BC 3
Sequences and Series Quiz
No Calculators allowed.
Show set-up/method clearly on all problems.
Name:
Key
n2  3n
#1(4 pts) Given  ak  2
for all n  1, determine whether the series
3n  2n  2
k 1
converges or diverges and if it converges, find its sum.
n
Sn 

a
k 1
k
n2  3n
1
 as n   .
2
3n  2n  2
3

By the definition of infinite series, this means  ak converges and
k 1

#2 (2 pts each)
Suppose the series
a
n 1
n

a
k 1
k
1
 .
3
converges to 5 and an  0 for all n  1.
Indicate whether each of the following statements is T (must be true) or F (false).
You may write a quick one sentence explanation, this may get you partial credit.
a.
T
The sequence ak must converge.
By nth term test, an must converge to 0.
b.
F
The partial sum S100 could have a value of 5.01.
Since all the terms of ak are positive, S n is increasing. So all terms
of S n are less than 5.
c.
T
The sequence of partial sums must converge to 5.

As in #1, this is what it means for the series
a
n 1
d.
F
n
to converge.
ak 1
k  ak
If the ratio test is applied to this series, then the value of lim
must be a finite positive number strictly less than 1.
ak 1
= 1, The series may converge (or diverge).
k  ak
If lim
e.
F
If bk  ak  0 for all k  1 , then

b
n 1
n
must converge.
#3(6 pts each) Determine whether each series converges or diverges. Explain reasoning
carefully and completely.

a.
n!
3
n 1
n
an1
 n  1! 3n
n 1
 lim n1   lim
.
n an
n 3
n ! n 3
Use the ratio test: lim

n!
diverges by the ratio test.
n
n 1 3
Since   1 , the series 

b.
n
e
n 1
n2
b 1

 1 b 
 1 1  1
lim
du

lim
 
=
dn
2



  lim
u
1 en
b   2e
b  2eu 1
b  2eb
2e  2e



1


Using Integral Test:
n

Since this integral converges, the series
n
e
n 1
n2
converges by the integral test.
 n 2  2n 

 3 2

n 1  3n  n  5 

c.
Limit Comparison Test: Let an 
lim
n
1
n
2
n  2n
 lim
n
1
. Then,
n
3n3  n2  5
n  2n
3
2
3
3n  n  5


 n 2  2n 
1
Since    divereges (by p-test with p = 1), The series   3 2

n 1  n 
n 1  3n  n  5 
diverges by limit comparison test.
3
2
#3 (continued) Determine whether each series converges or diverges. Explain reasoning
carefully and completely.

d.
1
 n 3
n 1
n
1
1
 n , for n  1.
n
n 3
3

1
1
Since  n is a geometric series with r   1 , it converges.
3
n 1 3

1
Therefore, 
converges by comparison test.
n
n 1 n  3
Direct Comparison: 0 

1
k!
k 1
with an error of at most .001. Explain carefully. You need only find the value of n, you
need not find the approximation S n for this value of n.
#4 (6 pts). Find a value of n such that S n approximates the value of the series

n

1
1
1


 S n  Rn .



k
k
k
k 1 3  k !
k 1 3  k !
k  n 1 3  k !
1


1
1 3n 1
1
Rn   k
  k 

.
1 2  3n
k  n 1 3  k !
k  n 1 3
1
3
1
1

 .001
Since 36  729,
6
2  3 1000

1
.
k 1 3  k !
Therefore, S 6 is within .001 of the actual sum 
k
3
k

1
, where f n is the Fibonacci sequence defined by:
k 1 k  f k  2
f1  f 2  1, and f n  f n 1  f n  2 for n  3 .
#5(3 pts) Evaluate
f
f k 1
fk 2  fk
1
1
1




.
f k  f k  2 f k  f k 1  f k  2 f k  f k 1  f k  2 f k  f k 1 f k 1  f k  2
So,


k 1

1
1
1


f k  f k 2 k 1 f k  f k 1 f k 1  f k 2
,
Then
n
 1

1
Sn   


f k 1  f k  2 
k 1  f k  f k 1
1   1
1   1
1 
 1







 1 1 1  2   1  2 2  3   2  3 3  5 
1
 1
f n 1  f n  2
Therefore,


1
lim Sn  lim 1 
 1
n 
n 
 f n1  f n 2 

So,
f
k 1
1
=1
k  fk  2
 1

1



 f n  f n 1 f n 1  f n  2 