Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 10 down vote accepted

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

share|cite|improve this answer

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 $L$.

share|cite|improve this answer

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 $\lim_n a_n = L$.

share|cite|improve this answer

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.