Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?.

share|improve this question
    
It would be better if you could accept the answers for your previous questions before people answer this particular question. –  Paul Nov 27 '11 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".) –  Martin Sleziak Nov 27 '11 at 14:41

1 Answer 1

I think your problem is that you work with incorrect definition of Cauchy sequence. The correct definition is the following:

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.