Given the following recurrence relation: $$ \begin{cases} x_{n+1} = {1\over 2}\left(x_n + {b^2\over x_n}\right) \\ x_1 = a > 0 \\ n\in \mathbb N \end{cases} $$ Show that $x_n$ is a bounded sequence.
This question follows after this one so perhaps it may help me solve the linked question.
No constraints on $b$ are given in the problem statement. I've started with $x_n$:
$$ x_{n+1} = {1\over 2}\left(x_n + {b^2\over x_n}\right) \\ x_n = {1\over 2}\left(x_{n-1} + {b^2\over x_{n-1}}\right) \implies \\ x_{n+1} = {1\over 2}\left({1\over 2}\left(x_{n-1} + {b^2\over x_{n-1}}\right) + {b^2\over x_n}\right) = \\ = {1\over 2}\left({1\over 2}\left({1\over 2}\left(x_{n-2} + {b^2\over x_{-2}n}\right) + {b^2\over x_{n-1}}\right) + {b^2\over x_n}\right) = \\ \dots\\ \frac{b^2}{2^1x_n} + \frac{b^2}{2^2x_{n-1}} + \frac{b^2}{2^3x_{n-2}} + \dots + \frac{b^2}{2^nx_{1}} + {a \over 2^n} = \\ = {a \over 2^n} + \sum_{k=1}^n \frac{b^2}{2^{n-k+1}x_k} $$
So i got:
$$ x_{n+1} = {a \over 2^n} + \sum_{k=1}^n \frac{b^2}{2^{n-k+1}x_k} $$
This seems to be correct while i've tested several first terms and they match the initial recurrence. Now I'm wondering how is this even bounded? BTW W|A suggests that:
$$ x_n = b \coth\left(2^{n - 1} \coth^{-1}{a\over b}\right) $$
Not sure where it's derived from. How do I show this sequence is bounded using precalculus Maths?
