# 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?

## 4 Answers

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

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 $L$.

Just use an $\epsilon$-$\delta$ argument.

Choose $N$ large enough so that if $n>N$, then $|a_{2n}-L| < \epsilon$ and $|a_{2n+1}-L| < \epsilon$. Then if $m> 2N+1$, you have $|a_m-L| < \epsilon$. Hence $\lim_n a_n = L$.

There's a nice generalization to this: let $\,\{A_i\}_{i\in I}\,$ be a finite partition of the naturals $\,\Bbb N\,$ , with $\,|A_i|=\aleph_0\,\,\forall\,i\in I\,\,\,,\,|I|<\infty$ , and s.t. for a sequence $\,\{x_n\}\,$ there exists a number $\,\alpha\,$ s.t. we have $$\lim_{m\to\infty}\{x_m\;\;|\;\;m\in I\}=\alpha\,\,,\,\,\forall\,\,i\in I$$ then $$\lim_{n\to\infty}x_n=\alpha$$ The other way around is also true, of course, so the above can be put in iff form.

• I think you have to assume the partition is finite. – Dejan Govc Aug 27 '12 at 19:40
• I think you are right, @DejanGovc, since otherwise it may not be possible to get a bound for the number $\,N\,$ from which all the indexes fulfill $\,|a_k-\alpha|<\epsilon\,\,,\,\forall\,k>N\,$...Thanks, I'll edit my answer – DonAntonio Aug 28 '12 at 1:47