gilbert HW03 (55035)
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1
(ii) while if
0 < ck bk,
Xk ck converges,
then the Comparison Test is inconclusive be-
X
001
10.0 points
If ak , bk , and ck satisfy the inequalities
0 < ak ck bk ,
for all k, what can we say about the series
(A) converges , (B) need not converge .
X
(A) :
cause
bk could converge, but it could diverge - we cant say precisely without further
restrictions on bk.
Consequently, what we can say is
X
k =1
ak,
(B) : k =1
b
k
002 10.0 points
Determine whether the series
if we know that the series
n =1
3 n(n + 1)(n + 5)
p
converges or diverges.
(C) :
X
ck
k =1
is convergent but know nothing else about ak
and bk?
1. (A) diverges , (B) converges
1. series is convergent
2. series is divergent correct
Explanation:
Note first that
2. (A) need not converge , (B) converges
lim
3. (A) converges ,
correct
2
X
(B) need not converge
4. (A) converges , (B) converges
5. (A) converges , (B) diverges
n
3
n
n(n + 1)(n + 5)
Thus by the limit comparison test, the given
series
X
2
n =
3
n(n + 1)( n + 5)
p
converges if and only if the series
X
n
6. (A) diverges , (B) diverges
Explanation:
Lets try applying the Comparison Test:
(i) if
0 < ak
ck,
Xk
c
k
= 1 > 0.
2
=1 n
converges. But by the p-series test with p = 1
(or use the comparison test applied to the harmonic series), this last series diverges. Consequently, the given
converges,
series is divergent
then the Comparison Test applies and says
that
X
ak converges;
003
10.0 points
.
HW03 gilbert (55035)
X
Determine whether the series
k
(k + 2)3
1. A and B only
1. series is divergent
2. A, B, and C
2. series is convergent correct
Explanation:
We use the Limit Comparison Test with
k
=
(k + 2)3
k
b
,
4
converge(s)?
converges or diverges.
ak
3
n = 15
k
k=1
n
(C)
X
2
k
1
=.
k
3
For
k
lim ak = lim
= 1 > 0.
k bk
k k + 2
Thus the series
3. B and C only correct
4. C only
5. B only
Explanation:
(A) Because of the way the nTH term is
defined as a quotient of polynomials in
the series, use of the integral test is
suggested. Set
f (x ) =
X
k
k =1 (k + 2)3
Then f is continuous, positive and
decreasing on [1, ); thus
X
X1
3
1
converges if and only if the improper integral
2
1
3
2n
2
n = 1 3n + 6
k
converges. But this last series is a
geometric series with
3x2 + 6
10.0 points
n =1
(B)
2n
3n 2 + 6
X
n =1
5
n
2
x
dx .
3x2 + 6
1
2
Now after substitution (set u = x ), we
see that
Which of the following series
2
In =
series is convergent .
(A)
dx
converges, which requires us to
evaluate the integral
n
X
x
< 1,
hence convergent. Consequently, the given
series is
004
3x + 6
k=1
|r| =
.
2
k
converges if and only if the series
2x
In
= 3
1
2
ln(3x + 6) n
1
=3
But
1
2
ln(3n
+ 6) ln(3 + 6) .
HW03 gilbert (55035)
Consequently, of the three infinite series,
2
ln(3n + 6)
as n , so the infinite series
only B and C converge
2n
3n 2 + 6
X
n =1
005
diverges.
n
5
X=1
2 n
2 n
2
=
=3
2
1
We apply the Limit Comparison Test with
n
43
=
tan 1 n
,
3
5+n
bn
1
=.
3
n
For
is a geometric series, but the summmation
starts at n = 15 instead of at n = 1, as it
usually does. We can still show, however,
that this series converges.
Indeed,
X
Explanation:
an
X
43
1. series is convergent correct
5
n =15
converges or diverges.
2. series is divergent
2
because the initial term in the series
2
is 5 . (C) The series
3
5+ n
n=1
5
tan1 n
X
is a geometric series whose common
2
ratio is r = 5 . Since 0 < r < 1, the
series converges; in fact, we know that
5
10.0 points
nX=1
.
Determine whether the series
(B) The series
3
n
nlim n
3tan
1
5+ n 3
n
=
nlim tan
1
n=
2.
