The limit of a sequence, theorems
Sufficient condition for convergence of a sequence
Some important limits
Sufficient condition for convergence of a sequence - The Cauchy criterion (general principle of convergence)
A sequence of real numbers, a1, a2, . . . , an, . . .  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 an 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 an + rwhere  r = 1, 2, 3, . . .  denotes any term that follows an,  then
| an + r - an | < e   for all  n > n0(e) r = 1, 2, 3, . . .
shows the condition for the convergence of a sequence.
If a sequence {an} of real numbers (or points on the real line) the distances between which tend to zero as their indices tend to infinity, then {an} is a Cauchy sequence.
Therefore, if a sequence {an} is convergent, then {an} 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 nth term,
If we go along the sequence far enough, say to the 100th 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 101st 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 100000th term
 and the following terms, the distance will be smaller than
Hence, since we can make the left side of the inequality  | an + r - an | < e  as small as we wish by choosing n large enough, then all terms that follow an (denoted  an + r,   r = 1, 2, 3, . . .  ), infinitely many of them, lie in the interval of the length 2e symmetrically around the point an. Outside of that interval there is only a finite number of terms. That is,
- e < an + r - an < + e   for all  n > n0(e) r = 1, 2, 3, . . .
or            an - e < an + r < an + 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 nth 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  an = | 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 n-th 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  an ® 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