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.


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!

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    $\begingroup$ You also have poor title skills... please choose titles which reflect the mathematical content of the question. $\endgroup$ – Asaf Karagila Oct 1 '12 at 12:30
  • $\begingroup$ Suggestions about other books: math.stackexchange.com/questions/138232/… $\endgroup$ – user940 Oct 1 '12 at 12:44

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.)

  • $\begingroup$ You've proven there's a sub-sequence whose limit is $\alpha$, but how to prove the whole sequence converges? $\endgroup$ – ymfoi Oct 1 '12 at 10:43
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    $\begingroup$ @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. $\endgroup$ – Ragib Zaman Oct 1 '12 at 10:48
  • $\begingroup$ A bounded sequence has limit iff has only one congestion point. $\endgroup$ – Berci Oct 1 '12 at 11:33
  • $\begingroup$ @Berci how can we move from all (−2)kα numbers are congestion points to α=0 must be ? $\endgroup$ – ymfoi Oct 5 '12 at 6:53

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.

  • $\begingroup$ By 2. and 3., $|\alpha|=\beta$. And? A proof that $\alpha=\beta=0$ seems to be lacking. $\endgroup$ – Did Oct 1 '12 at 12:46
  • $\begingroup$ @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. $\endgroup$ – Brian M. Scott Oct 1 '12 at 19:06
  • $\begingroup$ 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. $\endgroup$ – Did Oct 1 '12 at 20:09
  • $\begingroup$ @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. $\endgroup$ – Brian M. Scott Oct 1 '12 at 22:53
  • $\begingroup$ 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. $\endgroup$ – Did Oct 2 '12 at 4:58

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 $\lim_{n\to\infty}a_{n}=0$

  • $\begingroup$ You can use \lim to get $\lim$. $\endgroup$ – Brian M. Scott Oct 1 '12 at 10:26
  • $\begingroup$ But how do I prove its limit to exist? $\endgroup$ – ymfoi Oct 1 '12 at 10:26

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