Math 181 Exam II Practice Solutions
1)
This is a Maclaurin series of degree 2: use n=2. Use the Taylor’s remainder formula
M
M
n 1
n  `1
xa 
max x - a . You must find M and max x  a 1) M  max f ' ' ' ( x) when
(n  1)!
(n  1)!
0.4  x  0.5 . Since f ' ' ' ( x)  e x is an increasing function, its maximum is attained at the right end
0.5
point of the interval x = 0.5. M  e
2) By looking at the expansion, this is a Mclaurin series, so a=0.
e 0.5
The maximum distance from 0 is 0.5. Thus max x  a =0.5. Therefore, R2 
(0.5) 3
(3)!
Rn 
2)
You cannot use the comparison test since its terms are not all positive. Take the absolute value and


observe that
n2
holds as
(1) n sin n 
1
 2

2
(n  4)
n  2 ( n  4)
n2  4  n2 
1
n2 
2
n
2

2
Note that the second to the last inequality
n2
n2
for n  2 . Therefore, the series is absolutely convergent by the comparison
2
test. The series is convergent since absolute convergent implies convergent.
3)


1
1
1
1
1

( 
) . This is a telescoping series with sn  
. Thus the sum of the series


n 1
2 n 1
n  2 n( n  1)
n2 n
1
1
1
is lim sn  lim( 
)
n 
n  2
n 1 2
4)
Apply the ratio test and set the limit less than 1:
an 1
lim
n 
an
2n 1 x n 1
2n  1
2n 1

1

lim
n n
n  2 n
2(n  1)  1 2 x
 lim
n 
the continuous function theorem,
lim
n 
2n  1
2n  3
2n  1 x n 1
1
1 2 x 1 x 
(Note that using
n
2
2(n  1)  1 x
 lim
n 
2n  1
 1 ). Next test the end points. At one end
2n  3
1
2n ( )n

2 x
1
1
1
2 
point x  , 
is divergent by the comparison test with 
. At


2 n  2 2n  1 n  2 2 n  1 n  2 2 n  1
n
1 n
n
 2 ( )

(1) n
1  2n x n
2 
the other end point x   , 
is convergent by the alternating


2 n  2 2 n  1 n  2 2n  1 n  2 2 n  1
1 1
series test. Thus the interval of convergence is [ , )
2 2

n
n

5)
A) Note that for large values of n, ln n  n
power function). In particular, ln n  n


n2
3
n ln n
n3  1


n2
1
3
n n
1
10
n3
n 
2
3

 2
n2
1
n
32
30
1
10

for any power of n (log function grows slower than any
for large n. Using the comparison test,
1
is convergent. (note that the power
1
of n in ln n  n 10 was
10
chosen so that the last exponent in the denominator would be more than 1) B) Apply the comparison
test

n ln n

n3  1
n2

n
n3

1
(note that since ln n  1 for large n , it can be dropped from the
n
numerator)
a)
Use the formula

k 
(1  x) k     x n
n 0  n 
.
We write the function in the standard form before applying the formula.
1
1
1
  
x 2 3
x

3
f ( x)  16  x  16(1  ( ) )  16 x  3  (( ) 2 ) n 
  4
4
n 0  
n 
1  1
1
1
2
  x 2n
3
16   3 ( 2 n )  16 3 (1 
x2 
x 4  ...)
4
3(16)
9(4)
n 0   4
n 
3
2
3
Next use the alternating series estimate.
16
1
3

0.3
0
1
1 2
1
1
1
(1 
x 
x 4  ..)dx  16 3 (0.3 
(0.3) 3 
(0.3) 5 ...) . The first term with its value
48
1152
144
5760
less than
6) Use the formula that
10 4 is
1
1
0.3
1
1 2
1
1
(0.3) 3 . Thus 16 3  (1 
x 
x 4  ..)dx  16 3 (0.3 
(0.3) 3
0
5760
48
1152
144
c0  a0b0 , c1  a0b1  a1b0 , c2  a0b2  a1b1  a2b0 , c3  a0b3  a1b2  a2b1  a3b0
4 x 2 8 x3 16 x 4
(e )(cos 2 x)  (1  2 x 



