I have some questions involving in the answer of the question in this link.
The excerpts said that the convergence $$\binom{n}{k}p^kq^{n-k}\sim\dfrac{1}{\sqrt{2\pi npq}}e^{-x_k^2/2},$$ is uniform with respect to $k$. What is "with respect to $k$" means? Why is it uniform? I am confused with this because we have a uniform convergent sequence like $f_n(x)=x^n/n$ for $x\in(0,a)$ with fixed $a<1$. Do we have that this $x^n/n$ converges uniformly with respect to $x$ or what is it actually?
How can we show that this statement means that the left hand side converges uniformly? Since $k$ depends on $n$ in the sense that $n\rightarrow \infty$ and so does $k$ as in $(7.3.14)$. This really confuses me as the example $f_n(x)$ an $x$ and $n$ are independent. The uniform convergence concerns $2$ variables in function?
Furthermore, the uniform thing here is important as we can calculate the right in the sum $$\sum_{a< x_k\le b}\dfrac{1}{\sqrt{2\pi npq}}e^{-x_k^2/2}$$ instead of the left one (to prove $\int e^{-x^2/2}dx$ things)? It seems make sense (if it is uniform?) that the uniform summand can be exchanged between limit (of its asymptotic) and integral (in the range $(a,b)$) and we can obtain the result $(7.3.19)$. Please forgive my stupidities and thank you for your reply in advance.