# Sequence is convergent if it has a convergent subsequence

1. Oct 23, 2012

### gottfried

1. The problem statement, all variables and given/known data
Show that an increasing sequence is convergent if it has a convergent subsequence.

3. The attempt at a solution
Suppose xjn is a subsequence of xn and xjn→x.

Therefore $\exists$N such that jn>N implies |xjn-x|<$\epsilon$
It follows that n>jn>N implies |xn-x|<$\epsilon$

Therefore xn→x

The solution that I've been given is much more complicated I'm just wondering whether my simpler solution is also correct.

2. Oct 23, 2012

### jbunniii

It's true, but WHY does it follow? This is the key part of the proof, so you need to be explicit. You need to use the fact that $x_n$ is an increasing sequence. This would not be true for an arbitrary sequence.

3. Oct 23, 2012

### gottfried

The original question was an 'if and only if question' which means the reverse also had to be proved ie:

Show that if xn → x that any subsequence of (xn) also converges to x
My solution which is simliar to the answer given is

If xn → x then given any  $\epsilon$> 0 there is an N such that n > N implies |xn - x| < $\epsilon$ . Now
consider a subsequence (xjn). Then since jn $\geq$ n > N we have that for any n > N,
|xjn - x| < $\epsilon$  and so we conclude that the subsequence has the same limit

I believe this correct and this raises the question why does logic work in the one direction but no the other.

4. Oct 23, 2012

### jbunniii

It does work in both directions. If EVERY subsequence of $x_n$ converges to $x$, then $x_n$ converges to $x$. This is trivial, because $x_n$ is a subsequence of itself.

But the hypothesis in the first part is weaker: $x_n$ has *a* convergent subsequence. Without the additional assumption that $x_n$ is increasing, this would not be enough to conclude that $x_n$ converges.

A general sequence can have some subsequences which converge, and others which do not. For example, let $x_n = 0$ if $n$ is even, and $x_n = n$ if $n$ is odd. The subsequence consisting of even indices converges, and the subsequence consisting of odd indices diverges. And of course the sequence itself does not converge.

5. Oct 23, 2012

### gottfried

Thanks it makes sense.