2
6
24
4 x 2 16 x 4
)(1 


2
24
2x
16 x3 2
2
)  1 2x 
 (  4  ) x4
6
3
3
7)
a)
Divergent by the test of divergence: Using L’hopital, lim (1 
n 
2 n
)  e 2  0
n
4n
2 n
3n  4n
b) Use the comparison test:  n
behaves like n  ( ) , which is a convergent geometric
n
6
3
n 1 5  6



3n  4n  4n  4n
2
2


2
2
series.  n
and



 
  is convergent by the Geo test. Thus
n
n
6
n 1 5  6
n 1
n 1  3 
n 1  3 

n
n
3n  4n
is convergent by the comparison test.

n
n
n 1 5  6

8)
a)
1 1
1
1
1
S5  



 3.2317
1
2
3
4
5
b) S5 


6
 1
1
dx  S  S5  
dx . Since
5
x
x


6
 1
1
2
2
1
1
dx 
,
dx 
, , 3.2317 
 S  3.2317 
5
x
6
x
5
6
5
9) Let y 

a
n 0

n
x n . Then y '   a n nx n 1 The DE y 'xy  0  0 becomes
n 1




n 1
n 0
n 1
n 0
 an nx n1  x an x n  0  an nx n1   an x n1  0 Next let k = exponent: for the first term, k = n-1
(thus n= k+1), while k = n+1 the second terms (thus n = k-1)




n 0
k 0
k 1
 an nx n1   an x n1  0   ak 1 (k  1) x k   ak 1 x k  0 . For k =0, we have a1  0 For
n 1

1
a k 1 . Since the recursive relationship has a jump of 2 and
k 1
k 1
1
1
, all the odd terms are 0. For even terms, observe that a 2 n 
a 2 n4 , …
a 2 n2 , a2 n2 
2n
(2n  2)
1
1
1 1
1
1
1


a0  n
a0 . Since y0  1, a0  1 Thus
a4 
a2 , a2 
a 0 Thus a 2 n 
(2n) (2n  2)
(4) (2)
2 (n!)
(4)
(2)
k  1,  (a k 1 (k  1)  a k 1 ) x k  0  a k 1 

1 2n
x
n 1 2 n!
y  1 
n
10) a) To find the power series representation of
x2
, first find the power series representation
(1  x 2 ) 2
1
1
x2
2
2
x
x
,
then
replace
by
to
find
.
Finally
multiply
by
to
find
.
x
(1  x 2 ) 2
(1  x ) 2
(1  x 2 ) 2



1
d
1
d  n
1
n 1
2 n 1
.
Next
 (
)   x   nx
  n( x )   nx 2 n2 . Finally,
2
2 2
dx 1  x
dx n0
(1  x)
(1  x )
n 1
n 1
n 1
of


x2
2
2 n2

x
nx

nx 2 n b) To find the power series reorientation of ln( 1  x 2 ) , first find the


2 2
(1  x )
n 1
n 1
power series representation of ln( 1  x ) , then replace
x
by x
2
to find
ln( 1  x 2 ) .



1
x n1
dx    (1) n x n dx   (1) n  x n dx  (1) n
 c . To find the constant, let x=0.
1 x
n 1
n 0
n 0
n 0

0 n1
ln( 1  0)   (1) n
 c  0  c Thus the constant is 0 . Therefore,
n 1
n 0
2 n 1
2n2


2
n (x )
n x
ln( 1  x )   (1)
  (1)
n 1
n 1
n 0
n 0
1
2
6
24
11) f ( x)  2  f " ( x)   3 , f " ( x)  4 , f ' ' ' ( x)   5 . Evaluate each function at a= 1 and apply the
x
x
x
x
2
formula, we get the second degree Taylor polynomial 1  2( x  1)  3( x  1) . Use the Taylor’s
M
M
n 1
n  `1
remainder formula Rn 
xa 
max x - a . You must find M and max x  a , the
(n  1)!
(n  1)!
maximum distance from the center of expansion . 1) M  max f ' ' ' ( x) when 0.9  x  1.3 . Since
ln( 1  x)  
f ' ' ' ( x) 
24
is a decreasing function, its maximum is attained at the left end point x = 0.9. Thus
x5
24
. 2) max x  a , the maximum distance from the center of expansion is 0.3. Thus since
0.9 5
24
0.9 5 (0.3) 3
this is a second degree polynomial, R2 
(3)!
M 
12) Use the Taylor’s remainder formula
Rn 
M
M
n 1
n  `1
xa 
max x - a . Here n = 1 since the
(n  1)!
(n  1)!
approximation used is the first degree. You must find M and max
from the center of expansion. 1)
xa
M  max f ' ' ( x) when  0.1  x  0.2 .
, the maximum distance
Since
f """ ( x)  e x is an
increasing function, its maximum is attained at the right end point x = 0.2. Thus M  e
xa
the max
R2 
: since a = 0, max
xa
=0.2
0 .2
. 2) To find
Thus since this is a second degree polynomial,
e 0.2
(0.2) 2
(2)!

