Define a sequence $(a_n)$ as follows $a_1 = 1$ and for $n\geq 1,\;a_{n+1} = (-1)^n\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)$

Question

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?

Answer 1

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
• $\lim\limits_{k\rightarrow\infty} a_{2k}=-\sqrt{2}$

Answer 2

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 $~\lim\inf a_n=-\sqrt 2~$.