# Study convergence of $x_{n+1} = a\left(x_n + {1\over x_n}\right)$, for $x_1 = a$ and $a \in (0, 1)$

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 $$a\in(0,1)$$?

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.
• Thank you for the answer. Could you please elaborate on $1-{1\over 4a^2} \ge 0$. Here $a$ is a given initial condition and $a\in(0, 1)$, while you are restricting it to be $a > {1\over 2}$. Is it rigorous to make such assumptions? Would't that contradict the initial conditions on $a$? Also I'm trying to wrap my mind on the last part, you have stated that $x_n < \infty$ why would that mean the boundedness for $x_n$? – roman Feb 14 at 10:27
• @roman, this because $x_{n+1}-x_n\ge0$ and $x_{n}-x_{n-1}\ge0$. – xpaul Feb 14 at 15:22
Let $$f(x) = a (x + 1/x)$$, and suppose $$0 < a < 1$$. $$L = \sqrt{a/(1-a)}$$ is the only fixed point of $$f$$ in the positive reals, and it is an attractor since $$|f'(L)| = |2a-1| < 1$$. Moreover, there are no $$2$$-cycles: $$f(f(x))-x = \frac{(a+ (a+1) x^2)(a + (a-1)x^2)}{x(x^2+1)}$$ The endpoints of the immediate basin of attraction of an attracting fixed point can only be singular points, $$\pm \infty$$, non-attracting fixed points, and members of a $$2$$-cycle. Thus in this case the immediate basin of attraction must be $$(0, \infty)$$, i.e. the sequence $$x_n = f(x_{n-1})$$ starting anywhere in $$(0,\infty$$ will always converge to $$L$$.