given sequence $(a_n)$ is defined as $a_1=1$ , $a_{n+1}= (-1)^n(\frac{1}{2})(|a_n|+ \frac{2}{|a_n|})$ we had found that $$\lim_{k\to ∞} (a_{2k})= -\sqrt{2}$$ and $$\lim_{k\to ∞} (a_{2k+1}) =\sqrt{2}$$

From this how can we can conclude that limit $$\limsup_{n\to \infty} a_n = \sqrt{2}$$ and $$\liminf_{n\to \infty} a_n =-\sqrt{2}$$ ?

I know that limit superior of $(a_n)$ is supremum of set of all subsequential limits of $(a_n)$ and similarly limit inferior of $(a_n)$ is infimum of set of all subsequential limits of $(a_n)$. But here we had just find the two subsequential limits, that is limits of $(a_{2k})$ and $(a_{2k+1})$ so from this how we can conclude the value of limit inferior and limit superior?

Further is there is any easy way to find limit inferior and superior of sequences that are define inductively just like above.

Please help....


In this particular case, you can first narrow down your scope by focusing on $|a_n|$ only, i.e. $$ |a_{n+1}|= \left(\frac{1}{2}\right)\left(|a_n|+ \frac{2}{|a_n|}\right).$$

First note $ |a_{n+1}|= \left(\frac{1}{2}\right)\left(|a_n|+ \frac{2}{|a_n|}\right) \ge \left(\frac{1}{2}\right)2\sqrt{|a_n|\times \frac{2}{|a_n|}}=\sqrt{2}.$ Hence $|a_n|\ge\sqrt{2}$ for all $n \ge 2$. Hence $\frac{2}{|a_n|} \le |a_n|$, and $ |a_{n+1}|\le\frac{1}{2}(|a_n| + |a_n|) = |a_n|$.

We now know $|a_n|$ is monotonically decreasing and has an lower bound of $\sqrt{2}$. Hence $|a_n|$ converges. It is a simple calculation to show that $$ \lim_{n \rightarrow \infty}|a_n| = \sqrt{2}. $$

Next $$ \limsup_{n \rightarrow \infty} a_n \le \limsup_{n \rightarrow \infty} |a_n| = \lim_{n \rightarrow \infty}|a_n|=\sqrt{2},$$ and $$ \liminf_{n \rightarrow \infty} a_n \ge \liminf_{n \rightarrow \infty} \left(-|a_n|\right) = -\lim_{n \rightarrow \infty}|a_n|=-\sqrt{2}.$$

Your conclusion is reached since you've already found two converging subsequences.

I think the general way of finding $\limsup$ and $\liminf$ is to find converging subsequences and then bound the general sequence. Others can complement if I miss anything.


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