# 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)

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

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