Thus the given series
X
tan1 n
3
5+n
n=1
n =15
=
4
315
+4
16
3
+ 43
17
is convergent if and only if the series
This shows that the series is a geometric series
3
whose common ratio is r = 4 and whose
3 15
initial term is ( 4 ) . Since 0 < r < 1, the
geometric series converges, and we know that
43
15 +
4
3 16 + 43 17 + . . .
=
43
15
3
14
=4
3
4
+. . . .
15
X
n=1
1
3
n
is convergent. But by the p-series test,
this last series converges because p = 3
> 1. Con-sequently, the given
series is convergent .
.
006
10.0 points
HW03 gilbert (55035)
4
Find all values of r for which the
infinite series
X
(7n
n =1
converges.
1. r >
7
007
6
4n
8
+ 4)
r
Which one of the following series is convergent?
correct
8
1.
2. r > 2
2
2.
2
7
8
3.
n =1
5+ n
X
1
2n
(1) 2
5
+ n
X
8
r
8r6
4
(7n + 4)
n
correct
X
4.
Explanation:
After simplification we see that,
6
4n
4
so
X ( 1)
6 + n
n =1
n =1
3. r <
5. r <
n 1
2
1
4. r >
10.0 points
(7n
=
n
6
8r6
4)r
8 n+
4
7+ n
7
X
5.
8 r ,
4n
X
(7n
8
r> 0
+ 4)r
will converge if and only if the infinite series
X
1
n=1 n
8r6
converges. But by the Integral test we
know that the series
X
1
n=1 n
converges if and only if p > 1.
Consequently, the given series will
converge if and only if 8r 6 > 1, I.E., when
r>
8
n =1
.
n
n1 6
(1)
5
+
+
n
Explanation:
Since
X
n =1
X
35
(1) 2 + n
= n =1
5
2 + n
,
use of the Limit Comparison and p1
series Tests with p = 2 shows that this
series is divergent. Similarly, since
X
2n
n =1 (1)
5
2 + n
=
X
n =1
5
2 + n
,
the same argument shows that this
series as well as
X
p
7
5
+ n
4
6
n =1
3
(1) 2
as n . Thus by the Limit Comparison
test, the infinite series
n =1
2
5+ n
n=1
is divergent.
On the other hand, by the Divergence
Test, the series
n
X
n =1
(1)
n1
6
5
+
+ n
HW03 gilbert (55035)
is divergent because
5
Now
3
6 + n
n 1
5 + n 6 = 0 .
nlim (1)
f (n) = n
n =1
6 + n
+ 6 = f (n + 1)
3
n + 5 = 0 .
nlim f (n) = n
Consequently, by the Alternating Series
Test, the given series
.
To see that this series is convergent, set
1
b
n
lim
n 1
X ( 1)
>
for all n, while
This leaves only the series
+5
3
=
6+
n
n
converges .
.
009
Then
10.0 points
Determine whether the series
(i) bn+1 bn,
lim bn = 0 .
(ii)
2
2
2
2
2
3 4 + 5 6 + 7 ...
n
Consequently, by the Alternating Series
Test, the series
is conditionally convergent, absolutely
con-vergent or divergent.
n 1
X ( 1)
6+
n =1
1. series is absolutely convergent
n
2.
is convergent.
series is conditionally convergent cor-
rect
008 10.0 points
Determine whether the series
X
n =0
n
3
+ 5 cos n
3. series is divergent
Explanation:
In summation notation,
2
2+2
34
with
converges or diverges.
5
2+2
67
... =
( 1)n1f (n) ,
X
n =1
f (x ) =
1. series diverges
2
x +2
.
Now the improper integral
2. series converges correct
Explanation:
n
Since cos n = (1) , the given series
can be rewritten as the alternating series
X
(1)
3
n
n +5
=
n=0
with
f (x) =
X
n=0
3
x +5
1
f (x) dx =
x+2
1
dx
is divergent, so by the Integral Test, the
given
series
is
not
absolutely
convergent. the On other hand,
(1)nf (n)
2
f (n) =
.
2
while
n+2
2
>
n+1+2
lim
n
2
n+ 2
= f (n + 1) ,
= 0.
