Extreme value theory: show that $ \lim_{n\rightarrow \infty}a_n $ exists and is finite Well known facts in extreme value theory: 


*

*Let $\{X_i\}_{\forall i \in \{1,...,n\}}$ be i.i.d. random variables with cdf $F$. If there exists $\{a_n\}_{n\in \mathbb{N}}>0$, and $\{b_n\}_{n\in \mathbb{N}}\in \mathbb{R}$ such that $Z_n\equiv \frac{M_n-b_n}{a_n} \Rightarrow_n Z$, where $Z$ has distribution of the same type as Gumbel and $M_n\equiv  \max_{i\in \{1,...,n\}}X_i$, then we say that $F$ is in the domain of attraction of the Gumbel.  

*A necessary and sufficient condition for being in the domain of attraction of Gumbel is
$$
\exists \text{ }A:\mathbb{R}\rightarrow (0,\infty) \text{ s.t. } \lim_{s\rightarrow w(F)}\frac{1-F(s+v A(s))}{1-F(s)} = e^{-v}\text{ }\forall v \in \mathbb{R}
$$
where $w(F)$ is the right end point of $F$ and $A$ is called auxiliary function of $F$.
Below I will focus on distributions for which this necessary and sufficient condition holds. In this case:


*

*The norming constants can be taken
$$
b_n\equiv F^{-1}(1-\frac{1}{n})
$$
and
$$
a_n\equiv A(b_n)
$$
where $F^{-1}$ denotes quantile function.

*Auxiliary functions are not unique. If the pdf $f$ of $F$ exists, an auxiliary function is
$$
A(x)\equiv \frac{1-F(x)}{f(x)}
$$
Question:
Assume that $F$ is continuous and $X_i$ has unbounded support. Then,
$$
\lim_{n\rightarrow \infty}b_n\equiv \lim_{n\rightarrow \infty}F^{-1}(1-\frac{1}{n})=F^{-1}(1-\lim_{n\rightarrow \infty} \frac{1}{n})=F^{-1}(1)=\infty
$$
A proof that I'm considering uses also that
$$
\lim_{n\rightarrow \infty}a_n
$$
exists and is finite, in order to prove, in turn, that
$$
\lim_{n\rightarrow \infty} F(a_nt+b_n)=1 \text{ }\forall t \in \mathbb{R} 
$$
Could you help me to show that $$
\lim_{n\rightarrow \infty}a_n
$$
exists and is finite?
 A: If I understood your question correctly, I am not sure that you have to implicate whether $\lim_{n \rightarrow \infty}a_n$ is finite or not in proving that $\lim_{n\rightarrow \infty} F(a_nt+b_n)=1$. If you know that $F$ satisfies the condition you mention and $w(F) = \infty$ then
$$\lim_{s\rightarrow \infty}\frac{1-F[s+v A(s)]}{1-F(s)} = e^{-v}\text{ }\forall v \in \mathbb{R}$$
you also know that
$$\lim_{s\rightarrow \infty} [1-F(s)] = 0$$
Thus, $\forall v \in \mathbb{R}$
$$\lim_{s\rightarrow \infty}\left([1-F(s)]\frac{1-F[s+v A(s)]}{1-F(s)} \right) = 0e^{-v} = 0$$
But
$$\lim_{s\rightarrow \infty}\left([1-F(s)]\frac{1-F[s+v A(s)]}{1-F(s)} \right) = \lim_{s\rightarrow \infty}\Big(1 - F[s+v A(s)]\Big)$$
From these two, you get
$$\lim_{s\rightarrow \infty}F[s+v A(s)] = 1$$
That means that for every sequence of $s$ going to $\infty$ the limit will be the same. Setting $s=a_n$ and taking into account that $b_n = A(a_n)$, you get
$$\lim_{n\rightarrow \infty}F(b_n + v a_n) = 1$$
which is what you are trying to prove. This further gives that $\lim_{n\rightarrow \infty}(b_n + v a_n) = \infty$ which is not affected by whether $\lim_{n\rightarrow \infty}a_n$ is finite or not.
If, on the other hand, you try to prove for an arbitrary continuous distribution $F$ with unbounded support that $\lim_{n\rightarrow \infty}a_n$ is finite, then $f(x) = \frac{1}{2} e^{-\sqrt x}$ gives you a counterexample.
