Distribution limit with $O(1)$ term Let's say we have a sequence of random variables $X_n$ where $V = \text{var}(X_n)$. Suppose that (this is convergence in distribution)
$$V^{-1/2}(X_n - E(X_n)) \leadsto N(0,1)$$
I'm wondering, is it true that $$V^{-1/2}(O(1)X_n - E(X_n)) \leadsto N(0,1)??$$
Thanks
Edit: I realized this shouldn't hold for every $O(1)$. Instead, let's replace $O(1)$ for a nonstochastic sequence $a_n \rightarrow 1$. I think a simple proof could follow by Slutsky theorem: If $X_n \leadsto N(0,1)$ and $X_n - Y_n$ converges to $0$ in probability, then $Y_n \leadsto N(0,1)$.
So, if $V^{-1/2}(X_n - E(X_n)) - V^{-1/2}(a_nX_n - E(X_n))$ converges to $0$ in probability, that's it. But
$$V^{-1/2}(X_n - E(X_n)) - V^{-1/2}(a_nX_n - E(X_n)) = V^{-1/2}(X_n - a_n X_n).$$ Is it true that $(X_n - a_n X_n)$ converges to $0$ almost surely? (I think is easier to check that instead of convergence in probability). Specifically, is it true that
$$P \Big(\lim_{n \rightarrow \infty} (X_n - a_n X_n)=0 \Big) = P \Big(\lim_{n \rightarrow \infty} X_n - \lim_{n \rightarrow \infty} a_n \lim_{n \rightarrow \infty} X_n=0 \Big)?$$
 A: Actually, in this case it isn't too bad to show that the sequence you are interested in converges to $0$ in probability using Chebyshev's inequality. Assuming $a_n\xrightarrow{n\to\infty}1$,
\begin{align*}
P(V_n^{-1/2}(X_n-a_nX_n)\ge \varepsilon) &= P\big((1-a_n)X_n\ge \varepsilon\sqrt{E(X_n^2)}\big)\\
&\text{Apply Markov $P(Z\ge 1)\le E|Z|$ with $Z = \frac{(1-a_n)^2X_n^2}{\varepsilon^2E(X_n^2)}$}\\
&\le \frac{1}{\varepsilon^2E(X_n^2)}(1-a_n)^2E(X_n^2) = \varepsilon^{-2}(1-a_n^2)\xrightarrow{n\to\infty}0.
\end{align*}
By Slutsky's theorem (also called "the converging together lemma"), $$V_n^{-1/2}(a_nX_n-E(X_n))\stackrel{n\to\infty}{\leadsto} N(0,1).$$

Added: If $|X_n|<\infty$ a.s. are arbitrary random variables and $b_n$ is a sequence converging to $0$ sufficiently fast, then we will have $b_nX_n\to 0$ a.s. To see this, pick $b_n > 0$ so that $P(b_n|X_n| > 1/n) < 2^{-n}$. Then $\sum P(b_n|X_n|>1/n) < \infty$, and by Borel-Cantelli, $b_nX_n\to 0$ a.s.
This leads me to believe that if all we assume about the sequence $a_n$ is that $a_n\to 1$, then we can't conclude in general that $(a_n-1)X_n\to 0$ a.s. This is clearly equivalent to asking whether if $b_n\to 0$ we have $b_nX_n\to 0$ a.s.
