Proving Convergent Sequence Limit Equality

[e179285]

If( a_n) convergent sequence,prove that lim n goes to infinity a_n = lim n goes to infinity a_2n+1.


I think a_2n+1 is subsequence of (a_n ) and for this reason their limit is equal.

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

 

[DryRun]

If( a_n) convergent sequence,prove that lim n goes to infinity a_n = lim n goes to infinity a_2n+1.

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?

 

[DryRun]

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

 

[LCKurtz]

If( a_n) convergent sequence,prove that lim n goes to infinity a_n = lim n goes to infinity a_2n+1.


I think a_2n+1 is subsequence of (a_n ) and for this reason their limit is equal.

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

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.