Find $n$ such that $1-a c^{n-1} \ge \exp(-\frac{1}{n})$ I am trying to find the integer $n$ such that
\begin{align}
1-a c^{n-1} \ge \exp(-\frac{1}{n})
\end{align}
where $a>0$ and $c \in (0,1)$.
I know that finding it exactly is difficult. However, can one find good upper and lower bounds it.
It tried using lower bound $\exp(-x) \le 1-x+\frac{1}{2}x^2$. However, it didn't really work.

You could chain you're inequalities in the following way:
$$1 - ac^{n - 1} \geq exp\left (\frac{-1}{n} \right) \geq 1 - \frac{1}{n}$$
Then
$$-ac^{n - 1} \geq -\frac{1}{n}$$
$$\iff ac^{n - 1} \leq \frac{1}{n} $$
Now, we can move all of the constants on one side and all of the variables to the other:
$$\frac{a}{c} \leq \frac{1}{n c^n}$$
Clearly, the left side is constant. On the other hand, we know that because $c \in (0,1),$ then
$$\frac{1}{n c^n} < \frac{1}{c^n}$$
Therefore, we can chain in the following way:
$$\frac{a}{c} \leq \frac{1}{n c^n} < \frac{1}{c^n}$$
$$\iff \ln(\frac{a}{c} ) < n ln(\frac{1}{c})$$
$$\iff \frac{\ln(\frac{a}{c} )}{\ln(\frac{1}{c})} < n$$
