Asymptotics of an alternating sum involving the prefix sum of binomial coefficients 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 $c$.

 A: 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
$$\lim_{n\to\infty} \dfrac{\log |S_n|}{n} = 
\begin{cases} 
- \log |g(\alpha_+)|, & \text{if } c < 3-2\sqrt{2};\\
\frac{c-1}2\log(2), & \text{if } c\in [3-2\sqrt{2},0)\cup (0,3+2\sqrt{2}];\\
- \log |g(\alpha_-)|, & \text{if } c > 3+2\sqrt{2}.
\end{cases}$$
A: My two cents. We can express the inner sum of binomial coefficients $\sum_{j=0}^k{N\choose j}$ with $N=cn+k$ by means of the integral remainder formula for the Taylor expansion of $(1+x)^{N}$
(this one). By linearity the sums over $j$ splits into two sums, the first of whom vanishes:
$$\sum_{k=0}^{n} \left(-\frac{1}{2} \right)^k \binom{n}{k} 2^{cn+k}= 2^{cn}\sum_{k=0}^{n} \left(-1 \right)^k \binom{n}{k}=0,$$
so we are left with
$$
  S_n= -cn\sum_{k=0}^{n} \left(-\frac{1}{2} \right)^k \binom{n}{k} {cn+k\choose k}\int_0^1(1+x)^{cn -1}(1-x)^kdx=
$$
$$
 = -cn\int_0^1\Bigg[ \sum_{k=0}^{n}  \binom{n}{k}{cn+k\choose k} \left(\frac{x-1}{2} \right)^k\Bigg]\big(1+x\big)^{cn -1}dx.
$$
We recognize the sum into brackets as the $n$-th Jacoby polynomial $ P^{\alpha,\beta}_n(x)$, with $\alpha=0$ and $\beta=(c-1)n$:
$$P^{0,(c-1)n}_n(x)=\sum_{k=0}^{n}  \binom{n}{k}{cn\choose k} \left(\frac{x-1}{2} \right)^k\left(\frac{x+1}{2} \right)^{n-k}=\sum_{k=0}^{n}  \binom{n}{k}{cn+k\choose k} \left(\frac{x-1}{2} \right)^k,$$
and your sum can be written
$$S_n=-cn\int_0^1 P^{0,(c-1)n}_n(x)(1+x)^{cn-1}dx.$$
Note: the integral over $[-1,1]$ vanishes, since it is
$$\int_{-1}^1 P^{0,(c-1)n}_n(x)(1+x)^{n-1}(1+x)^{(c-1)n}dx,$$
and $P^{0,(c-1)n}_n(x)$ is orthogonal w.r.to the weight $(1+x)^{(c-1)n}$ to all polynomials of degree less than $n$. So the
integral on  $[-1,0]$ gives $-S_n$, whence we may also express
$S_n$ as $-cn/2$ times the $n$th Fourier coefficient of the function $(1+x)^{n-1}\text{sgn}(x)$ w.r.to the said scalar product on $[-1,1]$, namely
$S_n=-\frac{cn}2\Big\langle P^{0,cn-n}_n(x),\;(1+x)^{n-1}\text{sgn}(x)\Big\rangle.$
Here I stop; but since there is a huge knowledge on Jacobi polynomials, one may hope that the existing bounds or a some smart use of formulas may give a quick conclusion.