2
1
1

(

) . This is a telescoping series with


n 1
n  2 ( n  1)( n  1)
n2 n  1
1 1
1
1 1
1
3
. Thus the sum of the series is lim s n  lim (1   
sn  1   
)
n 
n 
2 n n 1
2 n n 1
2
n
n
n
n



3 2
3
2
14) 
is a sum of two convergent geometric series  n and  n , thus convergent.
n
5
n2
n 2 5
n 2 5
3
5
3
5
x
x
0.1 0.1
15) sin x  x 
is an alternating series. Using the alternating series

   (0.1) 

3! 5!
3!
5!
13) First use partial fractions

error estimate, find the first term whose absolute value less than 0.001 by trial and error. It can be
shown that
x3
0.15
is the first term less than 0.001. Thus sin x  x 
has an error less than 0.001
3!
5!
when x = 0.1 Thus the degree is 3.

(1) n  2 n1
(1)n  2 n1
16) 
is the McLaurin series for sin x with x   . 
 sin   0
n0 (2n  1)!
n 0 ( 2n  1)!


(1) n 2 2 n
(1) n 2 2 n
2
2
17)  2 n
is the expansion of cos x with x 
Thus  2 n
 cos
3
3
n  0 3 ( 2n)!
n  0 3 ( 2n)!

18) tan
1
x
find C, let x=0.
19)
1
1
dx  
)dx  
2
1 x
(1  ( x 2 )
0 2 n 1
tan 0  
c c  0.
n  0 2n  1
1
3  3n1
sn   3 
1 3
k 1
n
k

 ( x
n 0


) dx   ( x ) dx  (1)
2 n
2 n
n 0
n 0
n

x 2 n1
x dx  
c .
n  0 2n  1

2n
To
x 2 n 1
Therefore, tan x  
n  0 2n  1
n 1
sn 3
3  3n 1 3  lim
n 
  , divergent.
and lim sn  lim
n 
n  1  3
1 3

1

20) The center of expansion is at a=5. Since power series is convergent at x=8, the radius of
convergence is at least 3. x = 4 is 1 unit away from the center of expansion a = 5, it is in an interior
of the interval of convergent. Thus it is absolutely convergent.
21) Apply the ratio test and set the limit less than 1:
an1
2n ( x  1)n1 (n  1)
x 1
lim
 lim n1
1
 1  x  1  2  1  x  3 At one end point x  1 ,
n
n a
n 2
( x  1) (n  2)
2
n


(1) n (2) n
(2) n
1


is divergent by the comparison test with



n
n 1
(n  1) n 2 n  1
n  2 2 ( n  1)
n2 2

1
 2n .
). At the other

(1) n (2) n
(1) n

is convergent by the alternating series test. Thus the interval


n
n  2 2 ( n  1)
n  2 ( n  1)
of convergence is (1,3]
end point
x  3,

22) Conditional:


1

k2 k
k 2
1

is divergent, but
k2
k 2


k 2
(1) k
k2  k
is convergent by the alternate series
test
23)
(3x) 2 (3x)3


2
6
x3
6
9
3
)  1  (3x  x)  (3x 2  x 2 )  1  2 x  x 2  ..
2
2
2
24) a) First find the power series expansion of ln( 1  x) , then replace x by x . Finally integrate term
e3 x sin x  (1  3x 
)(1  x 
by term and find the first term with value less than 0.001 by trial and error.
ln( 1  x)  
1
dx  
1 x
constant is 0. Then



