Question: Let $\\{a_n\\}$ be a sequence that satisfies $|a_n| < 2$ , $|a_{n+2}-a_{n+1}| \leq \frac{1}{8}|a_{n+1}^2-a_{n}^2|$. Then
a) $\\{a_n\\}$ is a cauchy sequence
b) $\\{a_n\\}$ is a bounded sequence
c) $\\{a_n\\}$ is not a bounded sequence
d) $\\{a_n\\}$ is not a convergent sequence
My attempt: Since $|a_n| < 2$ for all n it is obvious that the sequence is bounded and option b is true and option c is false.
$|a_{n+2}-a_{n+1}| \leq \frac{1}{8}|a_{n+1}^2-a_{n}^2|\\ =\frac{1}{8}|a_{n+1}+a_{n}||a_{n+1}-a_{n}|$
$|a_{n+1}+a_{n}|\leq |a_{n+1}|+|a_{n}|\\\leq 2+2 =4$
$|a_{n+2}-a_{n+1}|\leq\frac{1}{2}|a_{n+1}-a_{n}|$
After these steps I don't know how to proceed. I pondered over this sum for a few days but there was no improvement beyond these steps.
Kindly help me in proving or disproving options a and d
For showing (a), let us fix $\epsilon > 0$ and find $N\in\mathbb{N}$ for Cauchy Sequence Guarantee. Assume you have two integers such that $N \leq n < n+m$.
Then, $$ |a_{n+m} - a_{n}| \leq |a_{n+m} - a_{n+m-1}| + \cdots + |a_{n+1} - a_{n}| \leq \frac{1}{2^{n+m-1}} |a_1 - a_0| + \cdots + \frac{1}{2^{n}} |a_1 - a_0| < \frac{2^m - 1}{2^{n+m-1}} \times 4 \,. $$
The last inequality used the fact $|a_i|<2$.
Therefore, to guarantee $|a_{n+m} - a_{n}| < \epsilon$ for any case, $(1/2)^{N-1} \times 4 < \epsilon$ should suffice.
We can set $N$ as $1 + \lceil \log_2 (4/\epsilon) \rceil $.
For (d), do you know the notion of completeness? If so, you can get (d) directly from (a).