HW03 gilbert (55035)
Consequently, by the Alternating Series
Test, the given series
6
diverges, and so by the Comparison Test, the
series
e1/n
X
is conditionally convergent
.
keywords: alternating series, Alternating
se-ries test, conditionally convergent,
absolutely convergent, divergent
010
too diverges; in other words, the given
series is not absolutely convergent.
To check for conditional convergence,
con-sider the series
X
10.0 points
where
e1/x
1/n
X
n =1
(1)
n+1
e
n
(1) f (n)
n=1
Determine whether the series
5n
n=1
f (x) =
5x
.
Then f (x) > 0 on (0, ). On the other hand,
5n
is absolutely convergent, conditionally
con-vergent or divergent.
f (x) =
1
3
5x
e1/x
e1/x
5x
2
1+ x
= e1/x
.
3
5x
2. absolutely convergent
Thus f (x) < 0 on (0, ), so f (n) > f (n + 1)
for all n. Finally, since
1/x
= 1,
lim e
3. divergent
we see that f (n) 0 as n . By the
1. conditionally convergent correct
x
Alternating Series Test, therefore, the series
Explanation:
Since
X
n =1
1/n
n+1
(1)
e
5n
X
1X
= 5 n =1 (1)
1/n
n
e
n=1
n
,
we have to decide if the series
X
n
1/n
n
=1 (1)
e
First we check for absolute convergence.
1/n
Now, since e
1 for all n 1,
5n
1
5n > 0 .
But by the p-series test with p = 1, the series
X
1
n=1
5n
is convergent.
Consequently, the given series is
conditionally convergent .
n
is absolutely convergent, conditionally
con-vergent, or divergent.
e1/n
(1)nf (n)
keywords:
011 10.0 points
Determine whether the series
X
n =2
(1)
n
4
n ln n
converges conditionally, converges
absolutely, or diverges.
1. series converges absolutely
HW03 gilbert (55035)
2. series converges conditionally correct
7
Consequently, by the Alternating series
Test, the given series is
3. series diverges
conditionally convergent
Explanation:
The given series can be rewritten as
012
10.0 points
X
n =2
(1)
n
n ln
4
= 4 nX=2 (1)nf (n) ,
n
P
To apply the root test to an infinite series
n an
the value of
= lim (an) 1/n
where
f (x) =
1
n
= (x ln x)
1
> 0 x ln x
has to be determined.
Compute the value of for the series
for all x > 1. On the other hand, by the
Chain and Product Rules,
X
1
for all x > 1.
Thus f (x) is positive and
decreasing for all x > 1. But the improper
integral
2
f ( x) d x =
2
1
x ln x
dx
does not converge. By the Integral Test,
therefore, the given series is not absolutely
convergent. It could still converge conditionally, however. To use the Alternating Series
Test to show that the series
n
(1) f (n)
X
1. =
is convergent, we have to check that
7
6 correct
36
2. = 7
35
3. = 6
4. = 7
5. = 67
Explanation:
After division,
5n + 6
1 + 6 ,
5n
=5
n
n=2
so
(i) f (n) > f (n + 1) for n 2,
1/n
(an)
=
lim f (x) = 0 .
x
Since f is decreasing for all x 2,
however, we see that
n2
=
f (n) > f (n + 1) .
x
1/n
5 1 + 56 n
1 = 0.
x ln x
7
6.
But
6
1/n 1 + 5 n1/n = 1
nlim 5
as n . Consequently,
On the other hand,
lim
5 n n + 6 67 n .
n =1
(ln x + 1) < 0
f (x ) =
(x l n x )2
(ii)
.
=
7
6
.
HW03 gilbert (55035)
8
Consequently, for the given series,
013 10.0 points
= 1.
To apply the ratio test to the infinite series
Xn an, the value
= lim
n+1
an
n
has
014 10.0 points
a
Decide whether the series
Computetobedetermined,fortheseries
X
n =1
1
3n + 7
X
1
n
sin
n =1
.
converges or diverges.
1. converges
1
10
1. =
2. diverges correct
1
2. =
Explanation:
The given series has the form
3
1
3. = 7
4. = 0
X
But then
an =
an ,
2
1 n+
2
n
n2 n .
n =1
n
1
n
sin
1
n
+1
e
lim
.
n
lim sin x = 1 ,
x
x0
1
n
sin n+1
1
n+1
while
lim
n
also. Thus
= nlim
1
n
1
1/n
|an|
nlim
n +
n2
= 2> 1 ,
n
= nlim 1 + n
= 1,
2 n
2
2
=e>2 .