 (1) n x n   (1) n  x n dx   (1) n
n 0
n 0
n 0

ln( 1  x 2 )   (1) n
n 0
0. 1
2
 ln( 1  x )dx  
0. 1 
 (1) n
0
n 0
0

  (1) n
n 0
x n1
 c . Since at x = 0, ln( 1  x)  0 , the
n 1
2n 2
x
. Finally,
n 1
2n2

x
dx   (1) n
n 1
n 0

x 2n2
x 2 n 3
n
dx

(

1
)
0 n  1 
(n  1)( 2n  3)
n 0
0.1
0.1
0
2 n 3
(0.1)
1
1
 (0.1) 3  (0.1) 5  
(n  1)( 2n  3) 3
10
The second term is less than o.oooo1. Thus you need only to take the first term by the alternating
series remainder theorem .
replace

0.1
0
x
by  x . We have e
2
e  x dx  
2
 x2
n

(1) 2 x 2
(1) n



n!
n!
n 0
n 0
0.1 
0
(0.1) 3
x
b) First find the power series expansion of e , then
3
n

( x 2 ) n  (1) n x 2


Thus
n!
n!
n 0
n 0
ln( 1  x 2 ) 

0.1
0
(1) n x 2 n 1
n  0 n!( 2n  1)

x 2 n dx 
0.1
0
(1) n (0.1) 2 n 1
(0.1) 3 (0.1) 5

(
0
.
1
)




n!(2n  1)
3
10
n 0

The third term is less than o.oooo1. Thus you need only to take the first two terms by the
alternating series remainder theorem .
25)

0.1
0
e  x dx  (0.1) 
2
(0.1) 3
3
1
1
2
, f " ( x)  
, f ' ' ' ( x) 
. Evaluate each function at a= 3 and
2
1 x
(1  x)
(1  x) 3
1
1
apply the formula, we get the second degree Taylor polynomial P2 ( x)  ln 4  ( x  3) 
( x  3) 2 . The
4
32
M
M
n 1
maximum Use the Taylor’s remainder formula Rn 
xa 
(max x - a ) n 1 . 1)
(n  1)!
(n  1)!
2
M  max f ' ' ' ( x) when x  3  0.1. Since f ' ' ' ( x) 
is a decreasing function, its maximum is
(1  x) 3
2
attained at the left end point x = 2.9. Thus M 
. 2) the maximum distance from the center of
3 .9 3
2
3.9 3 (0.1) 3
expansion is 0.1 Thus since this is a second degree polynomial, R2 
(3)!
f ( x)  ln( 1  x)  f " ( x) 
26) They are sequences, not series.

tan 1 n 2
a) Use the squeeze theorem: 0 
 . Since the left and right approach zero, the middle
n
n
sequence approaches limit 0 b)
lim ( n 2  n  n)  lim
n 
n 
n2  n  n n2  n  n
n
 lim
 lim
1
n 2  n  n n  n 2  n  n n 
n
n
n
 (1) n n  n and
n
3
3
3

lim
n 
1
1
1
 1
n

1
c)
2
n
1
 lim n
 0 . By the squeeze theorem, the sequence converges to
n
n


3
3 ln 3
0.
27)
(1) k ln k
is not absolutely convergent: since ln x  x for all x>1, ln(ln k )  ln k . Since

k 3 k ln(ln k ))





ln k
ln k
1
(1) k ln k
(1) k ln k

 , 
is divergent by the comparison test. 
is

k 3 k ln(ln k ))
k 3 k ln k
n 3 k
k 3 k ln(ln k ))
k 3 k ln(ln k ))

convergent by the alternating series test: since a) it is decreasing as
1
1 1
 k ln(ln k )  ln k[ln(ln k )  k
]
d
ln k
ln k k  0 and b) 0  ln k  ln k , since both the left and
(
) k
k ln(ln k )
k
dk k ln(ln k )
k ln(ln k )2
the right go to zero, the middle sequence goes to 0 by the squeeze theorems.
28)

1
a) 
is divergent and
n 1 n

1
is divergent and

n 1 n
b)

1

1
n 
n
is
divergent.
But
is convergent.



