$F(X_{(n-k_n)})\overset{n\to\infty}{\to} 1$? Let $X_1, X_2, \dots$ be iid (possibly heavy tailed) with their df $F$. Notation $X_{(k)}$ represents the $k-$th order statistic, i.e. $X_{(1)}=\min_{i\leq n} X_i$.
Let $k_n\in\mathbb{N}$ fulfill  $$k_n\to\infty, \frac{k_n}{n}\to 0 \text{, as } n\to\infty.$$
Is the following true
$$F(X_{(n-k_n+1)})\overset{n\to\infty}{\to} 1^{-}?$$
If not, what "reasonable" conditions should we assume so it will hold?
More advanced question for dependent $X_i$ is also posted here $F(X_{(n-k_n)})\overset{n\to\infty}{\to} 1$ for time series?
 A: One short thing we should note. The actual ordered random variables depend on the number of i.i.d. elements we are sampling. I'll write $X^n_{(i)}$ to indicate the $i$th order statistic from $\{X_j\}_{j=1}^n$.
The proof follows from the strong law of large numbers. Let $p \in (0,1)$ be such that there exists an $x \in \mathbb{R}$ such that $F(x) = p$ (note that such a $p$ always exists unless $X_i$ is non-random in which case the result is obvious). Define the i.i.d. Bernoulli random variables $B_i = \mathbb{I}\{X_i\leq x\}$, so that each $B_i = 1$ with probability $p$. Then by the strong law of large numbers,
$$\lim_{n\to\infty} \frac{\sum_{i=1}^n B_i}{n} = \mathbb{E}[B_1] = F(x) = p \text{ a.s.}.$$
Let $m_n$ be the smallest index such that $X^n_{(m_n)} > x$, and if $X^n_{(n)} \leq x$ then we say $m_n = n+1$. Note that $m_n = 1 + \sum_{i=1}^n B_i$. Then,
$$\lim_{n\to\infty} \frac{m_n}{n} = \lim_{n\to\infty} \left(\frac{1}{n} + \frac{\sum_{i=1}^n B_i}{n}\right) = p \text{ a.s.}.$$
Note that,
$$\lim_{n\to\infty} \frac{n-k_n+1}{n} = \lim_{n \to\infty} 1 - \frac{k_n}{n} + \frac{1}{n} = 1.$$
Thus, there exists an almost surely finite random integer $N$ such that $n-k_n + 1 > m_n$ on the event $\{N\leq n\}$. Thus,
$$\lim_{n\to\infty} \mathbb{P}(F(X^n_{(n-k_n+1)})\geq p) = \lim_{n\to\infty} \mathbb{P}(X^n_{(n-k_n+1)} \geq x) = \lim_{n\to\infty} \mathbb{P}(N\leq n) = 1.$$
Now we have two cases to consider.
Case 1: There exists a sequence $p_i \to 1-$ such that for every $i$, there exists an $x_i$ such that $F(x_i) = p_i$.
In this case, our proof holds for all i. For any $\epsilon > 0$, let $i$ be sufficiently large that $1 - p_i < \epsilon$. Then,
$$\lim_{n\to\infty}\mathbb{P}\left(|1 - F(X^n_{(n-k_n+1)})| > \epsilon\right) \leq \lim_{n\to\infty}\mathbb{P}\left(F(X^n_{(n-k_n+1)}) < p_i\right) = 0.$$
Case 2: We cannot find such a sequence. This only happens if there exists an $\overline{x} \in \mathbb{R}\cup\{\pm\infty\}$ such that $\mathbb{P}(X_i \geq \overline{x}) = \mathbb{P}(X_i =\overline{x}) = q$ for some $q > 0$. In this case, we can use a similar argument to the one above (again, assuming $X_i$ is random). Let $C_i = \mathbb{I}\{X_i = \overline{x}\}$. Then $\{C_i\}$ are i.i.d. Bernoulli random variables with expectation $q > 0$. Let $M_n = \max\{i: X^n_{(i)} < \overline{x}\}$. Then by the strong law of large numbers,
$$\lim_{n\to\infty} \frac{M_n}{n} = \lim_{n\to\infty} \frac{n - \sum_{i=1}^n C_i}{n} = 1 - q\text{ a.s.}.$$
But,
$$\lim_{n\to\infty} \frac{n-k_n+1}{n} = 1.$$
Thus, there exists an almost surely finite random variable $M$ such that $n-k_n+1 > M_n$ on the event $\{M \leq n\}$. Finally,
$$\lim_{n\to\infty}\mathbb{P}(F(X^n_{(n-k_n+1)}) = 1) = \lim_{n\to\infty}\mathbb{P}(n-k_n+1 > M_n) = \lim_{n\to\infty}\mathbb{P}(M\leq n) = 1.$$ $$\tag*{$\blacksquare$}$$
This concludes the proof that $F(X^n_{(n-k_n+1)}) \to 1$ in probability.
Note that this proof holds very generally. $X_i$ does not need to have any moments at all. Even if it is not integrable, this proof will still be valid. $X$ also does not need to be continuous or discrete. This will work very generally for any Borel measurable random variable in the extended reals ($\mathbb{R}\cup\{\pm \infty\}$).
Also, I proved convergence in probability here. That is, I showed that for any $p < 1$, $\mathbb{P}(F(X^n_{(n-k_n+1)}) > p) = 1 - o(1)$. To get almost sure convergence, we would need to show that the $o(1)$ term is summable, and then we could apply the Borel-Cantelli Lemma.
