28th March 2004
Notes by Omar Shouman
Calculus Lecture
Dr.Kuenzer
Sequences and Series1
Aim
ex  1 
x x 2 x3 x 4
    ...... "Infinite Polynomial"
1! 2! 3! 4!
Compare it with 1 1 1 1 1  ............   , you find out that this is different.
Because the latter sum does NOT have a real value. So we will have to clarify convergence
questions.
Section 1 Sequences
A sequence is a sequence of complex numbers or a list of numbers written in a definite order.
def
(am , am1 , am2 ,........)  (am ) n  m  (an ) n
Example 1
(i, i 2 , i 3 ,......)  (i n ) n1
 (i, 1, i,1, i, 1, i,1,........) periodic
This sequence is divergent (not convergent).
Example 2
1 1 1 1 1
( , , , , ,........)  ( 1 ) n1 This sequence converges to zero.
n
1 2 3 4 5
Definition
Let (an )nm be a sequence.
It converges to a 
:
If for any   0 there exists an N  m such that an  a   for all n  N
1
Chapter 12 in Stewart Calculus
1
Example
1
( ) n1 converges to 0 as n  
n
Why?
Using the definition, we get for   0 :
1
1
 0   as soon as n 

n
Thus, we may take for N (any integer) 
1

.
Definition
If (an )nm converges to a, we write lim an  a or an  a .
n
If (an )nm is not convergent, it is called divergent and lim an does NOT exist.
n
Limit Laws for Sequences
We assume that (an )nm ,(bn )nm both converge. (Let this hold for all the rules.)
1) For  ,   :
lim( an  bn )   (lim an )   (lim bn ) .
n
n
n
CAUTION
Note that lim(n  (n))  0 . Nevertheless, it is NOT EQUAL TO
n
lim n  lim(n) because both are divergent.
n
n
Therefore, only use this rule from the R.H.S to the L.H.S.
This means that the sum of the limits is equal to the limit of the sums only when the limits
exist and have a value. So you have to make sure that both terms on the R.H.S converge and
then conclude that the L.H.S is equal to the sum of the two terms.
2
2) lim(an .bn )  (lim an ).(lim bn ) .
n
n
3) lim(
an
n
4) If
bn
n
)  (lim(an )) (lim(bn )) , this holds only if lim(bn )  0 .
n
n
n
f : R  R is continuous (or on some other interval), then
lim f (an )  f (lim an )
n
n
EXAMPLES
1)
n 3  2n
( divide by n 3 )
n 2n 3  n
2
2
1 2
lim(1  2 )
n  n
n  1 0  1 .
 lim
n
1
1
2  2 lim(2  2 ) 2  0 2
n

n
n
lim
2)
lim e(
n 1 
n) n
because the function e x is continuous. We can use rule 4 to get:
n
lim ( n 1 
 en
n) n
Calculate the limit separately: (we multiply by ( n  1  n ) )
lim
( n  1  n )( n  1  n )
n 1  n
n 1 n
. n  lim
. n  lim
n
n 
n 1  n
n 1  n
1
1
1
1
1
 lim
 lim
 lim

 .
n 
n

n

n 1  n
n 1
1
1 0 1 2
1
1 1
n
n
n
n 
n 
Now, we substitute back:
lim e(
n
n 1 
n) n
1
=e 2 .
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Section 2 Series
In dealing with series, we are interested in the "infinite sum" of a sequence.
2.1 What is it?
Given (an )n  m a sequence of complex numbers,
Consider the new sequence of partial sums:
k
( an ) k m   am , am  am1 , am  am1  am2 ,...... This sequence will lead to the sum that
nm
we want to calculate.
Definition
 k

a

lim
an  if the limit exists


n

k 
n m
 n m 


In this case, we say
a
nm
n
is convergent.
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