How to work out the limit of this bounded sequence?

Question

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!

Answer 1

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.)

Answer 2

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

Answer 3

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.