Present Remotely
Send the link below via email or IM
CopyPresent to your audience
Start remote presentation Invited audience members will follow you as you navigate and present
 People invited to a presentation do not need a Prezi account
 This link expires 10 minutes after you close the presentation
 A maximum of 30 users can follow your presentation
 Learn more about this feature in our knowledge base article
Sequence
No description
by
Tweetmohamed hesham
on 5 December 2014Transcript of Sequence
Power series
A power series about x = a is a series of the form
Infinite series
An infinite series is the sum of an infinite sequence of numbers
Sequence
It is nothing more than a list of numbers written in a specific order.
Examples
Find whether these functions are converging or diverging
Bounded and Monotony
Convergency
S = {a ,a ,a ,..,a }
.
.
.
lim a
n
n>
is exist,
lim a
n>
n
tends to infinity,
then the sequence
converges
then the sequence
Diverges
If
If
. . . . . . . .
Sequence is increasing or decreasing is said to be
monotonic
.
Sequence is
bounded above
if a
n
<
\
M
Sequence is
bounded below
if
>
\
a
n
m
\
\
\
\
n=0
o
o
a
n
= a + a + a + a + ...
1
2
3
4
Convergency
Infinite Geometric series
Telescoping Series
(a  a )+ (a  a )+ (a  a )+
a+(a + a )+ (a + a )+ (a+ a )+
lim S
n>
0
0
n
= L
=
lim S
n>
0
0
n
0
0
The Series converges
The Series diverges
\
\
\
\
n=0
o
o
a
n
r
= a + ar + ar + ....
2
, if r < 1

a
1r
=
Infinite geometric series converges when :
r < 1
>
converges to
a
/
1r
1r1 > 1
>
diverges
1
2
2
3
3
4
1
2
2
3
3
4
4
The series will only converge
if lim S = L
n> 0
0
The n term test
th
lim a
n> 0
0
lim a
n> 0
0
n
n
= 0
= 0
\
Diverges
Diverges or converge
\
\
\
\
n=0
o
o
\
\
\
\
n=0
o
o
\
\
\
\
n=0
o
o
Tests of convergance
Integral Test
Ratio test
Alternating series test
Comparison tests
b a c are nonegative series
\
\
/
\
\
/
/
/
n+1
n

Answer
by telescoping
\
\
/
\
\
/
\
\
/
\
\
/
S
n
=
2
1
2
3
\
\
/
\
\
/
/
/
n+1
n



+
(
)
(
(
(
+....+
(
=
\
\
/
/
/
n+1
\
\
/
1

=
0
0
lim S
n>
0
0
n
so it diverges
/
1
2
n
Answer
lim a
n>
0
0
n
= 0
so it may diverge or converge
Find another way to solve this
S
n
= 1 +
1
2
1
4
+
+
....
\
\
A geometric series that its "r" is less than 1 so it will converge to
\
1
\
2
1
1

3n+5n
2
n+4n
2
/
/
Answer
lim a
n>
0
0
n
= 3
So it diverges
n term test
th
by using
Suppose a series
\
\
\
\
n=0
o
o
a
n
, where
a
n
positive and decreasing function
lim a
a>
0
0
n
dn
k
a
converges
\
\
\
\
n=k
o
o
a
n
converges
diverges
diverges
lim a
a>
0
0
n
dn
k
a
0
0
0
0
,Converges
,Diverges
lim
n>
0
0
a
n
n+1
a
/
/
/
/
/
/
< 1
> 1
Converges
Diverges
Root test
lim
n>
0
0
a
n
n
\
\
/
< 1
Converges
Diverges
> 1
if b diverges,
a diverges
if c converges,
a converges
lim
n>
0
lim
n>
0
0
0
lim
n>
0
0
a
n
n
b
/
a
n
n
b
/
a
n
n
b
/
= c > 0
= 0
0
0
=
, so a will be the same as b
and b converges ,
so a converges
and b diverges ,
so a diverges
The series
\
\
\
\
n=1
o
o
(1)
n+1
a
n
= a  a + a  a + ..
converges
if :
a
n
's
are all positive nonicreasing
lim
n>
0
0
a
n
= 0
Absolute and conditional convergence :
<
/
<
/
n
n
n
n
n
> converges absolutely if the corresponding series of absolute values converges, if it converges but the absolute doesn't so it converges conditionaly
\
\
\
\
n=1
o
o
a
Absolute convergence test
If
\
\
\
\
n=1
o
o
a
 