Consequently, the Root Test ensures
that the given series
1
diverges .
=1
sin n
015
n
n + 1 3(n3+n+1)7+ 7
= nlim
2
since
n
so
n+
n
,
in which case
n+1
lim
n
2
Explanation:
By algebra,
1
sin
sin 1
n+1
n+1
1=
1
sin
n
1 n + 2 n
|an|1/n =
5. = 1 correct
But
2
21n n +n 2 n
3n + 7
1 3n + 3 +
7
10.0 points
The terms of a series are specified
recur-sively by the equations
.
a
1
= 4,
a
n+1
9n + 1
=
11n + 7
an .
HW03 gilbert (55035)
Determine whether the series
X
9
Consequently, the Ratio Test ensures
that the given series
an
n=1
converges
.
converges or diverges.
1. converges correct
017
Determine which, if any, of the following
series diverge.
n
(5n)
2. neither converges nor diverges
3. diverges
Explanation:
From the recursive formula we see that
a
9n + 1
=9
(A)
X
n =1
(B)
an
n 11n + 7
11
Consequently, by the Ratio Test the
series converges.
n
5
X
(C)
= lim
n+1
n!
.
lim
10.0 points
(n + 6)
n =1
n
X
n
6n5+n 4
n
45
n
n =1
1. C only
016
10.0 points
2. A only
Decide whether the series
3. none of them
X
n=1
(2n)!
2n
(n!) 5
4. all of them
converges or diverges.
5. B and C
1. converges correct
6. A and C correct
2. diverges
7. B only
Explanation:
The given series has the form
8. A and B
X
n =1
an ,
an =
(2 n )!
(n !) 2 5n
.
But then
a
n+1
a
=
n! n! (2n + 2)!
5 (n + 1)! (n + 1)! (2n)!
2
4n + 6n + 2
(2n + 1)(2n + 2)
=
=
,
2
5(n + 2n + 1)
(n + 1)(n + 1) 5
lim
n
a
n
a
= 4.
5
for each of the given series.
(A) The ratio test is the better one to
use because
a
n+1
n
in which case
Explanation:
To check for divergence we shall use
either the Ratio test or the Root test
which means computing one or other of
1/n
a n+1
nlim a n ,
nlim |an|
an+1
n
=5
n! (n + 1) n+1
(n + 1)!
nn
.
HW03 gilbert (55035)
10
Now
n!
1
(n + 1)! =
n+1
3. B and C
,
4. A only
while
(n + 1)
n
n
Thus
n+1
= (n + 1)
a n+1
a
= 5 n +n 1
n
n
n
n +
n
1
5. C only
.
5e > 1
as n , so series (A) diverges.
6. none of them
7. all of them
8. A and C correct
Explanation:
We compute one of
(B) The root test is the better one to
apply because
1/n 5
|an|
= 0 n + 6
a
lim
n
as n , so series (C) diverges.
Consequently, of the given infinite series,
(A) The ratio test is the better one to use:
a
n+1
=
3 n + 18 n + 3
3
<1
an
8nn+1+38
as n , so series (A) converges.
(B) The ratio test is the better one to apply:
a
n+1
a
only A and C
n
an
for each of the given series.
as n , so series (B) converges.
(C) Again the root test is the better one to
th
apply because of the n powers. For then
5
5n
25
|an|1/n =
>1
4 6n + 4
24
1/n
lim (an)
,
n+1
n
=
(
47 nn
+ 1 n
+ 7)
4)(
+ 1 + 7)
(
diverge.
4)( n
7
as n ,
4
so series (B) diverges.
018
10.0 points
Decide which of the following series
con-verge.
8
n
n
(A) X n + 3 83
n =1
(B)
n
X
n +
n
74 47
n =1
(C)
X
n =1
1. A and B
2. B only
4
nn3 ++ 87
n
(C) The root test is the better one to apply:
4n + 7
(an)1/n =
,
3
0
n +8
as n , so series (C) converges.
Consequently, of the given infinite series,
only A and C converge
.