On a recursive sequence (exercise 8.9 Apostol)

Question

The exercise states: show convergence of the sequence ${a_n}$ knowing that:

$$|a_n| \le 2, \ \ \ |a_{n+2}-a_{n+1}| \le \frac{1}{8}|a_{n+1}^2 - a_{n}^2|.$$

The solution states:

$$|a_{n+2}-a_{n+1}| \le \frac{1}{8}|a_{n+1}^2 - a_{n}^2| = \frac{1}{8}|a_{n+1} - a_{n}||a_{n+1} + a_{n}| \le \frac{1}{2}|a_{n+1} - a_{n}| $$

since $|a_n| \le 2$ we know that $|a_{n+1} - a_{n}| \le (\frac{1}{2})^{n-1}$ ...

This last step is not very clear to me, could somebody explain it to me.

Thanks in advance

Answer 1

Since $-2\leq a_n\leq 2$

$$|a_{n+2}-a_{n+1}| \le \frac{1}{8}|a_{n+1}^2 - a_{n}^2| = \frac{1}{8}|a_{n+1} - a_{n}||a_{n+1} + a_{n}| \leq \frac{1}{8}|a_{n+1} - a_{n}||4|= \frac{1}{2}|a_{n+1} - a_{n}| $$

Next, call $b_n=a_{n+1}-a_n$ and rewrite the above.