n
converges, then converges
\
\
\
\
n=1
o
o
n
a
\
\
\
\
n=1
o
o
c (xa)
n
n
= c +c (xa) +..+
c (xa)
n
n
1
1
in which
center
(a) and the
coefficients
c
0
0
,c
1
,c
n
,....
,....
are constants
How to test a power series for convergence
Examples
Use root ratio to find the interval where the series converges absolutely.
 x  a  < R or a  R < x < a + R
Then, test for convergence at each endpoint. Use any kind of tests.
If the interval of absolute convergence is
 x  a  < R
so the series diverges for
 x  a  > R
\
\
\
\
n=0
o
o
(x+5)
n
\
\
\
\
n=0
o
o
3
n!

x

n
n
Answers
By applying ratio test :
lim
n>
0
0
a
n
n+1
a
/
/
/
/
/
< 1
lim
n>
0
0
(x+5)
(x+5)
n+1
/
/
n
/
/
/
/
< 1
=>
lim
n>
0
0
 x+5 
< 1
6 < x < 4
at (4) =>
\
\
\
\
n=0
o
o
at (6) =>
\
\
\
\
n=0
o
o
(1)
n
(1)
n
=> Diverges, by applying n term test
=> Diverges, by applying n term test
th
th
By applying ratio test :
lim
n>
0
0
a
n
n+1
a
/
/
/
/
/
< 1
=>
lim
n>
0
0
/
/
3
x
(n+1)!
/
/
/
/
/
/
n+1
n+1
3
.
n!
3
x
n
n
< 1
= 3x lim ( ) < 1
Converges for all x
1
n+1
/
n>
0
0
0
0
0
0
=
/
. . . . . . .
. . . . . . .
. . . . .
n
n
Suppose
1
2
3
4
and
The team members :
1Yara Tarek
2Mirna Magdy
3Amar Loay Jahin
4Mohamed Asem Torky
5Amr Mamdouh
6Mohamed Hesham
so let's have some examples
Thank You
n
1
2
3
4
= S
n
n
n
n
n
n
n
n
so
Full transcriptA power series about x = a is a series of the form
Infinite series
An infinite series is the sum of an infinite sequence of numbers
Sequence
It is nothing more than a list of numbers written in a specific order.
Examples
Find whether these functions are converging or diverging
Bounded and Monotony
Convergency
S = {a ,a ,a ,..,a }
.
.
.
lim a
n
n>
is exist,
lim a
n>
n
tends to infinity,
then the sequence
converges
then the sequence
Diverges
If
If
. . . . . . . .
Sequence is increasing or decreasing is said to be
monotonic
.
Sequence is
bounded above
if a
n
<
\
M
Sequence is
bounded below
if
>
\
a
n
m
\
\
\
\
n=0
o
o
a
n
= a + a + a + a + ...
1
2
3
4
Convergency
Infinite Geometric series
Telescoping Series
(a  a )+ (a  a )+ (a  a )+
a+(a + a )+ (a + a )+ (a+ a )+
lim S
n>
0
0
n
= L
=
lim S
n>
0
0
n
0
0
The Series converges
The Series diverges
\
\
\
\
n=0
o
o
a
n
r
= a + ar + ar + ....
2
, if r < 1

a
1r
=
Infinite geometric series converges when :
r < 1
>
converges to
a
/
1r
1r1 > 1
>
diverges
1
2
2
3
3
4
1
2
2
3
3
4
4
The series will only converge
if lim S = L
n> 0
0
The n term test
th
lim a
n> 0
0
lim a
n> 0
0
n
n
= 0
= 0
\
Diverges
Diverges or converge
\
\
\
\
n=0
o
o
\
\
\
\
n=0
o
o
\
\
\
\
n=0
o
o
Tests of convergance
Integral Test
Ratio test
Alternating series test
Comparison tests
b a c are nonegative series
\
\
/
\
\
/
/
/
n+1
n

