Chapter 9(5) Alternating Series Test Alternating Series Remainder
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Transcript Chapter 9(5) Alternating Series Test Alternating Series Remainder
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(ln1x )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
2x
Lim
2x2 5
x
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
r
This is an Alternating Geometric series
1
2
Alternating Series Test
Let an 0, then 1 an and
n 0
n
1
n 0
n 1
an converge if
Lim an 0 and an1 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
an1
1
Positive
1
1
n1
Positive
0
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
1
3
Check an > 0
an
Check Lim = 0
Lim 1n
1
0
Lim n2
2
0
n
n
Check terms
get smaller
2
Positive
a2 1 a3
or
2
4
1
Positive
1
4
Positive
This part fails
The Alternating Series Test (AST) can not be applied
Alternating Series Remainder
If an 1 an
Rn is the remainder
S is the sum
S n is the approximate sum
S Sn Rn an1
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
n 1 1
n!
Check an > 0
an
Check Lim = 0
Lim n1!
Check terms
get smaller
1
( n 1)!
1
2
1
Positive!
n
91
S6 144
.63194
1
1
1
1
6
1
24
1
120
1
720
Positive
0
1
n!
Series converges
by the (AST)
Range is
1
Rn a7 5040
.0002 .63174 S .63214
Absolute Convergence
If
an converges, then an converges
(absolute convergence)
If an converges but
a
n
an diverges
is conditiona lly 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
n1
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) n1
n
1 12 13 14
an Positive?
Lim = 0?
an+1 < an ?
n 1
( 1) n1
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) n1
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) n1
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