<s>[INST] An exercise from Apostol's book I am trying to solve following problem from Apostol's Mathematical Analysis. The problem could be very trivial, but I am not getting clue for it.

Let $\{a_n\}$ be a sequence of real numbers in $[-2,2]$ such that 
  $$|a_{n+2}-a_{n+1}|\le \frac{1}{8} |a_{n+1}^2-a_n^2| \,\,\,\, \mbox{ for all } n\ge 1.$$
  Prove that $\{a_n\}$ is convergent. 

Q. Any hint for solving this? (I was not getting the restrictions of interval and the factor $\frac{1}{8}$).
My try: since $a_i\in [-2,2]$ so $a_i^2\in [0,4]$. Thus, $|a_{n+1}^2-a_n^2|\le 4$ and so $|a_{n+2}-a_{n+1}|\le \frac{1}{2}$. After this, I couldn't proceed. 
Any HINT is sufficient.
 [/INST] Let $A_{m+1}^-:=\left|a_{m+1}-a_m\right|$ and $A_{m+1}^+:=\left|a_{m+1}+a_m\right|$ for convenience. Then for a positive integer $k$, \begin{align}A_{n+k+1}^-&\le\frac{1}{8}\left|a_{n+k}^2-a_{n+k-1}^2\right|=\frac{1}{8}A_{n+k}^+A_{n+k}^-\\&\le\frac{1}{8}A_{n+k}^+\cdot\frac18\left|a_{n+k-1}^2-a_{n+k-2}^2\right|=\frac1{8^2}A_{n+k}^+A_{n+k-1}^+A_{n+k-1}^-\\&\le\frac1{8^3}A_{n+k}^+A_{n+k-1}^+A_{n+k-2}^+A_{n+k-2}^-\\&\le\cdots\\&\le\frac{A_{n+k-(n+k-1)+1}^-}{8^{n+k-1}}\prod_{i=1}^{n+k-1}A_{n-k-i+1}^+\\&\le\frac{A_2^-}{8^{n+k-1}}\prod_{i=1}^{n+k-1}|2+2|\quad(\text{since}\, \left|a_j\right|\le2\quad\forall j\ge1)\\&=\frac{A_2^-}{2^{n+k-1}}\end{align} Therefore, as $k\to\infty$, we have $\left|a_{n+k+1}-a_{n+k}\right|\to0$ and convergence is shown.
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