Suppose that $|a_n|<2$ and $|a_{n+2}−a_{n+1}|\leq\frac{1}{8}|a_{n+1}^2−a_n^2|$, prove that $\{a_n\}$ converges. (Part 2)

Question

This is my second attempt. The first attempt is here

I included the estimate $|a_m−a_n|\leq |a_m−a_{m-1}| + |a_{m-1}−a_{m-2}| \dots |a_{n+1} a_n|$

I really appreciate any advice\insight you can offer

Question

Let $\{a_n\}$ be a sequence of real numbers. Suppose that $|a_n|<2$ and $|a_{n+2}−a_{n+1}|\leq\frac{1}{8}|a_{n+1}^2−a_n^2|$ for all $n\leq1$.

Prove that $\{a_n\}$ converges.

My second proof attempt

$$\begin{align} |a_{n+1}^2−a_n^2| &= |(a_{n+1}+a_n)(a_{n+1}−a_n)| \\ & = |a_{n+1}+a_n||a_{n+1}−a_n| \\ & \leq (|a_{n+1}|+|a_n|)|a_{n+1}−a_n| \\ & < 4|a_{n+1}−a_n| \end{align}$$

thus $|a_{n+2}−a_{n+1}|\leq\frac{1}{2}|a_{n+1}−a_n|$.

by triangle inequality, $|a_m−a_n| \leq \sum_\limits{i=n}^{m-1} |a_{i+1}−a_i|$

thus $$\begin{align} |a_m−a_n| &\leq \sum_\limits{i=n}^{m-1} \frac{1}{2^{i-n+1}} |a_{n+1}−a_n|\\ & < 2 |a_{n+1}−a_n|\\ & \leq 2\times\frac{1}{2^{n-1}}|a_2−a_1|\\ & =\frac{1}{2^{n-2}}|a_2−a_1|\\ \end{align}$$

thus for $\forall \epsilon > 0$ pick an $n_0$ s.t for $n,m \geq n_0 $ ,$ \ \epsilon > \frac{1}{2^{n_0-2}}|a_2−a_1|$

thus $|a_m−a_n| <\frac{1}{2^{n_0-2}}|a_2−a_1| < \epsilon$

thus $\{a_n\}$ converges.

This concludes my proof attempt

Answer 1

It's perfect. In fact you don't need most of steps, just knowing that $|a_{n+2}-a_{n+1}|\leq \frac{1}{2}|a_{n+1}-a_{n}|$ That's because every contractive sequence is a Cauchy sequence, hence converges