Proof of strengthening of Borel-Cantelli Lemma If $A_1$,$A_2$, $\dots$ are pairwise-independent events with $\sum_{n \in \Bbb N} \Bbb P(A_{n})=\infty$ , then show for$$Q_{n}=\frac{\sum _{j=1}^{n} \Bbb 1_{A_{j}}}{\sum _{j=1}^{n} \Bbb P(A_{j})}$$
we have $$ \lim_{n \to \infty}Q_{n}=1,a.s.$$ 
The hint said we need to establish convergence in probability first, then show almost surely convergence: First for the subsequence ${Q_{n_{k}}}$ defined by $$n_{k}=\inf\left\{n \in \Bbb N:\sum _{j=1}^{n} \Bbb P(A_{j}) \geq k^{2}\right\}, k \in \Bbb N$$ then proceed to the entire sequence.
My question is that, how do we proceed from the subsequence to the entire sequence? I've proved $Q_{n}$ converges to $1$ in probability and the subsequence ${Q_{n_{k}}}$ converges to $1$ almost surely. I guess we need to show $Q_{n}$ is a Cauchy sequence almost surely, then we get the result since a Cauchy sequence which contains a convergent subsequence converges. I am stuck on this step and could someone help me with it?
Thanks in advance!
 A: In order to pass from the convergence of the subsequence $(Q_{n_k})$ to the whole sequence $(Q_n)$, it suffices to show that $\sum_k \mathbb E\left[Y_k^2\right]$ is finite, where 
$$
Y_k=\max_{n_k+1\leqslant N\leqslant n_{k+1}-1}\lvert Q_N-Q_{n_k}\rvert.
$$
[By the Borel-Cantelli lemma, $Y_k\to 0$ almost surely.]
To do so, we first bound $\lvert Q_N-Q_{N_k}\rvert$. We use the notation $R_i:=\sum_{j=1}^i \mathbb P(A_j)$ and for $i\lt i'$, 
$S(i,i')=\sum_{j=i+1}^{i'}\mathbf 1(A_j)$. Then 
$$
Q_N-Q_{N_k}=\frac{S(1,N)}{R_N}-\frac{S(1,n_k)}{R_{n_k}}
=\frac{S(1,n_k)}{R_N}+\frac{S(n_k,N)}{R_N} -\frac{S(1,n_k)}{R_{n_k}}
$$
hence 
$$
\lvert Q_N-Q_{N_k}\rvert\leqslant S(1,n_k)\left(\frac 1{R_{n_k}}-\frac 1{R_N}\right)+ \frac{S(n_k,n_{k+1}-1)}{R_{n_k}}.
$$
Since 
$$
\frac 1{R_{n_k}}-\frac 1{R_N}=\frac{R_N-R_{n_k}}{R_{n_k}R_n}\leqslant 
\frac{R_{n_{k+1}-1}-R_{n_k}}{R_{n_k}^2}
$$
we derive the bound 
$$
Y_k^2\leqslant 2 S(1,n_k)^2\left(\frac{R_{n_{k+1}-1}-R_{n_k}}{R_{n_k}^2}\right)^2
+2\left(\frac{S(n_k,n_{k+1}-1)}{R_{n_k}}\right)^2.
$$
Now we use the pairwise independence to get 
$$
\mathbb E\left[S(1,n_k)^2\right]=\sum_{1\leq j\neq j'\leq n_k}
\mathbb P(A_j)P(A_{j'})+R_{n_k}\leqslant R_{n_k}^2+R_{n_k}
$$
and by the same reasoning, 
$$
\mathbb E\left[S(n_k,n_{k+1}-1)^2\right]\leqslant \left(R_{n_{k+1}}-R_{n_k}\right)^2+R_{n_{k+1}}-R_{n_k}.
$$
We thus got 
$$
\tag{*}\mathbb E\left[Y_k^2\right]\leqslant 2\left(R_{n_k}^2+R_{n_k}\right)\left(\frac{R_{n_{k+1}-1}-R_{n_k}}{R_{n_k}^2}\right)^2
+2 \frac{\left(R_{n_{k+1}}-R_{n_k}\right)^2+R_{n_{k+1}}-R_{n_k}}{R_{n_k}^2}. 
$$
Due to the definition of $n_k$, $R_{n_k}\geqslant k^2$ and $R_{n_{k+1}-1}
\lt (k+1)^2$ hence $R_{n_{k+1}}-R_{n_k}\leqslant 2k+1$ and the right hand side of (*) is bounded by $Ck^{-2}$. 
