How to find this limit using squeeze theorem? 
Let $\{a_n\}_{n=1}^{\infty}$ be a sequence of real numbers and let $\{p_n\}_{n=1}^{\infty}$ be a sequence of positive real numbers
  so that $$\lim \limits_{n \to \infty} \frac{1}{\sum_{k=1}^n p_k}=0.$$ Prove that if $\{a_n\}_{n=1}^{\infty}$ converges to $a\in \Bbb R$, then $$\lim \limits_{n \to \infty} \frac{\sum_{k=1}^np_ka_k}{\sum_{k=1}^n p_k}=a.$$

I am not being able to apply squeeze theorem: I tried $$\left\lvert \frac{\sum_{k=1}^np_k(a_k-a)}{\sum_{k=1}^np_k}\right\rvert\le\frac{\sum_{k=1}^n\left\lvert p_k \right\rvert \left\lvert a_k-a\right\rvert}{\sum_{k=1}^n p_k}\le\frac{\sum_{k=1}^N\left\lvert p_k \right\rvert \left\lvert a_k-a\right\rvert+\sum_{k=N+1}^n|p_k|}{\sum_{k=1}^n p_k}$$ for some suitable $N$. but I can't get rid of that index $n$ in the last inequality.
 A: (edit- My answer is morally the same as RRL's but not correct because you ask to use the squeeze theorem. Sorry!)
Pick $N=N(ε)$ such that for all $k>N$, $|a_k - a| < ε$. Then
\begin{align}
 \left|\frac{∑_1^n p_k (a_k-a)}{\sum_1^n p_k}\right| 
&\leq  \left|\frac{∑_1^N p_k (a_k-a)}{\sum_1^n p_k}\right|+  \left|\frac{∑_{N+1}^n p_k (a_k-a)}{\sum_1^n p_k} \right| \\
&\leq \underbrace{\left|\frac{∑_1^N p_k (a_k-a)}{\sum_1^n p_k}\right|}_{\rightarrow_{n} 0} +  \underbrace{\left|\frac{∑_{N+1}^n p_k}{\sum_1^n p_k} \right|}_{\rightarrow_n 1}ε  
\end{align}
So for large $n$, the quantity on the left is $ε$ small.
A: For any $\epsilon > 0$ and sufficiently large fixed $N$ we have $-\epsilon < a_n - a < \epsilon $ when $n > N$.  
With $P_n = \sum_{k=1}^n p_k$ we have 
$$\frac1{P_n}\sum_{k=1}^Np_k(a_k-a) -\epsilon \frac{P_n - P_N}{P_n} < \frac1{P_n}\sum_{k=1}^np_ka_k - a  < \frac1{P_n}\sum_{k=1}^Np_k(a_k-a) +\epsilon \frac{P_n - P_N}{P_n}.$$
Now take the limit as $n \to \infty$ and apply the squeeze theorem.
A: I would apply the Stolz-Cesàro theorem with $x_n = \sum_{k=1}^n p_k a_k$ and $y_n = \sum_{k=1}^n p_k$.