2
n 1
n 1 n
n 1 n

1
( ) is divergent, but

n
n 1
1
is convergent but

2
n 1 n
b)




n 1
n

1
1
 ( n  ( n ))   0
is convergent.
n 1
1
1
is divergent since using L’Hopital, lim n 2  1
2
n
n
n
29)
a)

2
 (1  k )
k
is divergence by the test of divergence. (use L’hopital to show that the limit of the
k 1
sequence is e
b)
2
1  cos k
: Use the theorem that the absolute convergence is convergence:

k2
k 1


1  cos k
2



2
2
k
k 1
k 2 k

. Thus by the comparison test, the series is absolutely convergent, which implies the original series
is convergent. (you need to take the absolute value in order to use the comparison test since it is
applicable only to the series with positive terms)
c)

k
k 3
2
k ln k

is convergent: Note that for large values of n, ln n  n for any power of n (log
 2k  3
1
function grows slower than any power function). In particular, ln n  n 10 for large n. Using the
1
1



k ln k
k 2 k 10
1
comparison test,  2
  2   13 is convergent. (note that the power of n in
k
k  3 k  2k  3
k 3
k  2 10
k
1
10
ln n  n was chosen so that the last exponent in the denominator would be more than 1)

( k!) 2
d) 
is convergent by the ratio test :
k 1 ( 2k )!
an1
(k  1)!(k  1)! (2k )!
(k  1)( k  1)
1
lim
 lim

 lim

n a
n
(2k  2)!
(k!)( k!) n (2k  2)( 2k  1) 4
n
e)
2k  k

k
k 1 3  k

is convergent by the comparison test : :
since it is a geometric series with common ratio
f)


k 1
e k
k 1
g)

1
 tan( n
k 1
2

2k
is convergent

k
k 1 3
2
.
3
is convergent by the comparison test:


k 1
is a geometric series with a common ratio
2k  k  2k
  k and

k
k 1 3  k
k 1 3

e k
k 1

 e
k
and
k 1

e
k
is convergent since it
k 1
1
which is less than 1.
e
) is convergent: Apply the limit comparison test with bn 
1
:
n2
1
2
1
sec 2 ( 2 )(  3 )
tan( ) 2
a
n  lim
n
n 1
lim n  lim
n  b
n 
n 
1
2
n
( 3 )
2
n
n
30)
a) Apply the ratio test and set the limit less than 1:
lim
n
an1
an
( x  1) 2 n 2 32 n1
( x  1) 2

1

 1  x  1  3  2  x  4 At one end point x  2 ,
n
9
32 n1 ( x  1) 2 n
 lim


(3) 2 n
(1) 2 n (3) 2 n


3 is divergent by the divergence test. At the other end point x  4 ,



2 n 1
32 n 31
n 1 3
n 1
n2


(3) 2 n

3 is divergent by the divergence test. Thus the interval of convergence is (2,4) b) )


2 n 1
n2 3 3
n2

Apply the ratio test and set the limit less than 1:
lim
n
an1
an
( x  1) n1 (3)(5) (2n  1)( 2n  3) (4)(9) (5n  1)
1
n ( x  1) n (4)(9) (5n  1)(5n  4) (3)(5) (2n  1)
 lim
( x  1)( 2n  3)
2
5
 1  ( x  1)  1  x  1 
n 
5n  4
5
2
5
Therefore, the radius of convergence is
2
lim
31) No. Use the test of divergence to show the sequence converges to 1.
32) First, it appears that the sequence is decreasing since factorial grows faster than polynomial.
We need to show
an1  an 
true.
a n1  a n for all n. Start with a n1  a n , then obtain an expression that is always
(n  1) 2
n2
(n  1)! 2
2

 n  1 
n  (n  1) 2  (n  1)n 2
((n  1))! (n)!
n!
 n 2  2n  1  n 3  n 2  0  n 3  2n  1 . The last inequality is true when n is greater than 2. Thus the
sequence is monotonically decreasing after the third term. Since the first two terms do not affect
convergence/divergence, we can conclude the sequence is monotone. B) Next show the sequence is
bounded below. But since the terms are positive, the sequence is bounded below by 0. Thus the
sequence is convergent by MCT.
n  1 n  cos n n  1
. Since the first and the third sequence converge to 1, the sequence


