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
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