Convergence in probability for Sum of RV divided by sequence 
Let $\left\{X_n\right\}_{n \geq 1}$ be such (arbitrary) random variables that $\mathbb{E}\left(X_n\right)=$ $0, n \geq 1$. Define $S_n=\sum_{k=1}^n X_k$ and assume that.
$$
\lim _{n \rightarrow \infty} \mathbb{P}\left(\frac{S_n}{\sqrt{n}} \leq x\right)=\Phi(x), \quad x \in \mathbb{R} .
$$
Show that $S_n/a_n \rightarrow 0$ in probability for any sequence $\left\{a_n\right\}_{n \geq 1}$, with the property that $a_n/\sqrt{n} \rightarrow \infty$.


Here are some of my thoughts:
We need to show that $\tfrac{S_n}{a_n} \rightarrow 0$ converges in probability, which means that.
$$
\forall \varepsilon>0, \quad P\left(\left|\tfrac{S_n}{a_n}\right|\geq\varepsilon\right) \stackrel{n \rightarrow \infty}{\longrightarrow} 0.
$$
We know that
$$
\frac{a_n}{\sqrt{n}} \rightarrow \infty,
$$
that is, the sequence $\left\{a_n\right\}_{n \geq 1}$ grows faster than $\sqrt{n}.$
\item We also know from the script that for the random variables in this task.
$$
\frac{S_n}{n}\stackrel{P}{\longrightarrow} \mathbb{E}[X_1]=0.
$$
According to Cebysev
$$
P\left(\left|\tfrac{S_n}{a_n}\right|\geq\varepsilon\right) \leq \frac{\operatorname{Var}(\tfrac{S_n}{a_n})}{\varepsilon^2} .
$$
Is the variance here simply $0$?

Any hints would be appreciated.
 A: You don't need Chebyshev's inequality (and the condition $E(X_n) = 0$) to get the result, simply use the definition of convergence in probability. Our goal is to show that for any given $\varepsilon > 0$, $P[|S_n| > |a_n|\varepsilon] \to 0$ as $n \to \infty$.
For arbitrary $\delta > 0$, choose $M$ sufficiently large so that $\Phi(M\varepsilon) > 1 - \delta/2$.  Since $\lim_{n \to \infty}|a_n|/\sqrt{n} = +\infty$,  there exists a positive integer $N$ such that for all $n > N$, $|a_n|/\sqrt{n} > M$. Therefore, for every $n > N$:
\begin{align}
& P(|S_n/a_n| > \varepsilon) = P(|S_n/\sqrt{n}| > |a_n|\varepsilon/\sqrt{n}) 
\leq P[|S_n/\sqrt{n}| > M\varepsilon] \\
=& P[S_n/\sqrt{n} > M\varepsilon] + P[S_n/\sqrt{n} < -M\varepsilon] \\
\to & 1- \Phi(M\varepsilon) + \Phi(-M\varepsilon) = 2(1 - \Phi(M\varepsilon)) < \delta.
\end{align}
Since $\delta$ is arbitrarily small, the result follows.
In fact, you can directly apply the Slutsky's theorem to this exercise:
\begin{align}
\frac{S_n}{a_n} = \frac{S_n}{\sqrt{n}}\times\frac{\sqrt{n}}{a_n} \to_d 0
\end{align}
as $S_n/\sqrt{n} \to_d \Phi$ and $\sqrt{n}/a_n \to 0$.
A: Presumably $\Phi$ in the OP is the standard normal distribution. In any event, that is not important and we can in general assume that
$\Phi$ is a probability distribution function and that $\Phi_n(x)=P[S_n/\sqrt{n}\leq x]\xrightarrow{n\rightarrow\infty} \Phi(x)$ at any continuity point $x$ of $\Phi$.  This means that $S_n/n$ converges in distribution to $\Phi$ (or the probability measure induced by $\phi$ rather); hence, the  characteristic function $\phi_n$ of $S_n/n$ converges to the characteristic function $\phi$ of the distribution $\Phi$ uniformly in compact subsets of $\mathbb{R}$. Notice that
$$\psi_n(t)=E\big[e^{iS_n t/a_n}\big]=\phi_n\Big(t\frac{\sqrt{n}}{a_n}\Big)$$
The claim in the OP will follow by Lévy-Bochner's continuity theorem once it is established that $\psi_n$ converges to $1$ for all $t\in\mathbb{R}$.

Lemma: suppose $\phi_n\subset\mathcal{C}(\mathbb{R})$ converges uniformly to $\phi$ (which necessarily is continuous) in compact subsets. If $\alpha_n\xrightarrow{n\rightarrow\infty}0$, then $\phi_n(\alpha_nt)$ converges uniformly to $\phi(0)$ in compact subsets of $\mathbb{R}$.

The continuity of $\phi$ implies that for any $\varepsilon>0$, there is $\delta>0$ such that if $|t|\leq\delta$, then $|\phi(t)-\phi(0)|<\varepsilon/2$. Let $[-L,L]$ a closed bounded subset in $\mathbb{R}$. There is $N\in\mathbb{N}$ such that if $n\geq N$, then $|\alpha|_n<\delta/L$ and $\sup_{|s|\leq\delta}|\phi_n(s)-\phi(s)|<\varepsilon/2$. Hence, for any $t\in[-L,L]$ and $n\geq N$,
\begin{align}
|\phi_n(\alpha_n t)-\phi(t)|&\leq |\phi_n(\alpha_n t)-\phi(\alpha_n t)|+|\phi(\alpha_n t)-\phi(0)|\\
&\leq \sup_{|s|\leq \delta}|\phi_n(s)-\phi(s)|+ \varepsilon/2< \varepsilon
\end{align}
This shows that $\phi_n(\alpha_n\cdot)$ converges to $\phi(0)$ uniformly in compact subsets of $\mathbb{R}$.
