BC 3
Sequences and Series Quiz
No Calculators allowed.
Show set-up/method clearly on all problems.
#1.
Name:
Complete (no work necessary):

a. An expression of the form
a
k 1
k
is called an infinite series
Corresponding to this, we have the sequence {Sn}, which is called the sequence
of partial sums.
n
a
The nth term of this sequence is given by Sn =
k 1
k
The sequence {an} is called sequence of general terms
If lim(an )  0 then the series
n 

a
k 1
k

b. The series
ar
k
converges to
k 1
diverges.
ar
if and only if -1 < r < 1.
1 r
c. Root Test: If lim  n an   L , L < 1, and an  0 for all n  1 , then
n 

a
k 1
converges.

#2 (2 pts each)
Look at the series
a
k 1
n
k
. Let Sn   ak 
k 1
n 1
for all n  1. .
n
a. Find a1 , a2 and a3 .
11
S1  a1 
2.
1
2 1
1
S2 
 S1  a2  a2   .
2
2
3 1
1
S3 
 S 2  a3  a3   .
3
6
1 1
So, a1 , a2 , a3  2,  , 
2 6

b. Determine whether
a
k 1

a
k 1
k
k
converges or diverges. If it converges, find the sum.
 n 1 
 lim( Sn )  lim 
  1 . That is, the series converges to 1.
n 
n 
 n 
k
#3(6 pts each) Determine whether each series converges or diverges. Explain reasoning
carefully and completely.

a.
3n

2
n
n 1 n  2
an1
3n1
( n) 2  2 n
n2
3 3
 lim

 lim
   L.
Use the ratio test: lim
2
n

1
n
2
n an
n (n  1)  2
n ( n  1)
2 2
3

3n
diverges by the ratio test.
2
n
n 1 n  2
Since L  1 , the series 
b.
 2
 1  
 n

n 1
n
2
2
Limit Comparison Test: Let an 
1
n
2
. Then,
1
lim
n
n
2
 2
1  
 n
2
n
2
 lim
n
Since 0  L   and
1
 2
1  
 n
2
1 L

1
converges (by p-test with p = 2), The series
2 

n 1
  n
 2
 1  
 n  converges by limit comparison test.
2
2

n 1

c.
 tan
n
1
( n)
n 1
Since lim  tan 1 (n)  
n 

2
0,

 tan
n 1
1
( n) diverges by the nth term test.
#3 (continued) Determine whether each series converges or diverges. Explain reasoning
carefully and completely.
n ! n n

n 1 (2n)!

d.
Use the ratio test:
an1
(n  1)! (n  1) n1 (2n)!
lim
 lim

n an
n
(2n  2)!
n ! n n
(2n)! (n  1)! ( n  1)( n  1) n


n
n (2n  2)!
n!
n
 lim
n
1
(n  1) ( n  1)  n  1 
 lim



 .
n (2n  2)(2n  1)
1
1  n 
1  1
 lim  1  
n 4 
n
e
  L 1
4
n
n ! n n
converges by the ratio test.
n 1 (2n)!

Since L  1 , the series 

1
1
approximates
the
value
of

k
k
k 1 k  5
k 1 k  5
with an error less than .001. You need only find the value of n, not S n . Explain your
reasoning clearly.
n
#4(6 pts) Find a value of n such that S n  

n

1
1
1


 S n  Rn .



k
k
k
k 1 k  5
k 1 k  5
k  n 1 k  5
1


1
1 5n 1
1
Rn  
  k 

.
k
1 4  5n
k  n 1 k  5
k  n 1 5
1
5
1
1

 .001
Since 4  54  4  625  1000,
4  625 1000

1
.
k 1 3  k !
Therefore, S 4 is within .001 of the actual sum 
k

#5 (3 pts) Evaluate the sum
k 1
n
3k
5
k
.
3k 3 6
3n
  2
 n
k
5 5
5
k 1 5
n
3k
6 9
3n
 n 1 .
Then, 5Sn   k  3   2 
5 5
5
k 1 5
3 3
3 3n
 n 1  n .
So, 5Sn  Sn  4Sn  3   2 
5 5
5
5
Repeating this process:
3 3
3
3n
5  4Sn  20 Sn  15  3   2 
 n  2  n 1
5 5
5
5
Therefore,
3 3n 3n
20Sn  4Sn  16 Sn  15  n 1  n  n 1
5
5 5
So, finally
1
3 3n 3n 

lim S n  lim 15  n 1  n  n 1 
n 
16 n 
5
5 5 
15

16
Sn  