Find convergence rate ($c$) and order of convergence ($\rho$) of the $\left\{\dfrac{1}{2},\,\dfrac{1}{8},\,\dfrac{1}{128},\,\cdots\right\}$

Question

$$x_n=\left\{\dfrac{1}{2},\,\dfrac{1}{8},\,\dfrac{1}{128},\,\cdots\right\}=\left\{\frac{1}{2^{n^2-n+1}}\right\}_{n=1}^\infty$$
$$\lim_{n \to \infty} \dfrac{1}{2^{n^{2}-n+1}} = 0.$$
$$ \frac{x_{n+1}}{x_n} = \frac{\frac{1}{2^{(n+1)^{2}-(n+1)+1}}}{\frac{1}{2^{n^{2}-n+1}}} = \frac{1}{{2}^{2n}} \rightarrow 0, \quad n \rightarrow \infty, \quad n \in \mathbb{N}.$$
Therefore $c=0$ (convergence rate), $\rho=1$ (order of convergence) and $x_n$ is superlinearly convergent.

Is this correct? If it is false, Can anyone offer some assistance, please?
Thank you!

Answer 1

You have correctly identified the convergence as superlinear. There two points to be wary of

1. Writing $$x_n = \left\{\frac{1}{2^{n^2-n+1}}\right\}_{n=1}^\infty$$ is wrong, because $x_n$ is the standard notation for a single element and the right-hand side specifies an entire sequence, i.e., the function $f : \mathbb{N} \rightarrow \mathbb{R}$ given by $$\forall n \in \mathbb{N} \: : \: f(n) = \frac{1}{2^{n^2-n+1}}.$$ While you meaning is perfectly clear to me, your enemy will not be willing to understand.
2. When you say that the order of convergence is $\rho = 1$, it is automatically understood that $$ \frac{x_{n+1}}{x_n} \rightarrow c \in (0,1), \quad n \rightarrow \infty, \quad n \in \mathbb{N}.$$ This you do not have, because $c = 0$. Instead, you should simply state that the sequences converges superlinearly to zero.