Prove the convergence of $\sum_{n=1}^\infty P(n^a X_n \geq \epsilon)$ where $X_n=\min \{Y_1,...,Y_n \}$, $Y_i$ i.i.d. Good day,
I have an old exam without solutions and there is the following exercise:

Let $(Y_n)_{n \in \mathbb{N}}$ be a sequence of independent identically distributed (i.i.d.) random variables which are uniformely distributed on $[0,1]$. Further define $X_n :=\min\{Y_1,...,Y_n\}$. Show:
(i) $F_n(t)=[1-(1-t)^n] 1_{[0,1]}(t)+1_{(1,\infty)}(t)$ is the distribution function of $X_n$ - got this one
(ii) $X_n$ is integrable and compute $E(X_n)$ - got this one
(iii) Prove the convergence of $\sum_{n=1}^\infty P(n^a X_n \geq \epsilon)$ for all $a \in [0,1)$, $\epsilon>0$
Hint: You can make use of the fact that $\log(1-x) \leq -x$ holds for all $x<1$
(iv) $n^a X_n \to 0$ almost surely - got this one if I use (iii)

Normally I'd show exercises like (iii) with Borel-Cantelli-Lemma but here it seems not useful (further I use Borel-Cantelli in (iv) to show almost sure convergence).
Therefore I want to compute $P(n^a X_n \geq \epsilon)$ directly and hope I get sth like $\frac{1}{n^2}$ to get convergence of the sum.
Okay, let's try:
$$P(X_n \geq n^{-a} \epsilon)=1-F_n(n^{-a} \epsilon)=\left(1-\frac{\epsilon}{n^a} \right)^n $$
It seems a bit like $(1+\frac{x}{n})^n \to e^x$. I don't know if this is correct up to now since I didn't use the hint.
I'd have to check: $\sum \left(1-\frac{\epsilon}{n^a} \right)^n < \infty$
Maybe with the root test: $\sqrt[n]{ |a_n|}=1-\frac{\epsilon}{n^a}<1=C$ and this implies convergence. But this feels not right.
Can somebody please tell me how to solve this (maybe as the hint implies with the logarithm.)
Thanks a lot, Marvin

By the elementary estimate $\log(1-x) \leq -x$, $x \in (0,1)$, we have
$$\log \left( 1- \frac{\epsilon}{n^a} \right) \leq -\frac{\epsilon}{n^a}<0$$
implying
$$\begin{align*} \left( 1- \frac{\epsilon}{n^a} \right)^n &= \exp \left( n \log \left[1- \frac{\epsilon}{n^a} \right] \right) \\ &\leq \exp \left( - n \frac{\epsilon}{n^a} \right) = \exp(-\epsilon n^{1-a}). \end{align*}$$
Since the exponential function is growing faster than any polynomial, we can find for any $N \in \mathbb{N}$ some constant $c>0$ such that $x^N \leq c \exp(x)$ for all $x > 0$, or, equivalently,
$$\exp(-x) \leq \frac{1}{c} \frac{1}{x^N}$$
for all $x > 0$. Applying this for $x = \epsilon n^{1-a}$ and choosing $N \in \mathbb{N}$ sufficiently large, it is not difficult to see that
$$\sum_{n \geq 0} \exp(-\epsilon n^{1-a}) <\infty$$
for any $a \in [0,1)$. Consequently, we find
$$\sum_{n \geq 0} \mathbb{P}(X_n \geq \epsilon n^{-a}) \leq \sum_{n \geq 0} \exp(-\epsilon n^{1-a}) < \infty.$$
