Why is $\sum_{t=0}^{n-1} \exp \{ -t \Delta^2 \} \geq \frac{1-e^{-1}}{\Delta^2}$ for $n \geq \frac{1}{\Delta^2}$? This is a follow-up question on this question. I was reading a paper about lower bounds for bandit problems (https://arxiv.org/abs/1302.1611). In Theorem 5, they prove a lower bound with an example problem with two arms. In the comments/answers of the previous question on stackexchange, it turned out that the theorem in the paper contains a few errors: the lower bound in Thrm. 5 should be $\frac{1-e^{-1}}{4\Delta}$, is should only hold for $n \geq 1/\Delta^2$, and the sum should start at 0.
I still don't understand the following step in the corrected proof:
$\sum_{t=0}^{n-1} \exp \{ -t \Delta^2 \} \geq \frac{1-e^{-1}}{\Delta^2}$ for $n \geq 1 / \Delta^2$.
I've tried to use:

*

*Jensen's inequality,

*Taylor expansion,

*infinite sums,
all leading to an upper bound in stead of a lower bound.


Use$$\sum_{t=0}^{n-1}\exp(-t\Delta^2)=\frac{1-e^{-n\Delta^2}}{1-e^{-\Delta^2}}\ge\frac{1-e^{-1}}{1-e^{-\Delta^2}}\ge\frac{1-e^{-1}}{\Delta^2}.$$
