# series

Document Sample

```					                                           Infinite Sequences and Series

A sequence is an ordered set of numbers a n   a1 a 2 a3 a 4 a5  a n  ,
where an denotes the nth term. The sequence converges if lim a n = L (and L is
n 

finite), otherwise it diverges.
Note: if lim a n = we say that the limit exists, but the sequence diverges.
n 

Theorem 1) Sandwich Theorem: if an < bn < cn, and lim a n = lim cn = L, then
n       n

lim bn = L also.
n

Theorem 2) If lim | an | = 0, then lim a n = 0.
n                          n 

Theorem 3) If f is a function defined for real numbers x, lim f ( x) = L and f(n) =
x 

an, then lim a n = L also. (This allows the use of l'Hopital's rule to find the limits
n 

of sequences.)
Theorem 4) A geometric sequence {xn} has limit zero for |x| < 1, but
n
lim x does not exist for |x| > 1 and x = -1. lim 1n = 1 when x = 1
n                                                              n 

A sequence is increasing if for all values of n, an < an+1; a sequence is
decreasing if for all values of n, an > an+1. A monotonic sequence is either
increasing or decreasing. A sequence is bounded above if there exists a number
M such that M > an and bounded below if there exists a number m such that
m < an. A bounded sequence is bounded both above and below.
Theorem 5) A bounded monotonic sequence converges.

Given a sequence {ak} , if we add the terms we get a series. An infinite
                        
series is denoted by:                ak Given a series
k 1
a
k 1
k   , we can define the partial sums
n
S n   ak These form a new sequence {Sn}. If {Sn} converges to S (that is
k 1

lim Sn = S) then we say the series converges and S is called the sum of the
n 

series. Otherwise the series diverges. Although it is sometimes possible to find
the exact value of a series, in general this is a very difficult problem. The first
step with a series is to determine whether it converges or diverges. If it does
converge, the sum can usually be estimated numerically.

Geometric Series:            x
k 0
k
converges if |x| < 1 and diverges otherwise.

The sum of this series is 1/(1 - x).
                                    n
1                                                    1
Harmonic Series:   k
k 0
The partial sums are usually denoted by H n  
k 1 k

Since Hn grows like ln(n), this series diverges.

Theorem 6) If         a
k 1
k     converges, then lim a k = 0.
k 

                                                                      
Theorem 7) If            ak and  bk both converge,                                    then     a      k    bk converges also
k 1                              k 1                                    k 1
                                                                                                   
and     ak  bk   ak   bk Also for any constant c,
k 1          k 1                  k 1
 c ak  c ak
k 1           k 1

Tests for Convergence


Nth Term Divergence Test If lim ak then                                         a
k 1
k   diverges.

The converse of this is false. Just because the lim a k = 0 the series
k 


a
k 1
k     may not converge. Write an example of this here: ____________

Integral Test: Let f be a positive, decreasing and continuous function for x > b.
                                                                                             
If ak = f(k) then  ak converges whenever                                                   f ( x ) dx converges and          a     k
b
k 1                                                                                            k 1

diverges whenever                        b
f ( x ) dx diverges.

P - series Test: 
1
k
k 1
p
converges for p > 1 and diverges for p < 1.

                                          
Comparison Test: If 0 < an < bn then: if                                      bk converges so does
k 1
a
k 1
k   ; if
                                                    

 ak diverges so does
k 1
b
k 1
k
Limit Comparison Test: If 0 < ak , 0 < bk, lim ak bk = c and c is positive and
k 

finite then:
                     

b
k 1
k   and   a  k 1
k    either both converge or they both diverge.
                                                               
If lim ak bk = 0 and
k 
 bk converges, so does
k 1
a
k 1
k

                                                           
If lim ak bk = 
k 
and     b
k 1
k       diverges, so does                       a
k 1
k

Common series used for comparisons are geometric series and p-series.


If 0 < ak, then                    (1)
k 1
k 1
a k  a1  a 2  a3  a 4  a5  a6    (1) n 1 a n  
is called an alternating series.
Alternating Series Test: (Leibniz’ Test)

If {ak} is decreasing and lim a k = 0 then
k 
 (1)
k 1
k 1
ak converges.

                                                                                             
A series         a
k 1
k   is called absolutely convergent if                                           a
k 1
k          converges.
                                                                
A series          ak may converge while
k 1
ak 1
k     diverges. Example? _________

                                                                    
Theorem 8) Convergence of                                             ak             implies convergence of                               a        k   . This means
k 1                                                                 k 1

that all of the tests above can be applied to { |a n| } if {an} contains both positive
and negative terms.

                                                                            
an 1
Ratio Test: Given a series                                       ak , let r  lim
k 1
n      an
If r < 1 then                a
k 1
k   converges

absolutely; if r > 1 then                       a
k 1
k       diverges and if r = 1 no conclusion can be drawn.

                                                                                         
Root Test: Given a series                                        ak , let r  lim n | an |
k 1
n 
If r < 1 then              a
k 1
k


converges absolutely; if r > 1 then                                             a
k 1
k   diverges and r = 1 no conclusion can be

drawn.
Power Series


The function f(x) =         a ( x  c)
k 0
k
k
is called a power series centered at c or

about c. Of course, this function only makes sense at those points x where the
series converges. Power series, if they converge at more than one point,
converge absolutely on an open interval centered at c. The distance from the
center c to the endpoints of the interval is called the radius of convergence R. If
the interval is the whole real line then R =  . If R is finite, then the power series
diverges for |x - c| > R. Convergence at the endpoints of the interval x = c ± R
must be checked separately.

Our goal is to find a power series representation for a given function f(x).
Suppose f(x) is the given function; we want to determine the {a k} so that:

f(x) =    a ( x  c)
k 0
k
k
for |x - c| < R. By evaluating f and all it’s derivatives at c, it

f ( k ) (c )
can be shown that ak must satisfy: ak 
k!
Thus the power series for f(x) can only be the Taylor series of f(x) about x = c.

f ( k ) (c )
f(x) =                  ( x  c) k  f (c)  f '(c)( x  c)  f ''(c)( x  c ) 2 / 2! 
k 0     k!
When c = 0, we have a special case of Taylor series called the MacLaurin
Series:
f(x)  f(0)  f ' (0)x  f ' ' (0)x 2             f ' ' ' (0)x 3         ...
2!                      3!
All of this assumes that the given function f(x) has a power series
representation. How can we be sure that this is the case? Here is an example of
a function which is not equal to its Taylor series except at x = 0: f(x) = e1 x for
2

x  0, and f(0) = 0. It can be shown that all of the derivatives of f at zero are zero.
Thus the Taylor series for f is the function T(x) = 0 for all x, while f(x) is never
zero when x  0.
Taylor's formula (with remainder) tells us when f(x) is equal to its Taylor
series. Suppose that f(x) has n+1 derivatives in a interval I containing the points
x and c. Then there exists a number z between x and c such that:
f(x) = f ( x)  f (c)  f '(c)( x  c)  f ''(c)( x  c) 2 / 2!   f ( n ) (c)( x  c) n / n !  Rn ( x)
where Rn ( x)  f ( n 1) ( z )( x  c) n 1 /(n  1)! Rn ( x) is called the remainder term. Note
that Rn ( x) looks just like the (n+1)st term in the Taylor series except that the
(n+1)st derivative is evaluated at a point other than c. Now f(x) = Tn ( x ) + Rn ( x)
and f(x) is equal to its Taylor series expansion provided that lim Rn ( x)  0 .
n

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 5 posted: 12/2/2011 language: English pages: 4
How are you planning on using Docstoc?