Please identify the flaw in my reasoning:
$\displaystyle \sum_{n=0}^\infty c_n4^n$ is convergent, so by the ratio test: $\displaystyle \lim_{n \to \infty}\left\vert\frac{a_{n+1}}{a_n}\right\vert = \lim_{n \to \infty}\left\vert\frac{c_{n+1}4^{n+1}}{c_n4^n}\right\vert < 1$.
$\displaystyle \lim_{n \to \infty}\left\vert\frac{c_{n+1}}{c_n}(4)\right\vert < 1 \Rightarrow \displaystyle \lim_{n \to \infty}\left\vert\frac{c_{n+1}}{c_n}\right\vert < \frac{1}{4}$
Now, applying the ratio test to $\displaystyle \sum_{n=0}^\infty c_n(-4)^n$:
$\displaystyle \lim_{n \to \infty}\left\vert\frac{a_{n+1}}{a_n}\right\vert = \lim_{n \to \infty}\left\vert\frac{c_{n+1}(-4)^{n+1}}{c_n(-4)^n}\right\vert = \lim_{n \to \infty}\left\vert\frac{c_{n+1}}{c_n}(-4)\right\vert = \lim_{n \to \infty}\left\vert\frac{c_{n+1}}{c_n}\right\vert(4)$
Since $\displaystyle \lim_{n \to \infty}\left\vert\frac{c_{n+1}}{c_n}\right\vert < \frac{1}{4}$, $\displaystyle \lim_{n \to \infty}\left\vert\frac{c_{n+1}}{c_n}\right\vert(4) < 1$.
Therefore by the ratio test, $\displaystyle \sum_{n=0}^\infty c_n(-4)^n$ is convergent.
Turns out the answer is that we cannot conclude $\displaystyle \sum_{n=0}^\infty c_n(-4)^n$ is convergent, so I'm trying to figure out where I took a wrong turn.