Given a recurrence relation:
$$
x_{n+1} = a\left(x_n + {1\over x_n}\right) \\
x_1 = a\\
n\in\Bbb N
$$
Show that:
$$
\begin{align*}
a \ge 1 &\implies \lim_{n\to\infty} x_n =+\infty \tag1\\
a \in (0, 1) &\implies \lim_{n\to\infty} x_n =\sqrt{\frac{a}{1-a}} \tag2\\
\end{align*}
$$
I think I've been able to prove case $(1)$ but faced difficulties in case $(2)$. Here is what i've done for $a\ge 1$. Suppose a limit exists, then it must equal to one of the fixed points:
$$
\exists \lim_{n\to\infty}x_n = L \implies L = aL + {a\over L}\\
\iff L = \frac{\pm \sqrt{4a(1-a)}}{2(a-1)} = \mp \sqrt{\frac{a}{1-a}}
$$
Since $x_n > 0$ the only possible finite limit is $L = \sqrt{\frac{a}{1-a}}$.
Consider the following expression:
$$
x_{n+1} – x_n = a\left(x_n – {1\over x_n}\right) – x_n =\frac{x_n^2(a-1)+a}{x_n} > 0
$$
Therefore $x_n$ in monotonically increasing:
$$
x_{n+1} \ge x_n
$$
Also for $a\ge1$:
$$
L = \sqrt{\frac{a}{1-a}}\notin \Bbb R
$$
Thus by monotonicity of $x_n$ the only possible options left is:
$$
\lim_{n\to\infty} x_n = +\infty
$$
I've tried to apply a similar reasoning to $(2)$ but unfortunately it lead nowhere. The difficulty is also in the fact that the sequence is not necessarily monotone. Here are two sandboxes i've been using while solving the problem.
How can I show that the sequence has a limit when $a\in(0,1)$?
Best Answer
You already showed $$ x_{n+1}\ge x_n. \tag{1}$$ Note that, by the AM-GM inequality, one has $$ x_{n+1}\ge 2a.$$ From $$x_{n+1}-x_n=a(x_n-x_{n-1}+{1\over x_n} - {1\over x_{n-1}})=a(1-{1\over x_nx_{n-1}})(x_n-x_{n-1})\le a(1-\frac{1}{4a^2})(x_n-x_{n-1})$$ one has $$ 1-\frac{1}{4a^2}\ge 0$$ or $a>\frac12$. Define $r= a(1-\frac{1}{4a^2})$ and then $r\in(0,1)$. So $$ x_n-x_{n-1}\le r^{n-1}(x_2-x_1)$$ $$ x_n=\sum_{k=2}^n(x_k-x_{k-1})+x_1\le\sum_{k=2}^nr^{k-1}(x_2-x_1)+x_1<\infty.$$ Namely ${x_n}$ is bounded. Now one can use the Bounded Monotonic Convergence Principle to get the limit.