$\sum_{j:0\leq n+jk\leq N}\binom{N}{n+jk}p^{n+jk}(1-p)^{N-n-jk}\to 1/k$ as $N\to\infty$ for any given $k\in\mathbb N$ and $n=0,1,\cdots,k-1$?

Question

We know that $$\sum_{j=0}^N\binom{N}{j}2^{-N}=(1/2+1/2)^N=1\qquad\text{ and}\qquad\sum_{j=0}^N\binom{N}{j}(-1)^j2^{-N}=(1/2-1/2)^N=0,$$ hence $\sum_{j:0\leq 2j\leq N}\binom{N}{2j}2^{-N}=1/2$. More generally, given $p\in(0,1)$, it seem plusible to me that we can have $$\sum_{j:0\leq n+jk\leq N}\binom{N}{n+jk}p^{n+jk}(1-p)^{N-n-jk}\to 1/k\text{ as }N\to\infty\text{ for any given }k\in\mathbb N\text{ and }n=0,1,\cdots,k-1.$$ But I have no idea how to prove this. Is his true? If so how can one prove it? Can we also give a rate of convergence in terms of $N$?

Answer 1

Let's rewrite the sum as $S_N(k,n,p):=\sum\limits_{j=0}^{N}\binom{N}{j}[j\equiv n\ (k)]p^j(1-p)^{N-j}$, where we use $$[j\equiv n\ (k)]:=\begin{cases}1,&j\equiv n\pmod{k}\\0,&\hfill\text{otherwise}\hfill\end{cases}=\frac1k\sum_{s=0}^{k-1}\color{blue}{\zeta_k}^{(j-n)s},$$ where, in turn, $\zeta_k=\exp(2\pi i/k)$ is a primitive $k$-th root of unity. Summing over $j$, we obtain $$S_N(k,n,p)=\frac1k\sum_{s=0}^{k-1}\zeta_k^{-ns}(1-p+p\zeta_k^s)^N.$$ The absolute value of the $s$-th term is $r_s^N$, where $r_s=\sqrt{1-4p(1-p)\sin^2(s\pi/k)}$.

Thus, when $N\to\infty$, only the term with $s=0$ "survives", giving the desired limit of $1/k$.