n
n
n
n  cos n
converges to 1 by the squeeze theorem b) Use L’hopital: let y  n (2n  1) . Take the ln of
n
ln( 2n  1)
ln( 2n  1)
1
both sides and bring down the exponent on the right: ln y 
. lim,
 lim,
 0.
n 
n  2n  1
n
n
33) a)
Since ln y converges to 0, by taking e,
lim,
n
n
2n  1  e 0  1
34)
Apply the ratio test:
lim
n
an1
an
(1)(3) (2n  1)( 2n  1) (2)(5) (3n  1)
2n  1 2

 lim
 , convergent
n (2)(5) (3n  1)(3n  2) (1)(3) (2n  1)
n  3n  2
3
 lim
35)
By trial and error, find the number n such that
Rn 
M
e 0.1 (0.1) n1
n 1
xa 
 0,01
(n  1)!
(n  1)!
36)
Yes, use the integral test:
1
 x(ln x)
2
dx  
1
1
1
1
du  lim (

)
is convergent
2
b


ln b ln 2
ln 2
u
37)
First find the partial sum : since it is a telescoping series, s n  sin
1
1
1
. s  lim s n  sin
 sin
n


2
2n  2
2
38)
(1) n ln n

is absolutely convergent: Note that for large values of n, ln n  n for any power of n

2
n 1
n 1

1
(log function grows slower than any power function). In particular, ln n  n 10 for large n. Using the
comparison test,

1
10


ln n
n
1


is absolutely convergent. (note that the power of n in



2
2
19
k 3 n  1
k 3 n
k  2 10
k
1
ln n  n 10 was chosen so that the last exponent in the denominator would be more than 1
39)

0.5
0
cos x dx  
2
0
n


(1) n ( x 2 ) 2
(1) n 0.5 4 n
(1) n x 4 n 1

x
dx




0
(2n)!
n 0
n  0 ( 2n)!
n  0 ( 2n)! ( 4n  1)
0.1 
0.5
0
(1) n (0.5) 4 n1
(0.5) 5 (0.5) 9

(
0
.
5
)




10
(24)(9)
n 0 (2n)!(4n  1)

The third term is less than o.oooo1. Thus you need only to take the first two terms by the
alternating series remainder theorem .

0.5
0
cos x 2 dx  (0.5) 
(0.5) 5
10
40)
Use the theorem that absolute convergence is convergence: It is easier to show the absolute
1  2 cos n
3
  2 and the bigger series is
2
2
2

1  2 cos n
convergent by the geometric series test. Thus 
is convergent.
2n
n 1
convergence since the comparison test can be used:

41)
Apply the ratio test and set the limit less than 1:
lim
n 
a n 1
an
 lim
n 
n  1( x  1) n 1
n ( x  1)
 1  x  1  1  x  1  1  0  x  2 At one end point x  0 ,

 (1)
n
n is
n 1
divergent by the divergence test. At the other end point x=2,


n is divergent by the divergence
n 2
test. Thus the interval of convergence is (0,2)
5
250 x
, f " ( x)  
, . Evaluate each function at a= 1 and apply
2
1  25 x
(1  25 x 2 ) 2
5
1
the formula, we get the first degree Taylor polynomial P1 ( x)  tan (5) 
( x  1) . The maximum
26
M
M
n 1
error: Use the Taylor’s remainder formula Rn 
xa 
(max x - a ) n 1 . 1)
(n  1)!
(n  1)!
250 x
M  max f ' ' ( x) when 0.9  x  1.2 . Since f " ( x) 
is a decreasing function (you can show it
(1  25 x 2 ) 2
42)
f ( x)  tan 1 (5 x)  f ' ( x) 
is decreasing by computing its derivative), its maximum is attained at the left end point x = 0.9.
250(0.9)
. 2) the maximum distance from the center of expansion is 0.2.
(1  25(0.9) 2 ) 2
250(0.9)
(1  25(.9) 2 ) 2
maximum error is R1 
(0.2) 2
(2)!
Thus M 
Thus the
43)
Use the integral test: If P = 1, it is divergent by the integral test. If p  1


