Sufficiency to prove the convergence of a sequence using even and odd terms

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?

Solution 1:

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 $\lim_{n\to\infty}a_n=L$.

Solution 2:

If you are familiar with subsequences, you can easily prove as follows. Let $a_{n_k}$ be the subsequence which converges to $\limsup a_n$. it is obviously convergent and contain infinitely many odds or infinitely many evens, or both. Hence, $\limsup a_n = L$. The same holds for $\liminf a_n$, hence the limit of the whole sequence exists and equals $L$.