Expressing Law of Large numbers in terms of binomial probabilities Suppose I let $p \in (0,1)$. Then for $n \ge 1$ and $0 \le k \le n$ let $${P}(k;n,p) = \dbinom{n}{k} p^k(1-p)^{n-k}$$.
How can I express the law of large numbers in terms of the probabilities $P(k;n,p)$. In my notes the law of large numbers is described like this:
Fix $p \in (0,1)$ for $n = 1,2,3,...$ Let $X_n$ have Binomial distribution with parameters $n$ and $p$. Then for any $\epsilon > 0$: $$ P(np-n\epsilon \le X_n \le np+n\epsilon) \to 1$$ I am unsure how to state the (weak) law of large numbers in terms of $P(k;n,p)$.
 A: Markov Inequality. Let $Y$ be a random variable with finite expectation $E(Y) = \mu_Y$ and $P(Y > 0) = 1.$ Then for $c > 0,$ we have $P(Y \ge c) ≤ \mu_Y / c$ and $P(Y < c) > 1 – \mu_Y / c.$ [See Wikipedia.]
Weak Law of Large Numbers (WLLN). If  $X_1, X_2, \dots, X_n$  are independent with  $E(X) = \mu$ and $V(X) = E[(X – \mu)^2] = \sigma^2,$  then the sample mean $\bar X_n = \frac 1 n \sum_{i=1}^n X_i$  has  $E(\bar X_n) = \mu$  and  $V(\bar X_n) = \sigma^2/n.$
Furthermore, in Markov's Inequality, letting  $Y = (\bar X_n – \mu)^2$  and  $c = \epsilon^2 > 0,$ we have
$$1 ≥ P[(\bar X_n – \mu)^2 < \epsilon^2] = 
P[|\bar X – \mu| < \epsilon] ≥ 1 – \sigma^2/ n\epsilon^2,$$  for any $\epsilon > 0.$
Because the right-hand side goes to 1 with increasing $n,$ we have
$$\lim_{n\rightarrow\infty} Q_n =\lim_{n\rightarrow\infty} P[|\bar X_n – \mu| < \epsilon] = 1,$$
and by definition,  $\text{plim}_{n\rightarrow\infty} \bar X_n = \mu.$ (We say that $\bar X_n$ converges in probability to $\mu.)$
WLLN for Binomial. For example, if $V_i$ are IID $\mathsf{Bern}(\theta),$ then
$S_n = \sum_{i=1}^n V_i \sim \mathsf{Binom}(n, \theta),$
and $p_n = \bar V_n = S_n/n$  is the
usual estimate of $\theta.$ 
So by the WLLN we have $\lim_n P[|p_n – \theta] < \epsilon] = 1$  and  $\text{plim}_n \, p_n =$ $\theta.$ 
We illustrate for  $S_n \sim \mathsf{Binom}(n,1/2),$  graphing values of 
$$Q_n = P[|p_n – \theta] < \epsilon] = 
P\left(\frac n 2 – n\epsilon < S_n < \frac n 2 + n\epsilon\right)$$
for $n = 1, 2, \dots, 1000$ and $\epsilon = 0.05$ (jagged black lines). 
By the WLLN $Q_n > 1 – \frac{1}{4n\epsilon^2}.$ This lower
bound approaches $1$ (shown as a red curve). 

B = 1000;  n = 1:B;  eps = .05
Q = pbinom(n/2 + n*eps, n, .5) - 
       pbinom(n/2 -n*eps, n, .5)
plot(n, Q, type="l", xaxs="i")
  abline(h=0:1, col="green2")
  curve(1-1/(4*x*eps^2), 1,1000, 
     lwd=2, col="red", add=T)

TBL = cbind(n, Q)
head(TBL)
     n      Q
[1,] 1 0.0000
[2,] 2 0.5000
[3,] 3 0.0000
[4,] 4 0.3750
[5,] 5 0.0000
[6,] 6 0.3125
tail(TBL)
           n         Q
 [995,]  995 0.9984893
 [996,]  996 0.9983063
 [997,]  997 0.9984726
 [998,]  998 0.9982879
 [999,]  999 0.9984558
[1000,] 1000 0.9984389

Notes: (1) According to the usual limit of a numerical sequence, it does not make sense to say that random sequences $\bar X_n$ or $p_n = S_n/n$ converge. The
values of the terms in the sequence are not known
in advance of doing an experiment. However WLLN is
framed in terms of a numerical sequence, such as $Q_n,$ of
probabilies that can be computed in advance (as
shown above in the binomial case). So 'limit in
probability' is defined in terms of the convergence
of $Q_n.$
(2) Notice that the lower bound provided by Markov's Inequality provides what is needed to
prove the WLLN. However, as can be seen in the figure, this bound is sometimes not a 'tight' one and
is not always useful as an approximation in applied
probability modeling.
