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I was asked this question:

Prove that $a_n$ converges if and only if:

$a_{2n},a_{2n+1},a_{3n}$ all converge

I thought this was an easy generic question until I read the hint which said: Note: It is not required that the three sub-sequences have the same limit. This needs to be shown

This is what is confusing me because I have found two sources stating something different:

Proposition 4.2. A sequence an converges to L ∈ R if and only if every subsequence converges to L.

and

Let $a_n$ be a real sequence. If the subsequence $a_{2n}$ converges to a real number L and the subsequence $a_{2n+1}$ converges to the same number L, then $a_n$ converges to L as well.

So my question is: for a sequence $a_n$ to converge does it's subsequences have to converge to the same limit? (I suspect not) and if the answer is no can you help me prove why?

Thanks in advance

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  • $\begingroup$ If a sequence converges its subsequences must converge to the same limit. This follows directly from the definition of the limit of a sequence. $\endgroup$ – John Douma Dec 1 '15 at 21:53
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If a sequence has two subsequences that do not both converge to the same limit, then the sequence does not converge.


This can be proven using the $\epsilon-\delta$ definition of convergence:

  • Let $a_n$ converge to $L$, and let $\{a_{n_k}\}_{k\to\infty}$ be a subsequence of $a_n$.
  • Let $\epsilon > 0$.
  • Then, because $a_n$ converges to $L$, there exists some $N$ such that if $n>N$, then $|a_n - L|<\epsilon$.
  • Because $n_k$ is an increasing sequence of integers (by definition of a subsequence), there exists such $K$ that $n_K > N$.
  • Then, if $k > K$, we have $n_k > n_K > N$, and from the previous point, we get that $|a_{n_k} - L|<\epsilon$, meaning that $a_{n_k}$ converges to $L$.

However, your case is special in that there is an overlap of sequences, for example $a_6$ is in the first and third sequence, and $a_9$ is in the second and third sequence. In fact, the third sequence alternates between the first two sequences, and you can use this to prove all three limits must be equal.

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  • $\begingroup$ Ok so for $a_{2n}$ and $a_{2n+1}$ I can say we are either in the odd or the even case and then $a_{3n}$ goes between the two cases. $\endgroup$ – babylon Dec 1 '15 at 22:10
  • $\begingroup$ @babylon Yup. And that, forces the sequences to converge to the same number. Then you need to prove that the whole sequence also converges to the same number. $\endgroup$ – 5xum Dec 1 '15 at 22:16
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I think the word "required" in the hint is confusing. Focus instead on the last sentence:

Note: It is not required that the three sub-sequences have the same limit. This needs to be shown

In other words, you need to show that as a result of the given hypotheses, the three sub-sequences do have the same limit. You need to do that precisely because it is a necessary condition for $a_n$ to converge.

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