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Let $\{a_n\}$ be a sequence and $k$ a natural number so that: $\{a_{nk}\}$ , $\{a_{nk+1}\},\ldots, \{a_{nk+(k-1)}\}$ - every sub-sequence converges to the same limit $L$.

I need to show that $\{a_n\}$ itself converges to $L$.

It is pretty obvious to me why, regarding the fact that from $N=k$, every $a_n$ belongs to one of the sub-sequences above, so we have $N$ that from it and on all numbers will close to $L$, close as we want.

I just don't remember how make it to a formal prove.

I tried to assume the opposite and get Contradiction, but It didn't work.

Thank you

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Hi -- you've asked quite a number of questions now. I think it's about time you familiarize yourself with how to format them so people don't have to keep wading through this mangled notation to read them. This might help you to get started:… – joriki Apr 4 '11 at 17:44
@joriki: i'm trying to paste what I've wrotr in but it doesn't work correctly, what seems to be the problem? is there any special way to paste it? – user6163 Apr 4 '11 at 18:08
@Nir: You can look at what other people do to format by right-clicking on a formula and selecting "View Source". In any case, you should know that it is extremely bad practice to use capital and lower case letters interchangeable; most mathematicians would not consider K and k to necessarily represent the same variable, nor n and N, yet you use them interchangeably (and do this often). – Arturo Magidin Apr 4 '11 at 18:10
@Nir: You need to enclose any $\TeX$ you enter in dollar signs, like this: \$x\$. More generally, as suggested in the thread I linked to above, look at other questions (and answers) by clicking on "edit" (you don't have to be the author; you don't even have to be allowed to edit them; you can always get the source by clicking on "edit"). – joriki Apr 4 '11 at 18:11
@Nir,joriki: You don't have to click on edit. Hover over the formula, right-click, and click on "Show Source"; it will not show you the dollar signs, but will show you the actual LaTeX code. Enclose it in $ signs to get the effect. @Nir: do keep your lower and upper case letters separate. $A\neq a$, $K\neq k$, $N\neq n$, etc. – Arturo Magidin Apr 4 '11 at 18:21
up vote 2 down vote accepted

You nearly wrote out the proof already. In short: use your info about the subsequences to go far enough along in all of them so that each is $\varepsilon$-close to $L$. Convert those subsequence indices into indices of the original sequence, and take their max. This lets you know how far in the original sequence you have to go to be sure that no matter which subsequence a term belongs to, that subsequence is already close enough to $L$. This can be done as a direct argument.

If we want to show that $\{a_n\}$ converges to $L$, then we need to show that for any $\varepsilon > 0$, $\{a_n\}$ eventually stays within $\varepsilon$ of $L$.

So, pick some such $\varepsilon > 0$ to start. We want to find some $N$ so that $|a_n - L| < \varepsilon$ for all $n > N$.

You've noticed already the key idea that every natural number (index) is of the form $k\cdot m + b$ for some integers $m$ and $b$, so that each term is a member of one of the subsequences. You've also noticed that within each of those subsequences, we are eventually as close to $L$ as we want. That is, for each of these $k$-many subsequences $\{a_{k\cdot n + b}\}$, there is an $N_b$ such that from the $N_b$-th term onward, we are within $\varepsilon$ of $L$. Note that the $N_b$-th term of such subsequences has index $k\cdot N_b + b$ in the original sequence.

So, we can write down the index for each subsequence after which the subsequence is close to $L$, and we know that all terms in the original sequence fall into one of these subsequences. You know somehow that we just need to ensure that we are far enough along our original sequence so that all of the subsequences are close enough.

I guess the one thing you're missing is that, we're lucky here, because there's only finitely many subsequences to worry about. So we can just take the max over the subsequences. That is, if we take $\max\{k\cdot N_b + b\}$ over the different choices of $b$, this will give us an index in the sequence $\{a_n\}$, after which, all terms are members of a subsequence $\{a_{k\cdot n + b}\}$ after the $N_b$-th term, hence close to $L$.

That should be enough to be considered a formal proof (maybe with a bit more explicitly giving the inequality rather than saying "close" where I did), and it's just a direct argument. I suspect your difficulty (since this is where I'm struggling to get good notation :P ) was somehow in the notation for talking about transitioning between terms "so far along" in the subsequences, and "correspondingly far along" in the original sequence.

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Thank you very much, very helpful answer. – user6163 Apr 4 '11 at 19:03

If $A_n$ converges to $M\neq L$. Then $A_{2n}$ converges to $M$ as well (contradiction!)

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Do you believe that if a sequence does not converge to $L$ then there exists $M\ne L$ such that it converges to $M$? – Did Apr 4 '11 at 17:55
@Didier:my mistake for not being careful reading the problem.The subsequences $\{a_{2n}\}$ and $\{a_{2n+1}\}$ converges to $L$. Hence for every $\epsilon >0$ there are $N_1,N_2$ such that $|a_{2n}-L|<\epsilon$ and $|a_{2n+1}-L|<\epsilon$ for $2n>N_1$ and $2n>N_2$. Hence for $N>\max\{N_1,N_2\}$ we have $|a_n -L|<\epsilon$ reregradless $n$ is odd or even. – user9077 Apr 4 '11 at 18:21
@user9077: The point is that you cannot assume that $a_n$ converges; of course, if it converges, then every subsequence converges to the same thing, so the limit of the sequence equals the limit of any subsequence. But you have to first establish that $a_n$ does in fact converge; you cannot merely assume it does. – Arturo Magidin Apr 4 '11 at 18:25

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