2, 0, 0, 0, 2, . . . }
4
as n . Consequently, the given sequence
converges with limit =
006
5
4
10.0 points
.
10.0 points
Determine whether the sequence {an}
cov-erges or diverges when
converges and, if it does, find its limit.
1. limit = 4
2. limit = 0
3. sequence diverges correct
4. limit = 1
ln (2n)
an =
5. limit = 2
ln (5n) ,
and if it converges, find its limit.
1. converges with limit = 1 correct
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
2. sequence diverges
sequence diverges
3. converges with limit = 0
4. converges with limit
=
ln 2
ln 5
because the values of an oscillate from 0
to 2 and back as n .
008
5. converges with limit
= 25
Rewrite the finite sum
Explanation:
Since
ln (2n)
ln (5n)
10.0 points
9
12
6
+
+
2+4
3+4
4+4
= ln 2 + ln(n) =
ln 5 + ln(n)
ln 2
+1
ln(n)
.
ln 5
ln(n) + 1
21
+ 15 + . . . +
5+4
7+4
using summation notation.
HW02 gilbert (55035)
5
X
1. k
=0
3k
k2
(1)
k+4
8
X
2. k
=3
1. convergent with sum
2. convergent with sum
3k
k2
(1)
k+4
X
5
16
16
correct
5
3. convergent with sum
5
3.
4
3k
16
3
3
4. convergent with sum 16
k =0 k + 4
7
5. divergent
X
4. k =2 k 3+k 4 correct
Explanation:
The infinite series
8
X
5. k =3 k 3+k 4
7
X
6. k
=2
1
k2
(1)
k+4
The numerators form a sequence
9,
12 ,
1
X
a rn1
n=1
a
2+4, 3+4, 4+4, 5+4, ... , 7+4.
Thus the general term in the series is of the
form
(i) converges with sum 1 r when |r|
< 1, (ii) and diverges when |r| 1 .
Consequently, the given series is
3k
k +4
k
where the sum ranges from k = 2 to k =
7. Consequently, the series becomes
16
convergent with sum
7
X
3k
010
k+ 4
10.0 points
Determine whether the series
k=2
in summation notation.
X
n =1
009
10.0 points
Determine whether the infinite series
41+
1
rn1
15 , . . . , 21 ,
while the denominators form a sequence
a=
X
is an infinite geometric series with a = 4
1
and r = 4 .
But a geometric series
Explanation:
6,
1
4 1 + 4 16 + 64 = n =1 a
3k
1
4 16 + 64
1
is convergent or divergent, and if
convergent, find its sum.
2
5n +n 4
is convergent or divergent, and if
convergent, find its sum.
1. convergent with sum =
2. divergent correct
9
2
5
.
HW02 gilbert (55035)
5
25
5. converges with sum
7
= 12
4. convergent with sum = 52
6. converges with sum
= 23
3. convergent with sum =
2
Explanation:
An infinite geometric series
5. convergent with sum = 9
r<
Explanation:
X
an is divergent when
The infinite series
(ii) diverges when |r| 1 .
n
Now
Note for the given series,
2n
a
=
n
2
5n + 4
=
5+n
4,
so
lim
n
sum = 1 r ,
while it
lim an 6= 0 .
a = lim
n
n
2n
X
k =1
1
5
k
X
1 k1
= k =1 5 5
1
1
is a geometric series with a = r = 5 < 1.
Thus it converges with
2
= 6= 0 .
5n + 4 5
Thus the given series is
divergent .
011
10.0 points
sum =
while
X
1
X
2k
= k =1
k =1 5 k
2 k1
5 52
2
is a geometric series with a = r = 5 < 1.
Thus it too converges, and it has
sum =
2
k =1
3.
Consequently, being the sum of two
conver-gent series, the given series
1 + 2k
5k
converges or diverges, and if it
converges, find its sum.
converges with sum =
1. series diverges
2.
4,
Determine if the series
X
n1
n =1 a r
(i) converges when | | 1 and has
a
n=1
lim an exists but
n
P
1
4+
2
3
11
= 12
012 10.0 points
11
12 correct
converges with sum =
Find the sum of the infinite series
X
2k
k =1 (cos ) , (0 < 2) ,
3
3. converges with sum = 4
5
4. converges with sum = 6
whenever the series converges.
