I am reading this paper (Theorem 3) and in some point it is stated that
$$\delta_{k+1} \leq \frac{3}{2} \left(\frac{\delta_k}{1-\delta_k}\right)^2 \leq \frac{8}{3} \delta_{k}^2$$
with $M \leq 2L$. Based on the above, the author states that $\delta_{k+1} \leq 1/4$ and $\delta_{k+1} \leq \frac{2}{3} \delta_k$. How did he come up with these bounds? Could you please give some intuitive explanations? Any help is highly appreciated, I am not familiar with series and their behavior.
Below I attach the part of the proof in this paper
Suppose $\delta_0\leq \frac14$ let us show that if $$\delta_{k+1} \leq \frac{3}{2} \left( \frac{\delta_k}{1-\delta_k}\right)^2\tag{1}$$ then for all $k$, $\delta_k\leq \frac14$. By induction, it is true for k=0, and if it is true for $k$, then $$1-\delta_k \geq \frac34$$, $$ \frac{\delta_k}{1-\delta_k} \leq \frac13$$ and therefore $$ \delta_{k+1} \leq \frac 23 \frac 19 < \frac14.$$ Now that this is true, using again $1-\delta_k \geq \frac34$ we find $$ \delta_{k+1} \leq \frac83 \delta_k^2 \leq \frac23 \delta_k. $$
On how $\delta_0$ was probably chosen (to put it in the answer and not in the comments). The relationship is of the form $$ \delta_{k+1} \leq f\left( \delta_k \right) $$ and the author wishes to show that the sequence was first bounded, then eventually decreasing following the rate of $f$ near the origin. The function $f$ is increasing near the origin, so if $\delta_k \leq a$, then $\delta_{k+1}=f(\delta_k) \leq f(a)$. This provides a bound provided $f(a)\leq a$. Here we see that $f(a)\leq a$ provided $a\leq a_0$, where $a_0=f(a_0)$ is given by $a_0=\frac{7-\sqrt{33}}{4}\approx 0.31$ a simple choice for $a$ is then $\frac{1}{4}$, as $\frac{1}{3}$ is slightly too large.