Let $x_1=1$ and $x_{n+1}=\frac{3}{x_n}$ for $n\ge 2$(correction:$n\ge 1$). Then $(x_n)$ diverges.
I have gone through the proof of convergent sequences generally in this format and they are proved using monotone convergence theorem, through mathematical induction it is proven that all the elements are below a said value and then proven to be increasing or decreasing and thereby convergent. I am confused on how to go about applying a similar approach over here, the sequences is most definitely bounded and has lower bound 1 and upper bound 3, but how do I prove the same and then later on prove that the function is not monotonic therefore not convergent?
Best Answer
That sequence is the sequence defined by $$ x_n=\begin{cases} 1&\text{ if $n$ is odd}\\ 3&\text{ if $n$ is even.} \end{cases} $$ It diverges, since it has subsequences converging to $1$ and subsequences converging to $3$.