I want to prove that the sequence $\{Q_{n}\}$ defined by
\begin{equation*} \left\{\begin{matrix} Q_{1}=x_{1}\\ Q_{n+1}=Q_{n}+x_{n+1} \end{matrix}\right. \end{equation*}
from where $\{x_{n}\}$ is defined by
\begin{equation*} \left\{\begin{matrix} x_{1}=\frac{1}{3}\\ x_{n+1}=1-\sqrt[3]{1-x_{n}} \end{matrix}\right. \end{equation*}
To prove that it is Cauchy, I thought of proving that this sequence is convergent and therefore it is a Cauchy sequence. To this end, we can see by induction that the succession is increasing ($Q_{n+1}\geq Q_{n}$ for all $n \in \mathbb{Z}^{+}$).
For the base step it is clear that $Q_{1}=\frac{1}{3} \leq Q_{2}=\frac{1}{3}+1-\sqrt[3]{1-\frac{1}{3}}$.
For the inductive step note that $Q_{n+2}=Q_{n+1}+1-\sqrt[3]{1-x_{n+1}}$ and you have to $x_{n+1}\in (0, 1)$ whereby $1-\sqrt[3]{1-x_{n+1}} > 0$ and thus $Q_{n+2}=Q_{n+1}+1-\sqrt[3]{1-x_{n+1}}\geq Q_{n+1}$.
According to this, the succession should be increasing. Now, I do not know if this sequence is really bounded, it is difficult for me to appreciate it. Any help? or an alternative way to prove that it is a Cauchy sequence? thank you
For every $n \in \mathbb{N}$, let $y_n = 1-x_n$. One has $y_1=\dfrac{2}{3}$, and $y_{n+1} = \sqrt[3]{y_n}$ for every $n \geq 1$.
Inductively, you get the closed form $y_n = \left(\dfrac{2}{3} \right)^\frac{1}{3^{n-1}}$, which gives that $$x_n = 1-\left(\dfrac{2}{3} \right)^\frac{1}{3^{n-1}}$$
In particular, you get the equivalent $$x_n \mathop{\sim}\limits_{+\infty} \dfrac{1}{3^{n-1}}\ln\left(\dfrac{3}{2} \right)$$
This is the general term of a convergent series, so by comparison, the series $\sum x_n$ converges, which means that the sequence $(Q_n)$ converges, so it is Cauchy.