Convergence of random variables and Law of large numbers 
Let $A_1,A_2,\cdots$ and $B_1,B_2,\cdots$ be two independent sequences of i.i.d random variables according to probability density function $p(x)$, with $\mathbb{E}[A_1]=\mathbb{E}[B_1]<\infty$ and $Var[A_1]=Var[B_1]<\infty$. For a fixed constant $c>0$, find the constant $k>0$ such that
$$\lim\limits_{n \to \infty}\mathbb{P}\left(\frac{1}{n}\sum\limits_{i=1}^{i=n}A_i\geq \left(1-\frac{c}{\sqrt{n}}\right)\left(1-\frac{k}{\sqrt{n}}\right)\frac{1}{n+k\sqrt{n}}\sum\limits_{i=1}^{i=n+k\sqrt{n}}B_i \right)=1             $$

My approach: $\frac{1}{n}\sum\limits_{i=1}^{i=n}A_i$ is a random variable and weak law of large numbers yields
$$\frac{1}{n}\sum\limits_{i=1}^{i=n}A_i\xrightarrow{p}\mathbb{E}(A_i)\,\,\,\,\,\,\,\,\,\,\,(1)$$
Similarly,
$$ \frac{1}{n+k\sqrt{n}}\sum\limits_{i=1}^{i=n+k\sqrt{n}}B_i \xrightarrow{p}\mathbb{E}(B_i)=\mathbb{E}(A_i)\,\,\,\,\,\,\,\,\,\,\,(2)$$
[I am not sure about this step:] Therefore, for all $k$
$$\left(1-\frac{c}{\sqrt{n}}\right)\left(1-\frac{k}{\sqrt{n}}\right)\frac{1}{n+k\sqrt{n}}\sum\limits_{i=1}^{i=n+k\sqrt{n}}B_i \xrightarrow{p}\mathbb{E}(B_i)=\mathbb{E}(A_i)\,\,\,\,\,\,\,\,\,\,\,(3)$$
Now, by (1) and (3)
$$\left[\frac{1}{n}\sum\limits_{i=1}^{i=n}A_i-\left(1-\frac{c}{\sqrt{n}}\right)\left(1-\frac{k}{\sqrt{n}}\right)\frac{1}{n+k\sqrt{n}}\sum\limits_{i=1}^{i=n+k\sqrt{n}}B_i \right]\xrightarrow{p}0$$
Thus, for all $k$
$$\lim\limits_{n \to \infty}\mathbb{P}\left(\frac{1}{n}\sum\limits_{i=1}^{i=n}A_i\geq \left(1-\frac{c}{\sqrt{n}}\right)\left(1-\frac{k}{\sqrt{n}}\right)\frac{1}{n+k\sqrt{n}}\sum\limits_{i=1}^{i=n+k\sqrt{n}}B_i \right)=1             $$
Is my approach correct? Is there a simpler precise method to prove? In the problem statement, if we had a function $f(n,c,k)$ instead of $\left(1-\frac{c}{\sqrt{n}}\right)\left(1-\frac{k}{\sqrt{n}}\right)$, such that $\lim\limits_{n\to\infty}f(n,c,k)=1$, could we use the same approach?
 A: Assume $\{A_i\}$ and $\{B_i\}$ are independent i.i.d. processes with $Var(A_1)=\sigma^2$. If $0<\sigma^2 < \infty$, then there is no such $k$. 
Proof:  Define $m=E[A_1]$. Fix $c>0$ and $k>0$ as arbitrary constants.  Define the following events: 
\begin{align}
\mathcal{C}_n &= \left\{\frac{1}{n}\sum_{i=1}^n A_i \geq (1-\frac{c}{\sqrt{n}})(1-\frac{k}{\sqrt{n}})\frac{1}{\lceil n + k\sqrt{n}\rceil}\sum_{i=1}^{\lceil n+k\sqrt{n}\rceil}B_i\right\} \\
\mathcal{D}_n &= \left\{\frac{1}{\lceil n+k\sqrt{n}\rceil}\sum_{i=1}^{\lceil n + k\sqrt{n}\rceil}B_i \geq m\right\}
\end{align}
Notice that $P[\mathcal{D}_n]\rightarrow 1/2$ by the central limit theorem (CLT).  Also, $\mathcal{D}_n$ is independent of the $\{A_i\}$ process. Hence, for all $n$ large enough to ensure $1>(1-\frac{c}{\sqrt{n}})(1-\frac{k}{\sqrt{n}})> 0$, we have: 
\begin{align}
P[\mathcal{C}_n|\mathcal{D}_n] &\leq P\left[\frac{1}{n}\sum_{i=1}^nA_i \geq (1-\frac{c}{\sqrt{n}})(1-\frac{k}{\sqrt{n}})m|\mathcal{D}_n\right]\\
&=P\left[\frac{1}{n}\sum_{i=1}^nA_i \geq (1-\frac{c}{\sqrt{n}})(1-\frac{k}{\sqrt{n}})m \right] \\
&=P\left[\frac{1}{\sqrt{n}}\sum_{i=1}^n(A_i-m) \geq - m(c+k) + \frac{ckm}{\sqrt{n}} \right] \rightarrow q
\end{align}
where we define $q=P[G \geq -m(c+k)]$ for $G$ a Gaussian with mean 0 and variance $\sigma^2$. Note that $0<q<1$. Then: 
\begin{align}
P[\mathcal{C}_n] &= P[\mathcal{C}_n|\mathcal{D}_n]P[\mathcal{D}_n] + P[\mathcal{C}_n|\mathcal{D}_n^c]P[\mathcal{D}_n^c] \\
&\leq  P[\mathcal{C}_n|\mathcal{D}_n]P[\mathcal{D}_n] + P[\mathcal{D}_n^c] 
\end{align}
and so
$$ \limsup_{n\rightarrow\infty} P[\mathcal{C}_n] \leq q(1/2)+(1/2)  < 1 \quad \Box$$
