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.
On
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: $$\frac{a}{c} \leq \frac{1}{n c^n} < \frac{1}{c^n}$$
$$\iff \ln(\frac{a}{c} ) < n ln(\frac{1}{c})$$ $$\iff \frac{\ln(\frac{a}{c} )}{\ln(\frac{1}{c})} < n$$
Since $\exp(-1/n) \sim 1 - 1/n $ while the left side converges exponentially to $1$, the inequality will be true for all sufficiently large $n$.
Somewhat explicitly, you want $$ a c^{n-1} \le 1 - \exp(-1/n)$$ so it suffices to have $$ c^{-n} \ge 2an/c $$ with $n \ge 1$ (note that $\exp(-1/n) \le 1 - 1/n + 1/(2n^2)$ so $\exp(-1/n) \le 1 - 1/(2n)$). Now $$ c^{-n} = (1 + (c^{-1}-1)^n) > \frac{n(n-1)}{2} (c^{-1}-1)^2 $$ so it suffices to have $$n-1 > \frac{2a}{c(c^{-1}-1)^2} = \frac{2ac}{(1-c)^2}$$