Let $X_n \sim \text{Exp}(c^n)$, with $c > 0$. Let $Y_n = \min\{X_1, \ldots, X_n\}$. Find the distribution of $Y_n$ and study its convergence in distribution. Finally, show that $(Y_n)_{n \in \mathbb N}$ converges almost surely to some random variable for every $c$.
With the use of the CDF and some algebra, we obtain $$Y_n \sim \text{Exp}\left(-\sum_{i = 1}^n c^i\right) = \text{Exp}\left(-\frac{c - c^{n + 1}}{1 - c}\right)$$ where last equality holds for $c \neq 1$.
For $0 < c < 1$, $$\lim_n F_{Y_n}(y) = 1 - \exp\left(-\frac{c}{1 - c}y\right)$$ and $c / (1 - c)$ is always positive, so $Y_n \xrightarrow{d} \text{Exp}(\frac{c}{1 - c})$. For $c \geq 1$, $Y_n \xrightarrow{d} 0$.
It's when I have to show almost sure convergence that I get stuck. For example, let $c \geq 1$. Then $$P(Y_n \to 0) = P(\{\omega \in \Omega \mid \lim_n Y_n(\omega) = 0\})$$ How do I show that this evaluates to $1$? Intuitively I would guess that positive values are progressively less and less likely, and so at infinity $0$ is the only possible outcome. But since $Y_n$ is a continuous variable, I cannot use the probability of getting $0$ since it's $0$.
I realize, though, that intuitively I'm thinking about the probability $P(Y_n > \varepsilon) \to 0$ for $\varepsilon > 0$, which is the convergence in probability. But the latter does not imply almost sure convergence...