You really want to prove by using monotonicity? It's possible to obtain the exact closed-form of $x_n$

@NN2 I was told in another question of mine that I have to prove that the sequence is increasing and bounded above first and then proof the limit, I actually managed to prove the limit without showing the monotonicity and bounded above part.

Here is the question I am referring to math.stackexchange.com/questions/4553181/… @NN2

I cannot see where you were "told to prove that the sequence is increasing and bounded above" in that post.

Just calculate $f(x)-x=\dfrac{x(B-x^2)}{D>0}$ since $x_1=1>0$ then all $x_n>0$ so this behaves as $(B-x^2)$. So increasing on the left of the limit and decreasing on the right.

$$x_n = \sqrt{B} \tanh \left(3^{n-1} \operatorname{arctanh}\left(\frac{x_1}{\sqrt{B}} \right) \right) \to \sqrt{B}$$ You make 2 transformation: $y_n = \frac{x_n}{\sqrt{B}}$ and $z_n = \tanh(y)$

@AnneBauval in the last comment of Thomas

@NN2 Please transform your comment into an answer. I will be happy to upvote it. This is a method that I have already seen in cousin contexts.

@NN2 is not there an easier way of proving it ...... I did not get your point in the proof actually.

@zwim but I think it is a decreasing sequence not increasing ...... so what stops it from decreasing below 0?

As long as $x_1>0$ the sequence is positive because your RHS is. Also calculate $f(x)-\sqrt{B}=\dfrac{(x-\sqrt{B})^3}{D>0}$ so your sequence is always the same side of $\sqrt{B}$.

But where is the proof of monotonicity in this proof? @zwim

Hint: except where it would involve division by zero,$$\frac{\left(\frac{x\left(x^2+3B\right)}{3x^2+B}\right)^2-B}{x^2-B}=\left(\frac{x^2-B}{3x^2+B}\right)^2.$$

Or you can show that for $y_n=\dfrac{x_n-\sqrt{B}}{x_n+\sqrt{B}}$ you get $$y_{n+1}=y_n^3$$ which proves the cubic convergence of this instance of the Halley method. // Or recognize that this iteration is the Newton method for $$f(x)=x^{3/2}-Bx^{-1/2}$$ and discuss the graph of $f$ above and below $\sqrt{B}$ to determine the locations of the roots of the tangent.

See math.stackexchange.com/questions/3364552/…, math.stackexchange.com/questions/787019/…

@zwim I do not see why the RHS is positive, can you explain this please?

About Halley's method mentionned by @Lutz Lehmann see here his own answer.