If $\{x_n\}_{n \geq 1} \subset \mathbb{R}, x_n \xrightarrow{n \to \infty}x^*$ with order of convergence $r > 1$, then $\{x_n\}_{n \geq 0} \subset \mathbb{R}$ converges to $x^*$ with superliniar convergence rate.
My proof: Because $x_n \xrightarrow{n \to \infty}x^*$ with order of convergence $r > 1$ we have that $\exists c > 0$ s.t. $\lim\limits_{n \to \infty} \frac{|x^*-x_n|}{|x^*-x_{n-1}|^r} = c$. So, we have that $\lim\limits_{n \to \infty}\frac{|x^*-x_n|}{|x^*-x_{n-1}|} = \lim\limits_{n \to \infty}\frac{|x^*-x_n|}{|x^*-x_{n-1}|^r} \cdot |x^*-x_{n-1}|^{r-1} = \lim\limits_{n \to \infty}\frac{|x^*-x_n|}{|x^*-x_{n-1}|^r} \cdot \lim\limits_{n \to \infty}|x^*-x_{n-1}|^{r-1} = c \cdot 0 = 0$
So, we have that $\{x_n\}_{n \geq 1}$ converges to $x^*$ with superliniar convergence rate.
My question: Is this proof ok? And if it's ok... how influence $n \geq 1$ or $n \geq 0$ the order of convergence and the rate of convergence? Because I don't understand why we first have $\{x_n\}_{n \geq 1}$ and need to proof for $\{x_n\}_{n \geq 0}$. Why we don't prove for $\{x_n\}_{n \geq 1}$ ?