Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 the sequence $\{a_n\}_{n=0}^\infty$ and the subsequences: $a_{3n}, a_{2n+1}, a_{2n}$ which converge. Prove that $a_n$ is a convergent sequence.

Thanks you very much.

share|cite|improve this question
all I could think about was to use the fact that if a sequence is convergent then all of its subsequences that converge must covnerge to the same limit of the sequence itself; maybe I could assume that these subsequences converge to different limits and get a contradiction? – Anonymous Apr 30 '12 at 17:50
Yes, you could approach it by contradiction, but there's an easier way. Let $L_{\text{odd}}=\lim_n a_{2n+1}$ and $L_{\text{even}}=\lim_n a_{2n}$, and use the convergence of $\langle a_{3n}\rangle$ to show that $L_{\text{odd}}=L_{\text{even}}$. – Brian M. Scott Apr 30 '12 at 17:53
up vote 3 down vote accepted

Hint: first show that the three subsequences have the same limit. (The subsequence $(a_{3n})$ has a further subsequence that is a subsequence of $(a_{2n})$, for instance.)

Then note that given $n>1$, $a_n$ is a term of one of the subsequences $(a_{2n})$, $(a_{2n+1})$. (So, given $\epsilon>0$, choose $N$ so that for any $n\ge N$, each of $a_{2n}$ and $a_{2n+1}$ is within $\epsilon$ of the common limit. Then... .)

share|cite|improve this answer
"The subsequence $(a_{3n})$ has a further subsequence that is a subsequence of $(a_{2n})$" - that's clear; however, why can I deduce that therefore $(a_{3n})$ and $(a_{2n})$ have the same limit? – Anonymous Apr 30 '12 at 18:06
@Anonymous Suppose $(a_{2n})$ converges to $L$. Then each of its subsequences converges to $L$. So some subsequence of $(a_{3n})$ converges to $L$. So $(a_{3n})$ converges to $L$ (since $(a_{3n})$ is convergent). – David Mitra Apr 30 '12 at 18:09
OK, awesome; therefore all the three subsequences have the same limit. let $n>1$, I understand that $a_n$ is a term of the subsequences $(a_{2n})$, $(a_{2n+1})$. how do I continue from here? I know that if n is even I can write: $|a_{2n}-L|<\epsilon$ and if n is odd then $|a_{2n+1}-L|<\epsilon$; how can I deduce now that $|a_n-L|<\epsilon$? – Anonymous Apr 30 '12 at 18:17
@Anonymous Finishing up the parenthetical comment at the end of my answer: For $n\ge N$, both $a_{2n}$ and $a_{2n+1}$ are within $\epsilon$ of $L$. But then every term $a_{2N}$, $a_{2N+1}$, $a_{2N+2}$, $a_{2N+3}$, $\ldots\,$ is within $\epsilon$ of $L$. Note those latter terms are a tail of $(a_n)$; so, for the sequence $(a_n)$, if $n\ge 2N$ ... . – David Mitra Apr 30 '12 at 18:32
finally got it; as always clear as crystal; thanks again for your great help. – Anonymous Apr 30 '12 at 18:49

Note that it suffices to show that, $$ \lim a_{2n} = \lim a_{2n + 1}.$$ As $ a_{6n} $ is a subsequence of $a_{2n}$ and $a_{3n}$. We have $ \lim a_{2n} = \lim a_{3n}$ And as $a_{3n} = a_{2n_m + 1} $. We have $$\lim a_{2n + 1} = \lim a_{3n} = \lim a_{2n}. $$

share|cite|improve this answer

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.