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](https://math.stackexchange.com/questions/3970671/if-frac1u-n-sum-k-1nx-k-rightarrow-x-then-we-can-find-a-sequence-v), but it seems the construction of $(v\_n)\_n$ is not trivial.


How is it possible to define $(v\_n)\_n$ ?



Finding a sequence from weak convergence
**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\*}


