BC 3
Advanced Tests for Covergence
Name:
Now that we see that convergence of improper integrals and series can be thought of as the same
problem under appropriate conditions, we immediately get the following theorem for series.
Limit Comparison Test: Suppose that 0  ak and 0  bk for all k 
and lim
n 
an
 L , with
bn
0  L   . Then

a
n 1

n
converges if and only if
n 1
If L  1 , then
If L  1 , then
a
If L  1 , then
and lim
n
converges .
n
diverges .

a
n 1
iii.
converges .

n 1
ii.
n
an 1
 L . Then
n  a
n
Ratio Test: Suppose that 0  ak for all k 
i.
b

a
n 1
n
may converge or diverge, so we must look for a different
method to determine convergence.
Root Test: Suppose that 0  ak for all k 
iv.
If L  1 , then
If L  1 , then
a
n
converges .
n
diverges .

a
n 1
vi.
If L  1 , then
k 

n 1
v.
and lim k ak  L . Then

a
n 1
n
may converge or diverge, so we must look for a different
method to determine convergence.
IMSA
Sequences and Series 4.1
Fall 14
The Grand Challenge: Using any of the tests that we have developed so far, determine whether
each of the series converges or diverges. Your reasoning should be easy to follow. Be careful to
refer to which test you are employing.
n!

n 1 (2 n )!

(3)
 3

1  
n
n 1 
(2)

2n
(4)

(5)
IMSA
n  sin 2 n

3
n 1 n  n


(1)
 5k  k k 


2k 
k  0  ( k  1)

5n

n 1 n !

(6)
 k 3  3k  21 

 5

3
k  2  5k  6k  k  1 
Sequences and Series 4.2
Fall 14

(7)

n 1

(9)

k 2
IMSA

 1 
 n  sin  n  
 

1
sin  
k

(8)
 k !
2
 (2k )!
k 1

(10)
 1 n 



n 1  5n 
Sequences and Series 4.3
n
Fall 14

1
converges, then find the value of this sum with an error of less than
k!
k 2
.0001. Explain your error analysis.
(11) Show that
2
k
(12) Find all values of x  for which the series


k 0
IMSA
xk
k!
Sequences and Series 4.4
Fall 14