Let $$f(n):=\sum_{i=k}^n {n \choose i } p^i (1-p)^{n-i}$$ where $k\geq 2$ is a fixed Parameter and $p=p(n) \in (0,1]$ depends on $n$ where $np\leq 1$. We consider $n \rightarrow \infty$.
I've found the following Derivation of an asymptotic Approximation of $f$, but I have no idea how to get to this:
$$f(n)={n \choose k} p^k (1+O(np))=\frac{(np)^k}{k!}(1+O(np+n^{-1}))$$
A hint might be that one can approximate the binomial coefficient by the poisson probability $$\hat{f}(n)=\sum_{i=k}^{\infty} \frac{(np)^i}{i!}e^{-np}$$ where $|f(n)-\hat{f}(n)|<p$ and $\hat{f}(n)$ is asymptotically distributed like $\frac{(np)^k}{k!}$.
If this Approximation is used, then I'd also like to know how one get to this asymptotic behavior $\frac{(np)^k}{k!}$ of the poisson.
A further hint might be on this site where it is written that $${n \choose i}p^i (1-p)^{n-i}= \frac{(np)^i e^{-np}}{i!}+o(1)$$ but even with this I am stuck.
I'm really stuck with each of the equations, I can't derive any of them. So I would really appreciate any hint.
Best Answer
For every Bernoulli random variable $X$ with parameter $p$ there exists some Poisson random variable $Y$ with parameter $p$ such that $$P(X\ne Y)\leqslant p(1-\mathrm e^{-p})\leqslant p^2.$$ Thus, for every binomial random variable $S$ with parameter $(n,p)$ there exists some Poisson random variable $Z$ with parameter $np$ such that $$P(S\ne Z)\leqslant np^2. $$ In particular, for every $k$, $f(n)=P(S\geqslant k)$ and $\hat f(n)=P(Z\geqslant k)$ are such that $|f(n)-\hat f(n)|\leqslant np^2$.
Furthermore, $\hat f(n)=\dfrac{(np)^k}{k!}g(n)$ where $$ g(n)=\mathrm e^{-np}\sum_{i\geqslant k}(np)^{i-k}\frac{k!}{(k+i)!}, $$ hence $$ \mathrm e^{-np}\leqslant g(n)\leqslant\sum_{i\geqslant k}(np)^{i-k}=\frac1{1-np}. $$ These bounds are nonasymptotic and valid for every $(n,p,k)$ such that $np\lt1$. Do these imply the asymptotics you are interested in?
For the approximation result mentioned in this answer, see R. J. Serfling, Some Elementary Results on Poisson Approximation in a Sequence of Bernoulli Trials (1976), freely available as preprint M374 on this Tech Reports page.
Edit: An improvement of the upper bound of $g(n)$, which might help the OP, is $$ g(n)\leqslant\mathrm e^{-np}\sum_{i\geqslant k}\frac{(np)^{i-k}}{(k+1)^{i-k}}=\frac{\mathrm e^{-np}}{1-\frac{np}{k+1}}. $$ Once the idea is understood, one can play with these bounds, and with their variants, for example, for every $(n,p,k)$ such that $np\lt1$, $$ 1-np\leqslant g(n)\leqslant1-c(k)np, $$ where $c(k)\to1$ when $k\to\infty$.