Check using Cauchy's convergence test if $a_{n}$ is convergent; $$a_{1}=\frac{1}{3}$$ $$a_{n+1}= \begin{cases} a_{n}^2, &\text{$a_{n}>\frac{1}{10}$} \\ 2a_{n}, & \text{else} \end{cases}$$
I need to show that $\forall m,n>\mathbb N$ Implies $\vert a_{m}-a_{n}\vert <\epsilon$
That's what I tried so far.
$$
\begin{split}
\vert a_{m}-a_{n}\vert
&= \vert a_{m}-a_{m-1}+a_{m-1}-a_{m-2}+a_{m-2}+...-a_{n+1}+a_{n+1}-a_{n}\vert \\
&\le \vert a_{m}-a_{m-1}\vert+\vert a_{m-1}-a_{m-2}\vert+...+\vert a_{n+1}-a_{n}\vert
\end{split}
$$
I don't know how to go from there and if this is even the right way..
It is actually not convergent. Try to prove:
[1] $a_n \gt \frac {1}{100}$ by induction on $n$.
[2] $|a_{n+1}-a_n| \gt \frac {1}{100}$ for every $n$.
[2] contradicts the Cauchy criterion (how?), thereby the sequence is divergent.