Answer
by telescoping
\
\
/
\
\
/
\
\
/
\
\
/
S
n
=
2
1
2
3
\
\
/
\
\
/
/
/
n+1
n



+
(
)
(
(
(
+....+
(
=
\
\
/
/
/
n+1
\
\
/
1

=
0
0
lim S
n>
0
0
n
so it diverges
/
1
2
n
Answer
lim a
n>
0
0
n
= 0
so it may diverge or converge
Find another way to solve this
S
n
= 1 +
1
2
1
4
+
+
....
\
\
A geometric series that its "r" is less than 1 so it will converge to
\
1
\
2
1
1

3n+5n
2
n+4n
2
/
/
Answer
lim a
n>
0
0
n
= 3
So it diverges
n term test
th
by using
Suppose a series
\
\
\
\
n=0
o
o
a
n
, where
a
n
positive and decreasing function
lim a
a>
0
0
n
dn
k
a
converges
\
\
\
\
n=k
o
o
a
n
converges
diverges
diverges
lim a
a>
0
0
n
dn
k
a
0
0
0
0
,Converges
,Diverges
lim
n>
0
0
a
n
n+1
a
/
/
/
/
/
/
< 1
> 1
Converges
Diverges
Root test
lim
n>
0
0
a
n
n
\
\
/
< 1
Converges
Diverges
> 1
if b diverges,
a diverges
if c converges,
a converges
lim
n>
0
lim
n>
0
0
0
lim
n>
0
0
a
n
n
b
/
a
n
n
b
/
a
n
n
b
/
= c > 0
= 0
0
0
=
, so a will be the same as b
and b converges ,
so a converges
and b diverges ,
so a diverges
The series
\
\
\
\
n=1
o
o
(1)
n+1
a
n
= a  a + a  a + ..
converges
if :
a
n
's
are all positive nonicreasing
lim
n>
0
0
a
n
= 0
Absolute and conditional convergence :
<
/
<
/
n
n
n
n
n
> converges absolutely if the corresponding series of absolute values converges, if it converges but the absolute doesn't so it converges conditionaly
\
\
\
\
n=1
o
o
a
Absolute convergence test
If
\
\
\
\
n=1
o
o
a
 
n
converges, then converges
\
\
\
\
n=1
o
o
n
a
\
\
\
\
n=1
o
o
c (xa)
n
n
= c +c (xa) +..+
c (xa)
n
n
1
1
in which
center
(a) and the
coefficients
c
0
0
,c
1
,c
n
,....
,....
are constants
How to test a power series for convergence
Examples
Use root ratio to find the interval where the series converges absolutely.
 x  a  < R or a  R < x < a + R
Then, test for convergence at each endpoint. Use any kind of tests.
If the interval of absolute convergence is
 x  a  < R
so the series diverges for
 x  a  > R
\
\
\
\
n=0
o
o
(x+5)
n
\
\
\
\
n=0
o
o
3
n!

x

n
n
Answers
By applying ratio test :
lim
n>
0
0
a
n
n+1
a
/
/
/
/
/
< 1
lim
n>
0
0
(x+5)
(x+5)
n+1
/
/
n
/
/
/
/
< 1
=>
lim
n>
0
0
 x+5 
< 1
6 < x < 4
at (4) =>
\
\
\
\
n=0
o
o
at (6) =>
\
\
\
\
n=0
o
o
(1)
n
(1)
n
=> Diverges, by applying n term test
=> Diverges, by applying n term test
th
th
By applying ratio test :
lim
n>
0
0
a
n
n+1
a
/
/
/
/
/
< 1
=>
lim
n>
0
0
/
/
3
x
(n+1)!
/
/
/
/
/
/
n+1
n+1
3
.
n!
3
x
n
n
< 1
= 3x lim ( ) < 1
Converges for all x
1
n+1
/
n>
0
0
0
0
0
0
=
/
. . . . . . .
. . . . . . .
. . . . .
n
n
Suppose
1
2
3
4
and
The team members :
1Yara Tarek
2Mirna Magdy
3Amar Loay Jahin
4Mohamed Asem Torky
5Amr Mamdouh
6Mohamed Hesham
so let's have some examples
Thank You
n
1
2
3
4
= S
n
n
n
n
n
n
n
n
so