Given a sequence $a_{n}$, if I know that the sequence of even terms converges to the same limit as the subsequence of odd terms:
$$\lim_{n\rightarrow\infty} a_{2n}=\lim_{n\to\infty} a_{2n-1}=L$$
Is this sufficient to prove that the $\lim_{n\to\infty}a_{n}=L$?
If so, how can I make this more rigorous? Is there a theorem I can state that covers this case?
Best Answer
You can prove it easily enough. For any $\epsilon>0$ there are $n_0,n_1\in\Bbb N$ such that $|a_{2n}-L|<\epsilon$ whenever $n\ge n_0$ and $|a_{2n+1}-L|<\epsilon$ whenever $n\ge n_1$. Let $m=\max\{2n_0,2n_1+1\}$; then $|a_n-L|<\epsilon$ whenever $n\ge m$, so $\lim_{n\to\infty}a_n=L$.