Math 162
Solutions: QUIZ V
Part I:
Select any three of the following four integrals. For each improper
integral that you select, determine convergence or divergence. Justify your
answers! (You may answer all four for extra credit.)
3
(a)

0
1
dx
(3  x) 4
Solution: This integral diverges since:
3

0
c
c
1
1
1
dx

lim
dx

lim

c 3  (3  x ) 4
c3 33  x 3 0
(3  x) 4
0


1
1

lim 

3
3 
c3




3
3

c
3
3

0


1  5 x  7 x15
(b) 
dx
x
0
1
Solution: Note that the dominant term in the numerator is 1 (not x15).
This integral converges because:
1  5x  7 x15 13
0

for 0  x  1
x
x
1
Using the p-test, we know that
test,

0
1
dx converges. Hence, by the comparison
x
2
1  5x  7 x

x
0
1

4
(c )
15
dx converges.

 tan 2 x dx
0
Solution: This integral diverges because:


4

c
tan 2 x dx  lim


c  0
4
0
tan 2 x dx  lim
c 
4
c
1
( ln | cos 2 x |) 
0
2
1
1
lim ln | cos 2c |  ln(cos 0)    lim ln | cos 2c |  
2 c 
2 c 
4
4
since cos(2c) 0+ as c  (/4)-.

(d )

0
1
dx
x1/ 4  x  x 2
Solution: This integral of mixed type converges because, by definition:


0
1
1
4
x  x  x2
1
dx 

0

1
1
4
x  x  x2
dx  
1
1
1
4
dx
x  x  x2
and each of these two integrals converges (by virtue of the Comparison Test):
For 0 < x ≤ 1:
0
1
1
4
x xx

2
1
x
1
4
3
and by the p-test for integrals of type II,
1
1

0
For x ≥ 1:
x
0
dx converges.
1
4
1
1
4
x  x  x2

1
x2
and by the p-test for integrals of type I,


1
1
dx converges.
x2
Part II: For each of the following sequences, determine convergence or
divergence. In the case of convergence, find the limit of the sequence. Briefly
justify each answer. (Select any 7 of the 8 sequences. For extra credit, you may
solve all eight.)
(a)
an 
n  2015 ln n
n!
Solution: Since n = o(n!) and ln n = o(n), {an} converges and its limit is 0.
(b)
 n  1789   4 
bn  3n  ln 
  1  
 n  1492   n 
n
Solution: Using the fact that the limit of the sum of two convergent sequences is
the sum of their limits, we have:
 n  1789 
 4
4
4
lim bn  lim 3  lim ln 
  lim 1    0  ln 1  e  e
 n  1492 
 n
n
n
4
Thus the sequence {bn} converges and its limit is e4.
(c )
cn 
sin 4n
n
Solution: Since -1 ≤ sin(4n) ≤ 1, we have:

1 sin 4n 1


n
n
n
Applying the Squeeze Theorem, we conclude that {cn} converges to 0.
(d )
nn
d n  14
n 1
Solution: Observing that dn ≥ nn/ (n14 + n14) = ½ nn/ n14 = ½ nn – 14   as
n  , we conclude that dn is unbounded, and thus divergent.
(e)
en = (-1)n cos(1/n)
Solution: First note that, as n  , cos(1/n)  cos 0 = 1.
Thus for large n, en is approximately (-1)n which is a divergent sequence.
(f)
(n 2  7n  1)3 (3n  77)(2n5  n  4) 2
An 
1  ln n  5n8 (n  1)9
Solution: By selecting the dominant terms, we have:
5
(n 2  7n  1)3 (3n  77)( 2n5  n  4) 2 (n 2 )3 (3n)(2n5 ) 2 12n17 12

 17 
1  ln n  5n8 (n  1)9
5n8 (n)9
5n
5
Hence we conclude that {An} converges to 12/5.
(g)
Bn = arctan (ln(n))
Solution: As n  , ln n  . and hence:
Bn  arctan(ln n) 

2
Thus the sequence {Bn } converges and its limit is /2.
Cn  n2  9n  31  n2  3n  5
(h)
Solution: Rationalizing this expression:
hn  n 2  9n  31 
n
n
2
2
 9n  31 
n 2  3n  5 
n  3n  5
2
 9n  31  n 2  3n  5
n 2  9n  31 
6n

3
nn
n 2  3n  5


n 2  9n  31 
n 2  3n  5
n  9n  31 
n  3n  5
2
2

6n  26
n 2  9n  31 
n 2  3n  5
Thus the sequence {Cn } converges and its limit is 3.
Mathematicians have tried in vain to this day to discover some order in the sequence of prime numbers, and we have reason
to believe that it is a mystery into which the human mind will never penetrate.
- Leohard Euler (1707-1783)