

1

2
3
n
2
n 1
1
bn   2
n 1 n
Larger Denom makes
fraction smaller
1

4
n 2 n(ln n)


2
Converges
by DCT
Continuous if n>1
Positive if n>1
Decreasing if n>1
b
1
1

Lim
n(ln n) 4 b 2 n(ln n) 4
b
Lim 3(ln1x)3  0  3( Ln1 2)3  3( Ln1 2)3
b

n 1
p-series with
p=2>1
converges
1
1

3n 2  2 n 2


2
Converges by Integral Test
2n
2n 2  5
Lim
x 
2x
2x2  5
2
0
2

Diverges by the nth term test
(Lim ≠ 0 then it diverges)

n
n 1

1
3
n

n 1
1
n7 / 2
p-series with p = 7/2 > 1
Converges
Chapter 9(5)
Alternating Series Test
Alternating Series Remainder
Absolute and Conditional Convergence
Rearranging an infinite series
Alternating series contain both positive and
negative terms – the signs alternate

 
n0
1 n
2
 1  12  14  18  161  321
This is an Alternating Geometric series
r  21
Alternating Series Test


Let an  0, then   1 an and   1 an convergeif
n0
n
n1
n0
Lim an  0 and an1  an for all n
n
If terms are positive, limit = 0, and terms get smaller,
then an alternating series converges
Determine convergence or divergence of:

 (1)  
n 1
n 1 1
n
Check an > 0
an  1n 
Check Lim = 0
Lim 1n 
Check terms
get smaller
n 
1
Positive
1

 Positive
0
an1  n11  1n  an
The series converges by the Alternating Series Test (AST)
Can the AST be used to show
convergence or divergence of:
2
1
        
1
1
2
2
1
2
2
3
Check an > 0
an 
Check Lim = 0
Lim 1n  1  0
2
Positive
1
3
or
2
4
1
Positive
1
4
 Positive
n 
Lim n2  2  0
n 
Check terms
get smaller
a2  1  a3
This part fails
The Alternating Series Test (AST) can not be applied
Alternating Series Remainder
If an 1  an
Rn is the rem ainder
S is the sum
S n is the approxim ate sum
S  Sn  Rn  an1
For a convergent alternating series, a range for the sum
can be found by taking n terms, finding the remainder
and using it to establish a range
Approximate
the sum from the first six terms:

 (1)
n 1
   11  12  16  241  1201  7201
n 1 1
n!
Check an > 0
an 
Check Lim = 0
Lim n1! 
Check terms
get smaller
1
( n1)!
1
Positive!
n 
0
 n1!
91
S6  144
 .63194
1
Rn  a7  5040
1

 Positive
Series converges
by the (AST)
Rangeis
 .0002 .63174 S  .63214
Absolute Convergence
If

an converges, then  an converges
(absolute convergence)
If  an convergesbut  an diverges
a
n
is conditionally convergent
Describe the convergence of each:


n0
Check:


n1
1
Pos
( 1)
n
n
( 1)n n!
2n
 11  12  24  86  
an > 0
Lim = 0
an+1<an

1
1

Not Decreasing so the AST Fails
1
2

1
3

1
4
 
 Positive!
Lim an 
n 
1
n1

1
n
1
n
0
Check Absolute Value


n1
Converges
by AST
( 1)
n
n

  n11/ 2
n1
Divergent p-series (p < 1)
Conditional
Convergence
Rearrangement of a series
If you rearrange a finite series the sum does not change
1  2  3  4  1  3  2  4  10
Absolute converging series can also be
rearranged with no change
Conditional converging series can be
rearranged to change the sum

Consider:

n1
( 1)n1
n
 1  12  13  14  
an Positive?
Lim = 0?
an+1 < an ?


n1
( 1)n1
n
Converges by the AST
 11  12  13  14  
This is the harmonic series which diverges
We have a conditional converging series


n1
( 1)n1
n
 1  12  13  14    Ln 2
1  12  13  14  15  16  17  18  
1  
1
2
1
4
 
1
3
1
6

1
8
 
1
5
1
10

 12   14  16   18  101   121 
1
2
1  12  13  14  15  16 
1
2
Ln 2
1
12
(Proved later)
Evens are negative

Put fractions
together with double
denominators
Combine insides
Factor out 1/2
From initial series

1

n
2
n 1
Lim Ln n  Ln     0

Geometric series
with r = ½ < 1
 Ln n
x 
n 2
Diverges by the nth term test
(Lim ≠ 0 then it diverges)
Converges
Conditional or absolute?


n 1
1
n
1
n1
p-series with p = ½ < 1
Diverges
an 
(1) n 1

n
n 1


Lim
x 
1
n
1
pos
1
x

 Pos
1

0
Since n 1 bigger
The series converges


n 1
n 1
(1)
n

1 p-series with p = ½ < 1

1/ 2
Diverges
n 1 n
The series converges conditionally