2
2
1. sum = sin cos
.
HW02 gilbert (55035)
2
2. sum = csc
2
3. sum = tan
014 (part 2 of 3) 10.0 points
(ii) is What an for n > 1?
4
a
2
4. sum = cot correct
a
Explanation:
For general the series
2.
k=1
(cos2 )k
a
4.
is an infinite geometric series with common
2
ratio cos . Since the series starts at k = 1,
its sum is thus given by
cos 2
2
1 cos
Consequently
cos 2
2.
= sin
P
a
5n 1
=
n
1. a1
= S n S n1 .
n1
,
5(n + 1) 6
=
n +1
n +1
6
= 5n+
1.
=5
4. a1
5
Sn =
= 2 correct
3. a1
correct
Consequently,
= 2
2. a1
n(n + 1)
n
But
is
n +1
(i) what is a1?
6
=
an
n=1
S
a
n2
Since Sn = a1 + a2 + + an , we see
that
.
n
= n(n + 1)
4
=
n
Explanation:
013 (part 1 of 3) 10.0 points
partial sum of
4
5 . an
6.
2
sum = cot
th
= n(n 1)
6
=2
n
n
3. a n
X
= n(n 1)
6
n
1.
2
5. sum = sec
If the n
6
=3
5. a1
an =
6
n
6
=
n +1
6
n(n + 1)
for all n > 1.
015 (part 3 of 3) 10.0 points
P
(iii) What is the sum
1. sum = 5 correct
= 3
2. sum = 6
Explanation:
Since a1 = S1,
a1
=2 .
3. sum = 4
n =1
an?
4. sum = 3
moreno (am49869) HW02 gilbert (55035)
series
X
5.sum=2
Explanation:
By definition
a,
n
n =3
sum = lim Sn =
n
5 n 1 .
lim
an = f (n) .
Clearly from this figure we see that
a3 = f (3)
n+1
n
7
<
3
2
f (x) dx,
= f (4)
<
4
3
f (x) dx ,
Let f be a continuous, positive,
decreasing function on [2, ). Compare
the values of the series
a5 = f (5)
<
5
4
f (x) dx,
17
a6 = f (6)
<
6
5
f (x) dx ,
Thus
a
sum = 5 .
016
4
10.0 points
X
A=
while
f (n)
n=3
and so on. Consequently,
and the integral
17
B=
A<B
f (x) dx .
.
2
1. A = B
keywords:
2. A < B correct
017
10.0 points
Determine whether the series
3. A > B
Explanation:
X
In the figure
k=1
4
k ln(5k)
is convergent or divergent.
1. series diverges correct
2. series converges
Explanation:
The function
a
a
3
2
a
4
3
a
5
4
f (x ) =
...
6
5
6
the bold line is the graph of f on [2, ) and
the areas of the rectangles the terms in the
4
x ln(5x)
is continous, positive and decreasing on
[1, ). By the Integral Test, therefore, the
series
4
X
k =1
k ln(5k)
moreno (am49869) HW02 gilbert (55035)
converges if and only if the improper integral
4
f (x) dx =
1
x ln(5x)
1
2. r =
exists. To evaluate this last integral, set
u = ln(5x). Then
x
1
2.47474747 .
1. r =
lim dx t 1 x ln(5x)
du =
Use this idea to find the rational number
r whose repeating decimal expansion is
dx
converges, I.E., if and only if
t
4
3. r =
dx ,
4. r =
in which case,
4 dx =
t
1
x ln(5 x)
4
LN(5t)
LN 5
u
5. r =
du
= 4 ln(ln(5t)) ln(ln 5))o .
But
Consequently,
4
lim
t
dx
1
245
250
250
49
101
99 correct
101
99
20
Explanation:
The decimal expansion
2.47474747
x ln(5x)
does not exist. The Integral Test thus
ensures that the given
series diverges
18
247
can be interpreted as the infinite
geometric series
47
47
47
+
+
+ ...
2+
2
4
6
10
10
10
lim ln(ln(5t)) = .
t
t
8
= 2+
245
r=
019
10.0 points
99 .
10.0 points
Which of the following series are convergent:
x = 5.78787878
can be interpreted as the infinite
geometric series
78
78
78
x = 5 ++
++ . . . .
