I feel the coefficient Cn has to be zero in order for the original series to converge, as the power series of 4^n will diverge as n - > ∞. Are there any other ways for this series to converge, and if so, will the convergence remain in an alternating series with bases of -2 and -4?
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If $c_n =\frac{(-1)^n}{n4^n} $, then $\sum_{=1}^{\infty} c_n 4^n$ converges (it's a alternating sum with the terms decreasing to zero), and $\sum_{=1}^{\infty} (-1)^n c_n 4^n$ does not converge (it's the well-known harmonic sum).
If $\sum_{=1}^{\infty} c_n 4^n$ converges, then $c_n 4^n \to 0$. This is necessary for the sum to converge. Therefore, for any $\epsilon > 0$, there is an $N$ such that $|c_n 4^n| <\epsilon $ for $n > N$. In particular, choosing $\epsilon = 1$, there is an $N_1$ such that $|c_n 4^n| <1 $ for $n > N_1$.
Therefore, for $n > N_1$, $|c_n (-2)^n| =|c_n 2^n| =\dfrac{|c_n 4^n|}{2^n} <\dfrac1{2^n} $. Since $\sum_{n=1}^{\infty} \dfrac1{2^n}$ converges, so does $\sum_{n=1}^{\infty} c_n (-2)^n$.
It is not true that the values of $C_n$ must be $0$. For example, if $C_n=(\frac{1}{8})^n$ then our series would be $\sum_{n=0}^\infty(\frac{1}{8})^n4^n=\sum_{n=0}^\infty(\frac{1}{2})^n$ which does converge.
It is not true that $\sum_{n=0}^\infty C_n (-4)^n$ must converge. For an example, suppose $C_n=(-\frac{1}{4})^n\frac{1}{n+1}$. Then $$\sum_{n=0}^\infty C_n 4^n=\sum_{n=0}^\infty\bigg(\frac{4}{4}\bigg)^n\frac{(-1)^n}{n+1}=\sum_{n=0}^\infty(-1)^n\frac{1}{n+1}=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k}$$ which converges. However, $$\sum_{n=0}^\infty C_n (-4)^n=\sum_{n=0}^\infty\frac{1}{n+1}=\sum_{k=1}^\infty \frac{1}{k}$$
which is the harmonic series, and diverges.