Let $c>1$.
Question.
What is the asymptotic behaviour of the sum
\begin{align}
S_n = \sum_{k=0}^{n} \left(-\frac{1}{2} \right)^k \binom{n}{k} \sum_{j=0}^{k} \binom{cn+k}{j}
\end{align}
as $n$ goes to infinity?
I don't need strong bounds; $\lim_{n \to \infty} \frac{\log \lvert S_n \rvert}{n}$ in terms of $c$ will do. Experimentally the limit exists for each $c$.
I tried generating functions, but the prefix-sum-of-binomials term is hard to handle. Also, the dominant term in the alternating sum seems to be $k = (1 – \alpha_c)n$ for some constant $\alpha_c$ which goes to $0$ as $c$ becomes larger.
See the figure below, graphing $\frac{\log_2 S_n}n$ for $n=100$, for different $c$.
Best Answer
The sum $\sum_{j=0}^{k} \binom{cn+k}{j}$ equals the coefficient of $x^k$ in $(1+x)^{cn+k}(1-x)^{-1}$, and by Lagrange–Bürmann formula it is also the coefficient of $t^k$ in $(1-2t)^{-1}(1-t)^{-cn}$. It follows that $S_n$ is the coefficient of $t^n$ in $-\frac12 (t-\frac12)^{n-1}(1-t)^{-cn}$.
Applying Lagrange–Bürmann formula again, we get the following generating function for $S_n$: $$\sum_{n\geq 0} S_n t^n = \frac{1-h(t)} {1-(c+1)h(t)+2ch(t)^2},$$ where $h(t)$ is the compositional reverse of $g(t):=\frac{t (1-t)^c}{t - 1/2}$, that is $h(t)$ satisfies $g(h(t))=t$.
It can be seen that g.f. decomposes to $$\frac{\sqrt{c^2-6c+1} - (3c-1)}{4c\sqrt{c^2-6c+1}}\frac1{\alpha_+-h(t)}+\frac{\sqrt{c^2-6c+1} + (3c-1)}{4c\sqrt{c^2-6c+1}}\frac1{\alpha_- - h(t)},$$ where $\alpha_{\pm} = \frac{c+1\pm\sqrt{c^2-6c+1}}{4c}$ are the zeros of $1-(c+1)x+2cx^2$. Noticing that poles $h(t)=\alpha_\pm$ correspond to $t=g(\alpha_\pm)$, we conclude that
$$\lim_{n\to\infty} \dfrac{\log |S_n|}{n} = \begin{cases} - \log |g(\alpha_+)|, & \text{if } c < 3-2\sqrt{2};\\ \frac{c-1}2\log(2), & \text{if } c\in [3-2\sqrt{2},0)\cup (0,3+2\sqrt{2}];\\ - \log |g(\alpha_-)|, & \text{if } c > 3+2\sqrt{2}. \end{cases}$$