2
4
6
10
10
10
By summing this infinite geometric series we
thus get the rational number having the given
repeating decimal expansion.
47 X 1
2 k =0 10 2 k.
Consequently,
.
A repeating decimal expansion such as
10
X
A.
2
n
2 /3
n =1
B.
X
3
n =1 n + 1
1
C.
1
1
1
1+ + +2 + +...
2
3
5
moreno (am49869) HW02 gilbert (55035)
9
is divergent.
1. A only
keywords: convergent, Integral test,
2. A and B only
020
3. all of them
10.0 points
Determine the convergence or
divergence of the series
4. C only
(A)
5. none of them correct
1+
1
1
1
1
4 + 9 + 16 + 25 + ... ,
and
6. B only
X
7. B and C only
8. A and C only
m =1
(B)
2
mem .
1. A convergent, B divergent
Explanation:
By the Integral test, if f (x) is a positive,
decreasing function, then the infinite series
2. both series convergent correct
X
f (n)
3. A divergent, B convergent
n=1
converges if and only if the improper integral
4. both series divergent
f (x) dx
1
converges. Thus for the three given series
we have to use an appropriate choice of f .
2
A. Use f (x) = x2/3 . Then
f (x) dx
1
Explanation:
(A) The given series has the form
1
1
1
1
X
1 + 4 + 9 + 16 + 25 + ... = n=1
1
n
(B) The given series has the form
X
f (m)
Then
f (x) dx
m=1
with f defined by
1
is divergent (log integral).
1
C. Use f (x) = x1/2 . Then
f (x) dx
1
.
This is a p-series with p = 2 > 1, so the
series converges.
is divergent.
3
B. Use f (x) = x + 1.
2
f (x) = xe
x2
.
Note first that f is continuous and
positive on [1, ); in addition, since
x
2
f (x) = e 2 (1 2x ) < 0
moreno (am49869) HW02 gilbert (55035)
for x > 1, f is decreasing on [1, ). Thus
we can use the Integral Test. Now, by
substitu-tion,
12
x dx =
2 ex
xe
2
t
1
and so
xex dx
2
1
=
converges if and only if the infinite series
X
1
2 e.
n=1
converges.
In the given example
1f
(x) = xp ,
Since the integral converges, the series
con-verges.
021
10.0 points
If the improper integral
1
(A)
(B)
(C)
(D)
(E)
n
X
n
1
p
f (x) = xp +1 ,
dx
converges, which of the following
statements is (are) always true?
X
which is a function both continuous and
pos-itive on [1, ). It will also be decreasing
on [1, ) if f (x) < 0 for all x > 1. But
xp
1
f (n)
t ,
1
10
so f will be decreasing provided p > 0.
On the other hand, the improper integral
p converges;
1
np +1 diverges;
1
X
n np 1 converges;
1
X
n np 1 diverges;
1
X
n np +1 converges.
n
1
x
1
dx
p
converges if and only if
1
n
nlim
x
1
p
dx .
exists. But
n
1
x
1
1 n
p dx = px p1
=
p
1
1
1 n p 1
.
1
Consequently, the improper integral
1. A, D and E
2. A and E only correct
x
1
3. B and D only
4. A, C and E only
1
dx
p
converges if and only if p > 1. Hence by
the Integral test, the infinite series
X
5. A only
1
p
n=1 x
Explanation:
To apply the Integral test we need to start
with a function f which is positive, continuous
and decreasing on [1, ). Then the integral test
says that the improper integral
f (x) dx
1
converges if and only if p > 1.
Now we can check which of the
statements is (are) always true.
(A) This is always true because of the
Inte-gral test.
moreno (am49869) HW02 gilbert (55035)
(B), (E) Since Pn >1 1 = p + 1 > 1, the
pn
Integral test ensures that
converges.
P+1
Thus (B) is false and (E) is true.
P
1
(C), (D) The series
nn
converges if
1
P
and only if p 1 > 1, I.E., when p > 2. Since
the convergence of the improper integral
1
1
xp
dx
guarantees only that p > 1, we see that
state-ments (C) and (D) are true for some
values of p and false for others.
Consequently, of the statements,
only A and E
are always true.
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11