
The limit of a sequence,
theorems 
Sufficient condition for convergence of a sequence 
The Cauchy criterion (general principle of convergence) 
Some
important limits 





Sufficient condition for convergence of a sequence
 The Cauchy criterion (general principle of convergence) 
A
sequence of real numbers, a_{1},
a_{2},
. . . ,
a_{n},
. . .
will have a finite limit value or
will be convergent if for no matter
how small a positive number e
we take there exists a term a_{n}
such that the distance between that term
and every term further in the sequence is smaller than e,
that is, by moving further in the sequence the
difference between any two terms gets
smaller and smaller. 
As
a_{n + r},
where
r = 1, 2, 3,
. . . denotes any term that follows a_{n},
then 

a_{n + r} 
a_{n}
 < e
for all n >
n_{0}(e),
r
= 1, 2, 3,
. . . 
shows
the condition for the convergence of a sequence. 
If
a sequence {a_{n}}
of real numbers (or points on the real line) the distances between which
tend to zero as their
indices tend to infinity, then {a_{n}}
is a Cauchy sequence. 
Therefore,
if a sequence {a_{n}}
is convergent, then {a_{n}}
is a Cauchy sequence. 

The Cauchy criterion
or general principle of convergence,
example 
The
following example shows us the nature of that condition. 
Example: We
know that the sequence 0.3,
0.33,
0.333,
. . . converges to the number 1/3
as 
1/3
= 0.33333
. . . . Let
write the rule for the n^{th}
term, 

If we go along the sequence far enough, say to the 100^{th} term, i.e., the
term with a hundred 3's in the fractional
part, then the difference between that term and every next term is equal
to the decimal fraction with the
fractional part that consists of a hundred 0's followed by 3's on the
lower decimal places, starting from the
101^{st} decimal place. That is, 

Therefore,
the absolute value of the difference falls under 


Then,
if we go further along the sequence and for example calculate the
distance between the 100000^{th} term 
and the following
terms, the distance will be smaller than 


Hence,
since we can make the left side of
the inequality 
a_{n + r} 
a_{n}
 < e as
small as we wish by choosing n
large enough, then all terms that follow a_{n}
(denoted a_{n + r},
r = 1, 2, 3,
. . . ), infinitely many
of
them, lie in
the interval
of the length 2e symmetrically
around the point a_{n}.
Outside of that interval there is only
a finite number of terms. That is, 

e <
a_{n + r} 
a_{n}
< + e for all
n >
n_{0}(e),
r
= 1, 2, 3,
. . . 
or a_{n}

e <
a_{n + r} < a_{n}
+ e. 
So,
the
terms of the sequence, starting from the (n +
1)^{th} term, form the
infinite and bounded sequence of numbers
and so, according to the above
theorem, they must have at least one cluster point that lies in that
interval. But they cannot have more than one cluster point since all
points that follow the n^{th} term lie inside the interval
2e length of
which is arbitrary small, if n
is already large enough, so that any
other cluster point will have to be outside of that interval. 
Thus,
the
above theorem simply says that if a sequence converges, then the terms of the sequence
are getting closer and closer to each
other as shows the example. 

Some
important limits 
(1)
Let examine convergence of the sequence given by a_{n}
=  a ^{n} 
a) if  a 
> 1 then we can write 
a  = 1 + h,
where h
is a positive number. 
So,
by the binomial theorem 

If
we drop all terms beginning from the third that are all positive since
the binomial coefficients are natural numbers, if n > 2
and h
> 0, the right side become smaller, so obtained is the Bernoulli's
inequality 
(1 + h)^{n}
> 1 + nh,
n > 2. 
When
n
®
oo
then 1 + nh
tends to the positive infinity too, since we can make 1 +
nh
greater than any 
given positive number N,
if only we take 

therefore will even more tend to infinity
 a ^{n}
which is 

greater.
Thus, 

or 




b) if
0 <  a  < 1 then we can write 



Since b
> 1 then 

such that  a ^{n}
< e
whenever 

however small e is, 

this
inequality can be satisfied by choosing n
large enough. Therefore, 

or 




(2) Let examine convergence of the sequence given by 


The
sequence 

the nth
term of which we can write as 


For
every  a  >
1 there exists a natural number m
such that m <
 a  < m + 1
and n
> m
then 

since 

it follows that a_{n}
® 0
or we write 



(3) Let examine convergence of the sequence given by 


a) If a
> 1 then the sequence 

is decreasing, that is 


Let 

then by the Bernoulli's
inequality a
> 1 + nh
so that 


Since the numerator a

1 is fixed number then,
if n ® 0
then h ® 0
too, therefore 


So we
can write 



b) If
0 < a < 1 then the sequence 

is increasing, that is 


For
example, 



If
we write 



so it
follows that 

Therefore,




c) If
a = 1 then 




Since
in all three cases above, a), b) and c) we've got the same result,
then we can write 











Calculus
contents B 



Copyright
© 2004  2020, Nabla Ltd. All rights reserved. 