a.s. convergence with CLT Let's say I have $S_n = \sum_{i=1}^n X_i$, where $X_{i} = 1$ with probability $p$ and $X_i=0$ with probability $1-p.$
$$\operatorname E(X_1) =p, \quad \operatorname{Var}(X_1)= p-p^2. $$
By the central limit theorem, $\frac{S_n-pn}{\sqrt{n}} \rightarrow Z \sim N(0,p-p^2)$ in distribution.
Can it also be said that this convergence is almost sure?
 A: My answer  refers only to the case $X_1,X_2,\dots,$ are IID. Not sure if that's what OP asked, but I hope it helps. 
In this case, the answer is in general NO with the only exception being the trivial case where the variance is $0$.
Let's make the discussion slightly more general. Suppose that $Y_1,Y_2,\dots$ are IID with mean $0$ (e.g. $Y_j =X_j - E[X_j]$) and variance $\sigma^2>0$. Let $S_n = Y_1+\dots+Y_n$. 
Now let  $Z_n = S_n /\sqrt{n}$. 
We will show that the probability that $Z_n$ converges (pointwise) is zero. 
Saying that $Z_n$ converges (pointwise) a.s. is equivalent to the sequence $(Z_n:n\in{\mathbb N})$ being Cauchy, a.s. That is, with probability $1$, for every $\epsilon$ there exists $N_\epsilon \in \{1,2,\dots\}$, and which may be random, such that $|Z_n - Z_m|<\epsilon$ for all $n>m\ge N_\epsilon$. 
Observe that $Z_n -Z_m = \frac{S_n-S_m}{\sqrt{n}} + S_m \left (\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{m} }\right)$. 
Then freezing $m$ and letting $n\to\infty$, the CLT gives 
$$ P(|Z_n -Z_m|\ge \epsilon) \to P(|N(0,\sigma^2)|\ge \epsilon)=1-\delta(\epsilon).$$ 
Note that $\delta(\epsilon)\to 0$ as $\epsilon \to0$. 
Therefore, for any fixed $m$, for $n$ large enough we have that  
$$ P(|Z_n-Z_m|\ge \epsilon) > 1-2\delta(\epsilon),$$
As a result
$$ P(|Z_n-Z_m|<\epsilon \mbox{ for all }n>m) \le 2\delta(\epsilon).$$ 
But this event contains $N_\epsilon \le m$. Therefore, 
$$ P(N_\epsilon \le m) \le 2\delta(\epsilon).$$
Taking $m\to \infty$, we have
$$ P(N_\epsilon <\infty) \le 2 \delta(\epsilon)\to 0.$$ 
Since $N_\epsilon$ increases as $\epsilon \to 0$, it follows that 
$$P(N_\epsilon < \infty\mbox{ for all }\epsilon>0) =0.$$ 
Therefore the probability pointwise convergence is zero.  
A: No.   
The closest approach (under i.i.d. sampling) to what you ask for is, I think, the Law of the Iterated Logarithm, which in your case implies $\limsup_{n\to\infty} (S_n-np)/\sqrt{2 n\log\log } = \sqrt{p(1-p)}$ almost surely and $\liminf_{n\to\infty} (S_n-np)/\sqrt{2 n\log\log } = -\sqrt{p(1-p)}$ almost surely.
That is, $(S_n-np)/\sqrt n$ fluctuates above and below what you want by an extra factor of $O(\sqrt{\log\log n})$.  The sample trajectories of the CLT re-centered and re-scaled $S_n$ are usually as large as a $N(0,1)$ variable should be, but they have occasional excursions.
Put another way, the LIL implies $$
\limsup_{n\to\infty}\frac{S_n-np}{\sqrt{np(1-p)}} = +\infty\text{  a.s.}$$
and
 $$
\liminf_{n\to\infty}\frac{S_n-np}{\sqrt{np(1-p)}} = -\infty\text{  a.s.}$$
which in turn implies $(S_n-np)/\sqrt{np(1-p)}$ does not converge almost surely to any real random variable. (Or even any extended real random variable.)
A: One can speak of almost sure convergence to a random variable, but there is no such thing as almost sure convergence to a distribution.
A: One can construct a sequence of random variables $\{Y_n\}$ and a random variable $Y_{\infty}$ defined on a common probability space s.t. $Y_n\overset{d}{=}n^{-1/2}(S_n-pn)$, $Y_{\infty}\sim N(0,p(1-p))$, and $Y_n\to Y_{\infty}$ a.s.
