Existence of the order of convergence of a sequence.

Question

Let $\{a_n\}_{n\in\mathbb{N}}$ be a real number sequence converging to $\alpha$.

It is said to have order of convergence $q\geq1$ if $\lim\limits_{n\rightarrow\infty}\frac{|a_{n+1}-\alpha|}{|a_n-\alpha|^q} > 0$ exists and is finite.

If $\lim\limits_{n\rightarrow\infty}\frac{|a_{n+1}-\alpha|}{|a_n-\alpha|^p} = 0$ and $\lim\limits_{n\rightarrow\infty}\frac{|a_{n+1}-\alpha|}{|a_n-\alpha|^q}$ diverges for $1\leq p < q$, is there the order of convergence $r$ between $p$ and $q$?

Answer 1

Solved by Ryszard Szwarc's comment:

What about $_=^{−^2}$ and $=0$ For $=1$ the limit is $0$. For any $>1$ the limit is $∞$.

$$\lim\limits_{n\to\infty}\frac{|a_{n+1}|}{|a_n|^p}=e^{(p-1)n^2-2n-1}$$