# Sequence Prove

1. May 1, 2012

### e179285

If( an) convergent sequence,prove that lim n goes to infinity an = lim n goes to infinity a2n+1.

I think a2n+1 is subsequence of (an ) and for this reason their limit is equal.

but ı don't know where and how to start..

2. May 1, 2012

### sharks

Your wording isn't very clear to me and doesn't make much sense. Is that exactly how the question appears in your book/notes?

3. May 1, 2012

### sharks

OK, your problem appears to be:
$$\lim_{n\to\infty} a_n = \lim_{n\to\infty} a_{2n+1}$$
You are already told that $a_n$ converges, so you have to show that $\lim_{n\to\infty} a_{2n+1}$ also converges to the same limit.

Every subsequence of a convergent sequence is convergent, with the same limit. So, if $a_n$ converges, then $a_{2n}$ and $a_{2n+1}$ are convergent as well.

Let $\lim_{n\to\infty} a_n = L$, then $\lim_{n\to\infty} a_{2n} = \lim_{n\to\infty} a_{2n+1} = L$

Last edited: May 1, 2012
4. May 1, 2012

### LCKurtz

Of course the limit of a subsequence is the same as the limit of the sequence. But it looks to me like that's what you are trying to prove, albeit in a special case. So prove it from the basics. Write down an $\epsilon - n$ definition of what it means for the original sequence to have a limit, then write down the same kind of statement for what you need to prove. Use what you are given to get what you need.