MATH 162
Solutions: TEST II
Friday the 13 March 2015
PART I (Answer all four problems.)
1.
Compute the value of the following improper integral:

2

0
cos x
dx
sin x
Solution:


2
2

0
cos x
cos x
dx  lim 
dx  lim 2 sin x
c  0
c  0
sin x
sin
x
c



2 lim  sin  sin c   2
c  0
2


 /2

c
2
2.
Albertine ponders the following recursively defined sequence:
c1 = 1,
cn1 
1
 3 for n ≥ 1
cn
(a) Find the values of c2 , c3 and c4.
Solution:
Setting n = 1:
c2 
1
 3 1 3  4
c1
Setting n = 2:
c3 
1
1
13
 3   3   3.25
c2
4
4
Setting n = 3:
c4 
1
4
43
3 3
 3.307
c3
13
13
(b) Assuming that the limit of cn as n   exists, help Albertine to find its
exact value.
Solution:
Assume that L = lim cn exists. Then:
1

lim cn1  lim   3 
 cn

and so:
L
1
3
L
3
Multiplying both sides by L yields: L2 = 1 + 3L. So: L2 – 3L – 1 = 0. Using
the quadratic formula:
L
3  13
 3.3027,  0.3027
2
We reject the negative root, since c1 > 0 and all subsequent terms of the
sequence are also positive (reasoning inductively).
Thus, if lim cn exists, this limit must be
3  13
2
3.
Determine convergence or divergence of the following improper integral.
Justify your answer:
2

1
x2
dx
( x 3  1) 2
(Hint: Integrate.)
Solution:
2

1
2
x2
x2
( x 3  1) 1 2
dx  lim  3
dx  lim 

2
c1
c1
c
( x 3  1) 2
(
x

1
)
3
c
lim 
c1
4.


1 3
(2  1) 1  (c  1) 1 does not exist
3
The life-span (in years) of a vampire bat can be modeled by a random
variable X with probability density function
ce  x /10 if x  0
f ( x)  
0 otherwise
4
(a) Find the constant c. (Hint: Every bat must die.)
Solution:

1   ce
 x / 10
c
dx  c lim
c 
0


0

c
e  x /10 dx  c lim   10e  x /10  
c 
0


c lim  10 e c /10  e 0  10c
c 
Hence c = 1/10
(b) Find the probability that a randomly chosen vampire bat will live longer
than 11 years. (Express your answer to the nearest hundredth.)
Solution:

1
P(bat lives longer than 11 years )   e  x /10 dx 
10 11
c
1
1
lim  e  x /10 dx  lim
10 c 11
10 c


c 
 (10)e  x /10  

11


(1) lim e c /10  e 11/10  e 11/10  0.33
c 
PART II Select any 4 of the following 5 sequences. For each selected
sequence, determine convergence or divergence. Briefly justify each answer. In
the case of convergence, find the limit. Calculator results will not earn full
credit. (You may answer all 6 to earn extra credit.)
1.
  
an  n sin 

 13n 
Solution: Let h = 1/(13n). Then n = 1/(13h) and as n  , h  0. Hence:
5
1
1  sin h 
1
 1 
an  n sin 
sin h  
as h  0


13  h  13
 13n  13h
Hence the sequence an converges to 1/13.
 
bn  1  
 n
2.
2n
Solution: Note that:
 
bn  1  
 n
2n
2
   n 
  1     e
 n  


 
2
 e2 
Hence the sequence bn converges to e2.
cn  n 4  19n 2   
3.
n 4  7 n 2  n  13
Solution: Rationalizing the “numerator” yields:
cn 


 n 4  19n 2    n 4  7n 2  n  13 

n  19n    n  7n  n  13 
 n 4  19n 2    n 4  7n 2  n  13 


4
2
4
2
26n 2  n    13
n 4  19n 2    n 4  7n 2  n  13
Hence the sequence cn converges to 13.

26n 2
n4  n4
 13
6
4. d n  ln(  n  2015 )  ln( 5n   )  arctan n
Solution: The sequence converges:
d n  ln(  n  2015  )  ln( 5n   )  arctan n 
ln
 n  2015
 
 arctan n  ln 
5n  
5 2
4
ln n
   sin  n
5. f n  13  44
  cos  
n
n
n
Solution: The sequence converges:
ln n/ n → 0
cos (/n) → cos(0) = 1
sin 4  n
0
n
Thus
4
ln n
   sin  n
f n  13  44
  cos  
 13  0    0
n
n
n

13  
7
PART III Select any 4 of the following 5 series. For each selected series,
determine convergence or divergence. Justify each answer. (You may answer
all 5 to earn extra credit.)

1.
 arctan( n  1)  arctan( n)
n 1
Solution: Since this series is telescoping, we will consider the sequence of partial
sums:
s1 = arctan(0) – arctan(1)
s2 = (arctan(0) – arctan(1)) + (arctan(1) – arctan(2) = – arctan(2)
s3 = (arctan(0) – arctan(1)) + (arctan(1) – arctan(2)) + (arctan(2) – arctan(3))
= – arctan(3)
We infer that, in general, sn = -arctan(n).
Now sn = -arctan n → -/2 as n → ∞. So the series is convergent.

