Finding a sequence from weak convergence Let $(X_n)_n$ be a sequence of independent random variable, $(u_n)_n$ a sequence of positive numbers, such that $$\frac{1}{u_n}\sum_{k=1}^nX_k \Rightarrow X$$
where $X$ is not degenerate.
Prove that either $(u_n)_n$ converges or there exist an increasing sequence $(v_n)_n$ such that $\lim_n v_n =+\infty,\lim_n v_n/u_n=1,$ for all $n \in \mathbb{N},u_n \leq v_n.$
This question was asked before Finding a sequence from weak convergence, but it seems the construction of $(v_n)_n$ is not trivial.
How is it possible to define $(v_n)_n$ ?
 A: 1.  by the symmetrization, we may discuss the sequence $ \{X_n,n\ge 1\} $ of independent random variables with following representation only:
\begin{equation*}
    X_k=Y_k^\prime-Y_k^{\prime\prime},
\end{equation*}
where $ \{Y_k^\prime,Y_k^{\prime\prime},k\ge 1\} $ is a sequence of independent random variables and $ Y_k^\prime\stackrel{d}{=}Y_k^{\prime\prime}, k\ge 1 $. For these $ X_k $, the characteristic function of $ X_k $,    $\phi_{X_k}(t)=|\phi_{Y_k^\prime}(t)|^2$, hence $ 0\le \phi_{X_k}(t) \le 1 $.
If $ S_n=\sum\limits_{1\le k\le n}X_k$, then
\begin{equation*}
0\le \phi_{S_{n+1}}(t)=\phi_{S_n}(t)\phi_{X_{n+1}}(t) \le \phi_{S_n}(t)\tag{1}
\end{equation*}
2. If $ \{u_n, n\ge 1\} $ does not converge( to a finite limit), then
\begin{equation*}
    \lim_{n\to\infty}u_n=+\infty. \tag{2}
\end{equation*}
Proof:  If $ \{u_n, n\ge 1\} $ does not converge and (2) is false, then there exists a subsequence $ \{u_{n_k}, k\ge 1\} $ with
\begin{equation*}
    \lim_{k\to\infty} u_{n_k}\to u<\infty  \tag{3}
\end{equation*}
Meanwhile, using convergence of type theorem we can deduce that
\begin{equation*}
    S_{n_k}\implies uX     \tag{4} 
\end{equation*}
from $ \frac1{u_{n_k}}S_{n_k}\implies X $ and (3). Furthermore, from (1) and (4), we could get
\begin{equation*}
    S_n\implies uX  \tag{5}
\end{equation*}
too. Using convergence of type theorem  again, from (5) and $ \frac1{u_n}S_n\implies X $ we can deduce $ \{u_n, n\ge 1\} $ converge to $ u $, which is a contradiction and (2) is true.
3. Now we are ready to construct $ \{v_n, n\ge 1\} $. Let
\begin{align*}
    v_n&=\max_{1\le j\le n} u_n, \quad n\ge 1. \tag{6}\\
    n_1&=1,\\
    n_{k+1}&=\min\{n>n_k:u_n>v_{n_k}\}, \qquad k\ge 1.
\end{align*}
Then
\begin{gather*}
        u_n\le v_n\le v_{n+1}, \qquad n\ge 1.\\
        n_{k+1}>n_k, \qquad v_{n_{k+1}}=u_{n_{k+1}}>v_{n_k}=u_{n_k}, \qquad k\ge 1. \tag{7}
\end{gather*}
Now from (7) we have
\begin{equation*}
    \frac{S_{n_k}}{v_{n_k}}=\frac{S_{n_k}}{u_{n_k}}\implies X,\quad\text{as}\quad k\to\infty \tag{8}
\end{equation*}
Using convergence of type theorem(random version)  again, from (8) we have
\begin{gather*}
    \lim_{k\to\infty}\frac{v_{n_{k+1}}}{v_{n_k}}=1, \tag{9}\\
    \mathrm{pr}-\lim_{k\to\infty}\frac{S_{n_{k+1}}-S_{n_k}}{v_{n_k}}=0.\tag{10}
\end{gather*}
Using Lévy's inequality and (10) we have
\begin{align*}
    &\mathsf{P}\Big(\max\limits_{n_k\le n\le n_{k+1}}|S_n-S_{n_k}|>\varepsilon v_{n_k}\Big)\\
    &\quad \le 2 \mathsf{P}\Big(|S_{n_{k+1}}-S_{n_k}|>\varepsilon v_{n_k}\Big)\to 0, \quad 
    \text{as}\quad k\to\infty, \quad \forall \varepsilon>0. \tag{11}
\end{align*}
Now prove the following:
\begin{equation*}
    \frac{S_n}{v_n}\implies X. \tag{12}
\end{equation*}
For $ n_k\le n\le n_{k+1} $,
\begin{align*}
    &\Big|\dfrac{S_n}{v_n}-\frac{S_{n_k}}{v_{n_k}}\Big|\\
    &\quad \le \dfrac1{v_n}\Big|S_n-S_{n_k} \Big| +\dfrac{S_{n_k}}{v_{n_k}}\Big|\dfrac{v_{n_k}}{v_n}-1\Big|\\
    &\quad \le\dfrac{\max\limits_{n_k\le n\le n_{k+1}}|S_n-S_{n_k}|}{v_{n_k}}|
    +\dfrac{S_{n_k}}{v_{n_k}}\Big|\dfrac{v_{n_k}}{v_{n_{k+1}}}-1\Big|\\
    &\quad\overset{\mathsf{P}}{\to}0,\qquad (\text{by (9,11) })
\end{align*}
Hence (12) is true from (8). Using convergence of type theorem again, from (12) we have
\begin{align*}
    \lim_{n\to\infty}\frac{u_n}{v_n}=1.
\end{align*}
