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There's a bounded sequence $\{{a_n}\} $ s.t. $\lim\limits_{n \rightarrow \infty}(a_{2n} + 2a_n) = 0$. Try to prove $\lim\limits_{n \rightarrow \infty}a_n = 0$. (You must prove the existence of its limit before getting the exact value)

The above problem is located in the chapter about "limsup and liminf", so you may use their property while proving.

EDIT

By the way, I find it hard to do the exercises at the end of each section of my textbook though I DO UNDERSTAND almost everything in that section. So I'm eager to know some other books that can help me to solve problems like the one described above. That would be a great help, thanks!

Riemann
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ymfoi
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    You also have poor title skills... please choose titles which reflect the mathematical content of the question. – Asaf Karagila Oct 01 '12 at 12:30
  • Suggestions about other books: http://math.stackexchange.com/questions/138232/teaching-yourself-analysis –  Oct 01 '12 at 12:44

3 Answers3

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Let $\alpha$ be a congestion point of $(a_n)_n$, i.e. there is a subsequence $(a_{n_k})_k$ that tends to $\alpha$, we are to show that $\alpha=0$. Now $$0 = \lim_{k\to\infty} (a_{2n_k}+2a_{n_k}) = \lim_{k\to\infty} a_{2n_k}+2\alpha$$ So, we got another congestion point by $\lim_k a_{2n_k} = -2\alpha$. This repeated means that all $(-2)^k\alpha$ numbers ($k\in\mathbb N$) are congestion points, so, because $(a_n)$ is bounded, $\alpha=0$ must be.

(Else, I tried to build a counterexample, and it is also possible if $a_n$ is not bounded.)

Berci
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  • You've proven there's a sub-sequence whose limit is $\alpha$, but how to prove the whole sequence converges? – ymfoi Oct 01 '12 at 10:43
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    @ymfoi The limit superior and limit inferior of $(a_n)$ exist and are finite because $(a_n)$ is bounded. They are also "congestion points" of the sequence (the usual terms I've seen are "limit point" or "accumulation point"). Berci showed any congestion point must be $0.$ Since the limit superior and limit inferior are examples of this, they both agree and are zero. Thus, the limit exists, and is zero. – Ragib Zaman Oct 01 '12 at 10:48
  • A bounded sequence has limit iff has only one congestion point. – Berci Oct 01 '12 at 11:33
  • @Berci how can we move from all (−2)kα numbers are congestion points to α=0 must be ? – ymfoi Oct 05 '12 at 06:53
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Since the sequence is bounded, $\limsup_na_n$ and $\liminf_na_n$ are both finite. If they are equal, the limit exists, so you need only show that $\liminf_na_n=\limsup_na_n$.

  1. Let $\alpha=\liminf_na_n$ and $\beta=\limsup_na_n$. Using the fact that $\lim_n(a_{2n}+2a_n)=0$, show that $\alpha\le 0\le\beta$.

  2. Suppose that $\beta\ge|\alpha|>0$. Let $\langle a_{n_k}:k\in\Bbb N\rangle$ be a subsequence converging to $\beta$; clearly $\langle 2a_{n_k}:k\in\Bbb N\rangle$ converges to $2\beta$. Show that $\lim_ka_{2n_k}=-2\beta$ and get a contradiction.

  3. If $|\alpha|>\beta$, let $\langle a_{n_k}:k\in\Bbb N\rangle$ be a subsequence converging to $\alpha$, and derive a similar contradiction.

Brian M. Scott
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  • By 2. and 3., $|\alpha|=\beta$. And? A proof that $\alpha=\beta=0$ seems to be lacking. – Did Oct 01 '12 at 12:46
  • @did: (1) It isn’t a proof; it’s an extended hint. (2) Belgi’s hint had already covered that part. (3) Combine $|\alpha|=\beta$ with my first point, and you have your proof anyway. – Brian M. Scott Oct 01 '12 at 19:06
  • Sorry but I do not follow: (1) If your answer is only an extended hint, you might want to mention the fact. (2) Belgi assumes the limit exists hence explicitly omits this part. (3) Your post fails to get rid of cases like $\alpha=-1$ and $\beta=1$, such that $|\alpha|=\beta$ but which must be excluded if one wants to show that the sequence converges. – Did Oct 01 '12 at 20:09
  • @did: (1) It’s obviously not complete, since many details are obviously missing, so it can only be a hint. (2) I show that the limits exist, providing what is missing from Belgi’s answer. (I also show that they’re $0$, as it happens.) (3) It doesn’t mention them explicitly, but anyone who actually carries out my second point has taken care of that case. However, I will restore the original version that made that explicit. – Brian M. Scott Oct 01 '12 at 22:53
  • As you know well, you now modified drastically your answer by replacing silently condition $\beta\gt|\alpha|$ in point 2. by $\beta\geqslant|\alpha|\gt0$. Contrary to what you claim in your last comment, the previous version was incomplete, even as an indication, since it did not show that the limit exists (the modified version is complete, but to see that it is requires more sophistication than anybody asking the question can have). I am baffled by this modus operandi (to leave implicit some crucial steps of the proof, to modify silently the answer to pretend comments do not apply). Very odd. – Did Oct 02 '12 at 04:58
  • @did: I consider it a minor modification that simply makes the hint more transparent. And I most certainly did not ‘silently’ modify the answer: as I said in my previous comment, I restored an earlier version. It happens to be a version that I originally decided not to post in favor of the version to which you objected, so you hadn’t seen it before, but I clearly indicated that I was changing the answer. I find your reaction utterly baffling, unless you are simply looking for reason to find fault. – Brian M. Scott Oct 02 '12 at 05:07
  • Why do you tweak the facts? There are two versions, not three. You modified version 1, which was wrong, after I mentioned (and showed) it was wrong. You did it silently since you never mentioned what the crucial modification was (not even that there was a modification). Version 2 is correct (although unnecessarily subtle). In my book, a modification which makes correct a previously wrong proof is not minor. The rest are cheap psychological speculations, which do not concern me. – Did Oct 02 '12 at 05:36
  • @did: I don’t tweak facts. You never saw version 1, because I changed my mind about it before I posted. What you call version 1 was in fact version 2, which was not wrong; it was merely intentionally less complete, leaving one significant step unmentioned for the OP to deal with. Version 3, which you call version 2, was a return to version 1 because I decided that version 2 was in fact too subtle. I find your insistence that version 2 was incorrect even after a clear explanation altogether incomprehensible. Your claim that I never mentioned making a modification is flatly false. – Brian M. Scott Oct 02 '12 at 05:53
  • Where did you mention the oh-so-subtle transition from $\beta\gt|\alpha|$ to $\beta\geqslant|\alpha|\gt0$? No further comment. – Did Oct 02 '12 at 06:17
  • @did: ‘However, I will restore the original version that made that explicit.’ – Brian M. Scott Oct 02 '12 at 07:02
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Hint: After you have proven $\lim_{n\to\infty}a_{n}$ exist note that $$\lim_{n\to\infty}a_{n}=\lim_{n\to\infty}a_{2n}$$ thus $$\lim_{n\to\infty}(a_{2n}+2a_{n})=\lim_{n\to\infty}a_{2n}+2\lim_{n\to\infty}a_{n}=0$$ and conclude $\lim_{n\to\infty}a_{n}=0$

Brian M. Scott
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Belgi
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