2.

n 1
7 n 
42n
Solution: Applying the ratio test
an1
an
7(n  1)
2 ( n 1)
4 2 n 7 (n  1)
4

 2 n2

7 n
7n 
4
4 2n

1  n 1
1

   1
16  n 
16
we find that the series converges since r < 1.
8


3.
n 1
 n  


n


n
Solution:
Since ln x < x , we have:
0
  ln x
x3

 ln x
x3

x
x3

1
x2
Thus, invoking both the p-test and the Comparison Test, our original
integral converges.
4.
314.314314314…
Solution: This is the geometric series: 314 + (314)10 -3 + (314)10 -6 + …
Since r = 10 -3 < 1, our series converges.
Its sum is
314
314
314000


1  0.001 0.999
999

5.

n 2
1  sin 4 (n)
n ln( n13 )
Solution: Consider the following inequality:
1  sin 4 (n)
1

0
n ln( n13 ) 13n ln n
Using the p-test, we see that the smaller series diverges and hence our series
diverges as well.
9
PART IV. Select any three of the following four problems. You may answer all
four for extra credit. For each improper integral below, determine convergence
or divergence. Justify each answer!


( A)
1
Solution:
2015  ln x
dx
x3
Since ln x < x for x > 1
0
2015  ln x 2015 x  x
1

 2016 3
3
3
x
x
x
Now using the Comparison Test, and the p-test for p = 3, we see that our
improper integral converges.

1  x 2e x  (ln x) 2015
dx
0
  e4 x
( B)
Solution: Observe that
1  x 2 e x  (ln x) 2015 e x  x 2 e x  e x
0


  e4 x
e4 x
( 4 )
x
2
1 x 1
3x
3e
 ( 4 ) x  ( 4 ) x  3
( 4 ) x
e
e
e
2
2
1
e
( 4 )
x
2
1
3
e
(1 / 2 )
x
2
1  xe2 x xe2 x  xe2 x 2 xe2 x 2 x 2e x 2
0

 4x  2x  2x  x
1  e4 x
e4 x
e
e
e
e
3
1
ex/4
10


Now
e  x dx converges.
0
Thus, invoking the Comparison Test, our original integral converges.

(C )

0
9   x 6  13 x
dx
8
5  4 x  x
Solution: Observe that
9   x 6  13 x 9 x 6   x 6  13x 6
1
0


(
22


)
x8
x2
5   4 x  x8
Thus, invoking the Comparison Test, our original integral converges.

( D)

2
4   3 cos 4 (4 x  1)
dx
x  
Solution:
Observe that
4   3 cos 4 (4 x  1)
1
1 1


0
x  
x    x 1   x
Thus, invoking the Comparison Test, our original integral diverges.
PART V. Select any 3 of the following 4 problems. You may answer all four
for extra credit. For each numerical series below, determine convergence or
divergence. Justify each answer.
11

(A)

n1
n 5 2 n ( n !) 2
(2n)!
Solution: Applying the ratio test to this positive series:
(n  1) 5 2 n1 (( n  1) !) 2
an1
(n  1) 5 2 n1 (( n  1) !) 2 (2n)!
(2n  2)!



5 n
2
5
n
2
n
2
(
n
!
)
an
n
2
( n !)
(2n  2)!
(2n)!
2
 (n  1) ! 
1
(n  1) 2
 n 1
 n 1

 2


 (2) 

n
n
!
(
2
n

2
)(
2
n

1
)
n
2
(
n

1
)(
2
n

1
)






5
5
(n  1) 2
1 1
 n 1
 15   r


2 2
 n  (n  1)( 2n  1)
5
Since r < 1, we conclude that our positive series converges.

(B)

n 1
1


1  
 n
n2
Solution: Applying the nth root test to this positive series:
(an )1/ n 
1
 
1  
 n
n

1
  1
e
Since  < 1, we conclude that our positive series converges.
12

(C)

n1
n5
en
2
Solution:
Applying the ratio test to this positive series:
(n  1) 5
n
( n 1) 2
an1
 n 1 e
e


 ( n1)2 
5
n
an
 n  e
2
en
5
2
 n 1 1
5

 2 n1  1 (0)  0
 n  e
5

(D)

n 1
 n3


 n  
 

Solution: Since
 n3
 1 3/ n 
33

 
  1  1,
 n  
1  / n 
33
33
We may invoke the n th Term Test for Divergence to conclude that our
original series diverges.

 n3

  1  1 we apply the nth Term Test for Divergence to
Since 
 n  
conclude that our series diverges.
13
EXTRA CREDIT (University of Michigan midterm problem)
14
I tell them that if they will occupy themselves with the study of
mathematics they will find in it the best remedy against the lusts of the
flesh.
Thomas Mann, THE MAGIC MOUNTAIN