I currently have the problem to establish that $ \underset{n \rightarrow \infty}{\lim} \sum_{k=0}^{\lfloor n\cdot l\rfloor} \binom{n}{k} (s)^k(1-s)^{n-k} = \begin{cases} 0, & \text{if}~l< s\\ 1, & \text{if}~l\geq s \end{cases} $
I know I have to use the law of large numbers and it looks a lot like the binomial distribution. For that I know that: $ \underset{n \rightarrow \infty}{\lim} \sum_{k=0}^{n} \binom{n}{k} (s)^k(1-s)^{n-k} =\underset{n \rightarrow \infty}{\lim} (1-s+s)^n=1 $
The $l$ term in the upper bound of the sum absolutly confuses me. Can anyone help me how to show this in a mathematically nice way?
We will address the only interesting case, that is, $0<s<1$. Let $(X_i)_{i\geqslant 1}$ be an i.i.d. sequence where $\mathbb P(X_1=1)=s$ and $\mathbb P(X_1=0)=1-s$ and let $S_n=\sum_{i=1}^nX_i$. Then by the central limit theorem and the fact that the limiting distribution has a continuous c.d.f, we infer that $$\tag{*} \sup_{t\in\mathbb R}\left\vert \mathbb P\left(\frac{S_n-\mathbb E[S_n]}{\operatorname{Var(S_n)}} \leqslant t\right)-\mathbb P(N\leqslant t)\right\rvert\to 0, $$ where $N$ has a standard normal distribution.
We are looking for the limit of $\mathbb P(S_n\leqslant \lfloor n\ell\rfloor)$. Using the fact that $\mathbb E[S_n]=ns$ and $\operatorname{Var}(S_n)=ns(1-s)$, we are reduced, in view of (*), to find $$ \lim_{n\to\infty}\mathbb P\left(N\leqslant \frac{\lfloor n\ell\rfloor -ns}{\sqrt n\sqrt{s(1-s)}}\right). $$ If $\ell<s$, then $\frac{\lfloor n\ell\rfloor -ns}{\sqrt n\sqrt{s(1-s)}}\to -\infty$ and the limit is indeed $0$.
If $\ell=s$, then $\frac{\lfloor n\ell\rfloor -ns}{\sqrt n\sqrt{s(1-s)}}\to 0$, because $\lvert \lfloor n\ell\rfloor -n\ell\rvert\leqslant 1$ hence the limit is $1/2$.
If $\ell>s$, then $\frac{\lfloor n\ell\rfloor -ns}{\sqrt n\sqrt{s(1-s)}}\to \infty$ hence the limit is $1$.