An exercise from Apostol's book

It suffices to show that $(a_n)_n$ is a Cauchy sequence.

By the the given condition, for $n\geq 1$, since $|a_n|\leq 2$, $$\begin{align}|a_{n+2}-a_{n+1}|&\le \frac{1}{8} |a_{n+1}^2-a_n^2| =\frac{1}{8}|a_{n+1}+a_n| |a_{n+1}-a_n|\\&\leq \frac{2+2}{8} |a_{n+1}-a_n|=\frac{1}{2} |a_{n+1}-a_n|. \end{align}$$ It follows that $$|a_{n+2}-a_{n+1}|\leq \frac{1}{2} |a_{n+1}-a_n|\leq \frac{1}{2^2} |a_{n}-a_{n-1}|\leq \cdots\leq \frac{1}{2^n} |a_{2}-a_{1}|.$$ Hence if $n>m\geq 1$ then $$|a_n-a_m|\leq |a_{2}-a_{1}|\sum_{k=m-1}^{n-2}\frac{1}{2^k}\leq \frac{|a_{2}-a_{1}|}{2^{m-2}}\to 0$$ as $m\to\infty$.

HINT. Let $d_n=|a_{n+1}-a_n|.$ Then

$$ \begin{align} d_{n+1}=|a_{n+2}-a_{n+1}| & \le |a_{n+1}^2-a_n^2|/8 \\ & = |a_{n+1}-a_n| \cdot |a_{n+1}+a_n|/8 \\ & \le |a_{n+1}-a_n| \cdot 4/8 \\ & = d_n/2. \end{align} $$

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.