Is $\mathbb{P}(S_n/n>p+a)$ a decreasing function of n? In my question, I have i.i.d random variables $X_1,...,X_n$ $\sim \text{Bern}(p)$ (Bernoulli distribution with parameter $p$). 
Denote $S_n = \sum_{i=1}^{n} X_i$. Does $\mathbb{P}(S_n/n>p+a)$ decrease when $n$ increases, where $a$ is a fixed positive value?
Update: or when $n$ is big enough?
 A: No, it does not decrease, but oscillates. 
Here is a graph of the probability in question for $p=1/2, a=1/10$, for $n=1$ up to $n=100$. A rigorous proof can be achieved by analysing the probability in question, given by 
$$P\{\frac{S_n}{n}>p+a\}=\sum_{k>n(p+a)}{n\choose k}p^k(1-p)^{n-k}$$
which follows from the fact that $S_n/n$ takes the value $k/n$ with probability ${n\choose k}p^k(1-p)^{n-k}$.

A: I'd like to add very simple observation which works for every sequence with bounded variance.
 Let $p_n = \mathbb{P}\left({S_n \over n} > p + \alpha\right)$. Let's calculate $\mathbb{D} {S_n \over n} = {1 \over n^2} \mathbf{Var} (S_n) = {1 \over n} \mathbf{Var} (X_1)$. Thus, by the Chebyshov's inequality $p_n \le {\mathbb{D} {S_n \over n} \over \alpha^2} = \mathbf{Var}(X_1) {1 \over n \alpha^2}$. Thus $p_n$ is bounded by a decreasing sequence, hence it contains a decreasing subsequence. 
A: This probability does decrease as $n$ grows.
A heuristic justification is that as $n \rightarrow \infty$, $S_n/n \rightarrow p$, in probability. That is for any $\epsilon > 0$
$$\lim_{n\rightarrow \infty} \mathbf{P}(|S_n/n - p| > \epsilon) = 0.$$
I'm sure its possible to prove this for small $n$ using direct calculation using inequalities for binomial coefficients (I'll leave that to someone else).
Since the sum of $n$ independent Bernoulli $p$ variables has a binomial $\text{Bin}(n,p)$ distribution, an argument for the limiting case is that the De Moivre - Laplace Theorem asserts as $n \rightarrow \infty$
$$ \text{Bin}(n,p) \sim N\big(np,np(1-p) \big)$$
Therefore (using properties of expectations and variance)
$$\left( \frac{1}{n}S_n -p \right) \sim N \left( 0, \frac{p(1-p)}{n}\right).$$
So that in the limit $n \rightarrow \infty$ your question is
$$\lim_{n\rightarrow \infty} \mathbf{P}\left(  N \left( 0, \frac{p(1-p)}{n}\right) > a \right) = 0.$$
One can of course show that the probability for the normal distribution above is decreasing in $n$, though you need to be careful with the fact that the Binomial approximation to the Normal holds only in the limit.
