# Cauchy sequence and convergence

A sequence is said to be Cauchy sequence if for given any integer n, there exists a positive real number R, such that for any n1, n2 > n, mod{n1th term - n2th term}

1,0,1,0,1,0,1,0........

Now for any integer n, whenever n1,n2 >n mod{n1th term - n2th term} < or equal to 1. So as per the definition of a Cauchy sequence we can say that this sequence is a Cauchy sequence, however, this is not a convergent sequence. How come? This implies there is gap in my understanding, can anyone kindly point out where am I wrong?.

Edited version:

A sequence $(a_n)$ is said to be Cauchy sequence if for given any integer $n$, there exists a positive real number $R$, such that for any $n_1, n_2 > n$, $|a_{n_1}-a_{n_2}|<R$.

We can prove that every Cauchy sequence is a convergent sequence.

Now let us consider the following sequence, $1,0,1,0,1,0,1,0,\ldots$

Now for any integer $n$, whenever $n_1,n_2 >n$ we have $|a_{n_1}-a_{n_2}|\le 1$ .

So as per the definition of a Cauchy sequence we can say that this sequence is a Cauchy sequence, however, this is not a convergent sequence. How come?

This implies there is gap in my understanding, can anyone kindly point out where am I wrong?.

• It would be better if you could accept the answers for your previous questions before people answer this particular question.
– Paul
Nov 27, 2011 at 14:21
• Primeczar: Pleas have a look at the edited version, whether this is what you wanted to ask. (I was not sure what you meant by "mod".) Nov 27, 2011 at 14:41

A sequence $(a_n)$ is said to be Cauchy sequence if for any given $\varepsilon>0$, there exists a positive integer $n$, such that for any $n_1, n_2 > n$ the inequality $|a_{n_1}-a_{n_2}|<\varepsilon$ holds.
Now if you choose $\varepsilon=\frac12$, you can see that your sequence $1,0,1,0,1,0,\ldots$ is not a Cauchy sequence.