I am reading this paper and in some point it is stated that
$$\sum_{i=0}^{\infty} \ln(1-\delta_i)\geq -\sum_{i=0}^{\infty} \frac{\delta_i}{1-\delta_i}\geq -\frac{1}{1-\delta_0}\sum_{i=0}^{\infty}\delta_i \geq -1 $$
with $\delta_i = \frac{L||\nabla f(x_k)||}{\lambda_n^2(f''(x_k))}$.
Could you please help to understand how this inequality is lower bounded by -1? Generally, any help to understand the general behavior of $\sum_{i=0}^{\infty} \ln(1-\delta_i)$ is highly appreciated.
Note: In the paper you linked, the authors first prove that $\delta_0 \leq \frac{1}{4}$ and that $\delta_{k} \leq \frac{2}{3} \delta_{k-1}$, from which it is immediate that $\frac{\delta_i}{1 - \delta_i} \leq \frac{\delta_i}{1 - \delta_0}$ and that all $\delta_i < 1$.
Since all $\delta_i < 1$, you can use the following inequality for the log:
$$ \log(1 + x) \geq \frac{x}{1 + x}, \quad \forall x > -1 \Rightarrow \log(1 - \delta_i) \geq \frac{-\delta_i}{1 - \delta_i} \geq -\frac{\delta_i}{1 - \delta_0}, \forall i. $$
The last step is due to the fact that $\delta_i \leq \left(\frac{2}{3}\right)^i \delta_0$. In particular,
$$ \sum_{i=0}^{\infty} \delta_i \leq \delta_0 \sum_{i=0}^{\infty} \left( \frac{2}{3} \right)^i = \frac{\delta_0}{1 - \frac{2}{3}} = 3 \delta_0. $$
Therefore,
$$-\frac{1}{1 - \delta_0} \sum_{i=0}^{\infty} \delta_i = - \frac{3\delta_0}{1 - \delta_0} \geq -1, $$
after taking into account the range of $\delta_0 \leq \frac{1}{4}$.