

Sequences
and limits 
The limit of a sequence 
The definition of the limit of a sequence 
Convergence of a sequence 
Verifying the convergence of a sequence from the definition, examples 





The limit of a sequence 
The definition of the limit of a
sequence 
A
number L is
called the limit of a sequence {a_{n}}
if
for every positive number e there exist a natural number 
n_{0}
such that if
n
> n_{0},
then 
a_{n}

L

< e. 
That
is, a
number L is the limit of a sequence
if the distance between the term a_{n}
and L
becomes arbitrary small
by choosing n
large enough (see the examples above). The n_{0}
denotes the value of the index n
starting from which the distance a_{n}

L
becomes smaller than the given e. 
Since the value of n_{0}
depends on the size of e
it is usually written as n_{0}(e). 

Convergence of a sequence 
Therefore,
if a sequence {a_{n}}
has a limit L
we write 


or 
a_{n}
®
L
as n
®
oo



and
we say that that a sequence a_{n}
has the limit L
as n
tends to infinity. 
A
sequence {a_{n}}
is
convergent if it has a limit. Otherwise, we
say the sequence is divergent. 
Thus,
the sequences, (3),
(4)
and (5),
from the example above, all converge or tend to the limit 1. 

Example: Let
examine the limit of the sequence (3)
given by 



Solution:
Prove that 



The
sequence 

tends
to the limit 1
as the distance 

can
become 

arbitrary small by
choosing the natural number n
sufficiently large. 
For
example, if we choose
n = 100
the distance between a_{n}
and 1
is 1/n
= 0.01
that is, 
for
all n > 100
the distance  a_{n}

1  < 0.01. 
Therefore,
n_{0}(e)
= 101 meaning, starting from the 101^{st}
term further, the distance of the remaining terms of the
sequence and 1,
is always less than 0.01. 

Verifying
the convergence of a sequence from the definition, examples 
Example: Find
the limit of the sequence (2)
given by 



Solution:
Let
prove that 



Using
the definition of limit we must find a natural number
n_{0}(e)
such that  a_{n}

0  < e
for all n >
n_{0}. 
Therefore,
if n >
n_{0} then
1/n 
0  = 1/n  = 1/n < 1/n_{0} < e. 
Suppose
we wish to make the difference (or the distance) between the a_{n}^{th}
term and
the limit L
to be less than e
= 0.001 = 1/1000. 
Then,
as 1/n < e
or n
> 1/e
it follows n
> 1000
that is, starting from n_{0}
= 1001
the distance a_{n}

L becomes smaller than the given e. 

Observe
that the absolute value of terms of the sequences that converge to zero
become arbitrary small as n tends
to infinity that is,
 a_{n}
 < e
for all n >
n_{0}(e). 
For
example such sequences are, 

therefore 




The
same way we can prove that the sequence (5)
above,
given by 

converges
to 1. 

As
can be seen on the number line above, the
terms of the sequence alternate from left to right approaching closer and closer to
1 as n
tends
to infinity. 
That
is, the sequence 

alone alternately converges from left to right approaching closer and 

closer to 0 as
n
tends to infinity. 

Example: Find
the limit of the sequence,
0.9,
0.99, 0.999, . . . , 0.999 . . . 9 . . . , .
. .

Solution: The
terms of the sequence can be written as


Therefore,





Example:
The
sequence


has
the limit
3/2,
starting
from which term 
a_{n} 
L  < 0.01.



Solution: Substitute
given values into the inequality 
a_{n} 
L
 < 0.01,


Check the result by plugging
n_{0}(e)
= 125
into 
a_{n} 
L  < 0.01.









Calculus
contents B 



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