We get $L^2$ convergence, but do we get a.s. convergence?

Assume you have a sequence of independent Bernoulli random variables$X_i$ each with probability $p_i$. Let $c_i$ be a sequence of real numbers and $ m,M$ be a real numbers such that$0 < m <c_i<M< \infty $. Define

$$Y_n=\frac{\sum\limits_{i=1}^n c_i X_i -\sum\limits_{i=1}^n c_i p_i}{\sum\limits_{i=1}^n c_i}.$$

Then $Y_n$ converges to zero in $L^2$, but do we have the same a.s.?

Proof that $Y_n$ converges to zero in $L^2$:

$E[Y_n^2]=E[Y_n^2]-E[Y_n]^2=V(Y_n)=\frac{\sum c_i^2 p_i(1-p_i)}{(\sum c_i )^2}\le \frac{nM^2}{n^2m^2}=\frac{M^2}{m^2n},$

and this last quantity goes to zero as $n$ goes to infinity.