1
x
1 (b 2  1)1 p
21 p
and this is convergent if 1  p  0  P  1 (need to have the
dx

lim

b  2
1 p
2(1  p)
( x 2  1) p
negative exponent so that
(b 2  1)1 p would be in the denominator. Therefore, convergent iff P  1
44)
Apply the ratio test and set the limit less than 1:
lim
n
an1
an
(2n  1)!( x  4) n1
( x  4)( 2n  1)( 2n)
 lim
  . This is greater than 1 except at the center
n
n (2n  1)!( x  4)
n
1
 lim
of expansion x = 4. Converges only at x = 4.
45)

(
n 1
n 1 3 n
 ( ) ) is divergent since
3n
4

3
( ) n is convergent and

n 1 4

(
n 1
n 1
) is divergent.
3n
d
1
1
x2
.
So
write
as
(
)
dx 1  x
(1  x) 2
(2  x 2 ) 2
x2
x2
x2
1


(
)
2
2 2
x
x2
(2  x )
4
(2(1  )) 2
(1  ) 2
2
2
46) a) First observe that
Thus to find a power series representation of (
a) Find a power series representation of
1
,
x2 2
(1  )
2
1
(1  x) 2
x2
2
b) Replace x by
c) Multiply by
x2
)
4
x2
4
1
d
1
d  n  n 1
 (
)   x  nx , x  1
a)
(1  x) 2 dx 1  x
dx n  0
n 0

1
x 2 n 1  x 2 n  2 x 2

n
(
n n 1 ,
1 x  2
 ) 
2
 x2  2 n 0 2
2
n0
(1   )
 2
b)
c)
(
x2
)
4
1
x2  x2n  2  x2n
, x 2
 n
n
x 2 2 2 n  0 2n 1 n  0 2n
(1  )
2
B) Note that
positive for
Thus
1
 1  x dx   ln( 1  x)  c for
x 1
(the absolute value in the ln can be dropped since 1-x is
x  1.
ln( 1  x)  
1
dx  
1 x


n 0
n 0
x n 1
c.
n 0 n  1

 xndx     xndx  
x n 1
, x 1
n 0 n  1

ln( 1  x)  

( x 2 ) n 1
x2n  2
 
, x 2  1  x  1
n0 n  1
n 0 n  1

Finally ,
ln( 1  x 2 )  
Let x=0 to get c=0. Thus
47) Use the comparison test to determine whether this series is convergent or divergent. Since


n2
1
n
n

1
and
2
n

n2


n 2
1
is convergent, then
2
n


n 2
1
n
must also be convergent.
n
48) Use the comparison test to determine whether this series is convergent or divergent. But we
must take the absolute value since terms may be negative. Then use the theorem that absolute
convergence implies convergence. Since


cos n
1

and

2
 2n n 1 2n 2
 4n
n 1

1
 2n
n 1
2
is convergent,

cos n
2
 2n
 4n
n 1
is convergent.
49)

4
 (1  n )
n
is divergent due to the Test of Divergence.(use L’Hopital’s rule and show the limit is
n 1
e4).
50) Use the alternating series error test. Find the first number which is less than .001, which is
Thus, it is estimated to be 1 
1
.
125
1 1
1


8 27 64
51) Separate the natural log function and use the Telescopic method to determine whether this
series is convergent or divergent. Take the limit of the partial sums lim ln( 1)  ln( n  1) which is  .
n 
Thus, this series is divergent.
52)

0.3
1  x2
0
(1  x)
1

1
2
 1
1
3
5
 
   2 x 2 n  1  x 2  x 4  x 6  .... 
2
8
16

n 0 
n 


1
2
dx : use the Binomial formula to expand (1  x ) :
2

0.3
0
1
1  x2
0.3
dx   (1 
0
1 2 3 4 5 6
x  x  x  ...)dx 
2
8
16
1 3 3 5
5 7 0.3
1
3
5
x 
x 
x ) 0  (0.3  (0.3) 3 
(0.3) 5 
(0.3) 7  ... . The first term with its value
6
40
112
6
40
112
0
.
3
3
1
1
less than 0.001 is
(0.3) 5 . Thus 
dx  0.3  (0.3) 3
2
0
40
6
1 x
(x 