Give $x_0 \in \mathbb{R}^*$. Show that $\{x_k\} \subset \mathbb{R}$ converge $r$-superlinearly for $x^∗=0$, where $x_k$ is defined by $x_{k+1}=(1−\beta_k)x_k$ and $\beta_k=1−2^{-k}$ if $k=i^2$ for some integer $i$, and $\beta_k=2$ otherwise.
Obs: $\{x_k\}$ converge $r$-superlinearly if there is $\{v_k\}$ such that $||x_k−x^∗||≤v_k$ and $v_k$ converge q-superlinearly for $0$, this is, $\lim_{k\to \infty} \frac{v_{k+1}}{v_k}=0.$
My problem is defining the sequence $v_k$ that satisfies those conditions.
Given $x_0 \in \mathbb{R}$ the sequence $\{x_k\}$ can be written as $$( x_0, \frac{x_0}{2^{1^2}}, - \frac{x_0}{2^{1^2}}, \frac{x_0}{2^{1^2}}, \frac{x_0}{2^{1^2}2^{2^2}}, -\frac{x_0}{2^{1^2}2^{2^2}}, \frac{x_0}{2^{1^2}2^{2^2}}, - \frac{x_0}{2^{1^2}2^{2^2}}, \frac{x_0}{2^{1^2}2^{2^2}},\frac{x_0}{2^{1^2}2^{2^2}2^{3^2}}, -\frac{x_0}{2^{1^2}2^{2^2}2^{3^2}}, \frac{x_0}{2^{1^2}2^{2^2}2^{3^2}},...)$$
I thought so separate into two cases: if $k=i^2$ for all $i \in \mathbb{N}$, and if $k \neq i^2$ for all $i$, and last case $k=i^2 +j$ with $1 \leq j \leq (i+1)^2-i^2$. Need an exponent $x(k)$ such that $ i^2+(i-1)^2+...+1^2 \geq x(k)$ and $2^{x(k+1)-x(k)} \to 0$ for $k \to \infty$, so defined $v_k = \frac{|x_0|}{2^{x(k)}}$.