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Define a sequence $(a_n)$ as follows $a_1 = 1$ and for $n\geq 1$ $$a_{n+1} = (-1)^n\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)$$ then what will be it's $\limsup (a_n)$ and $\liminf (a_n)$ and $\sup(a_n)$, $\inf(a_n)$.

I tried to determine the nature of $a_{2n+1}$ and $a_{2n}$. I got $a_{2n+1}$ is increasing and $a_{2n+1} \geq \text{something}$ where $a_{2n}$ is decreasing and $a_{2n}\leq \text{something}$. So I could not draw any conclusion.

Can anyone please help me?

rtybase
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cmi
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    This sequence satisfy $a_{2n+1}\geq \sqrt 2$ and decreasing for $n\geq 1$, $a_{2n}\leq -\sqrt 2$ and increasing for $n\geq 1$, $\limsup a_n = \sqrt 2$, $\liminf a_n=-\sqrt 2$, $\sup a_n = \frac12 (\frac 32+ \frac 43)$, $\inf a_n=-\frac 32$. – Sungjin Kim Jul 27 '18 at 22:01
  • To prove these, use induction. – Sungjin Kim Jul 27 '18 at 22:02
  • How $(a_{2n})$ becomes increasing sequence for $n\geq 1$? Can you please elaborately explain? I calculated and I got $a_2 = -3/2$ and $a_4 = -201/104$. So $a_2 \ge a_4$. @i707107 – cmi Jul 28 '18 at 01:36
  • It is not possible for $a_{2n}$ to be less than $-3/2$. I don't think your $a_4$ is correct. – Sungjin Kim Jul 28 '18 at 02:08

2 Answers2

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Hint. Taking absolute value of $$a_{n+1} = (-1)^n\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)\Rightarrow |a_{n+1}|=\left|(-1)^n\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)\right| \Rightarrow \\ |a_{n+1}|=\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)$$ Using this question and answers provided we have $$\lim\limits_{n\rightarrow\infty} |a_n|=\sqrt{2}$$ and $(|a_n|)_{n\in\mathbb{N}}$ is decreasing.

It shouldn't be difficult to see that $\lim\limits_{k\rightarrow\infty} a_{2k+1}=\sqrt{2}$ and $\lim\limits_{k\rightarrow\infty} a_{2k}=-\sqrt{2}$, because

  • $0<\sqrt{2}<a_{2(k+1)+1}<a_{2k+1}$, thus $(a_{2k+1})_{k\in\mathbb{N}}$ has a limit.
  • $a_{2k}<a_{2(k+1)}<-\sqrt{2}<0$ (because $|a_{2k}|>|a_{2(k+1)}|$), thus $(a_{2k})_{k\in\mathbb{N}}$ has a limit.

In both cases we can use the fact that $\lim\limits_{n\to\infty}a_k=a \Rightarrow \lim\limits_{n\to\infty}|a_k|=|a|$ to conclude (e.g. proof by contradiction) that:

  • $\lim\limits_{k\rightarrow\infty} a_{2k+1}=\sqrt{2}$ and
  • $\lim\limits_{k\rightarrow\infty} a_{2k}=-\sqrt{2}$
rtybase
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I think the following will help the readers to understand the solution:

We know $$\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)\ge\sqrt{|a_n| \times \frac {2}{|a_n|}}=\sqrt 2$$ $$\implies \frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)\ge\sqrt 2$$ Given that $~a_1 = 1~$ and for $n\geq 1$ $$a_{n+1} = (-1)^n\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)$$ When $~n~$ is even, $$a_{n+1} = \frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)\ge\sqrt 2$$ Again when $~n~$ is odd, $$a_{n+1} = -\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)\le-\sqrt 2$$ Which gives $~\lim\sup a_n=\sqrt 2~$ and $~\lim\inf a_n=-\sqrt 2~$.

nmasanta
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