Convergence of an infinite series? 

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?
 A: 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
$\sum_{n=1}^{\infty} c_n (-2)^n$.
A: 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.
