For what $a\neq0$ does $a_{n+1}=1/a_{n}+a_{n}/2$ the sequence converge

Solution 1:

It is enough to consider the problem for $a>0$ as if we have two sequences with $a_{0} = a$ and $b_{0} = -a$ with the same above recurrence relation then $a_{n} = - b_{n}$.

From

$$(x-\sqrt{2})^2 \geq 0$$

we have

$$\frac{1}{x}+\frac{x}{2} \geq \sqrt{2}\text{ for }x \geq 0$$

Thus $a_{1},a_{2},...$ are all atleast $\sqrt{2}$. Furthermore

$$ \frac{1}{x}+\frac{x}{2} \leq x \text{ for }x \geq \sqrt{2}$$

Hence we have

$$a_{1} \geq a_{2} \geq ...\geq \sqrt{2}$$

Thus $a_{1},a_{2},...$ is a monotonically decreasing sequence which conveges to some value $c$ atleast $\sqrt{2}$ by the monotone convergence theorem. Now $c$ must satisfy the equation $c = \frac{1}{c}+\frac{c}{2}$; thus $c \in \{-\sqrt{2},\sqrt{2}\}$, but $c \geq \sqrt{2}$ thus $c = \sqrt{2}$.

Hence we have the following;

If $a > 0$ the sequence converges to $\sqrt{2}$

If $a < 0$ the sequence converges to $-\sqrt{2}$