Your $c_n$ are fine. A power series has the form
$$
\sum_{n=1}^\infty c_n(x-a)^n
$$
where the coefficients $c_n$ are scalars that do not depend on $x$, they can depend on $n$.
Your series is
$$
\sum_{n=1}^\infty {(-1)^{n+1}\over n 2^n}(x-2)^n
$$
Maybe it's illustrative to write out this series more explicitly
$$\textstyle
{1\over2}(x-2)^1+{-1\over 2\cdot 2^2}(x-2)^2+{1\over 3\cdot 2^3}(x-2)^3+\cdots.
$$
The coefficients are the numbers
$$
\textstyle
{1\over2},\ {-1\over 2\cdot 2^2},\ {1\over 3\cdot 2^3},\ \cdots;
$$
or, in general, the coefficients are:
$$
c_n={(-1)^{n+1}\over n2^n}.
$$
To find the radius of convergence, you can compute (note the absolute values)
$$
\lim_{n\rightarrow\infty}{ |c_{n+1}|\over |c_n|}
=\lim_{n\rightarrow\infty}{{1\over (n+1) 2^{n+1}}\over{1\over n2^n} }
=\lim_{n\rightarrow\infty}{n\over (n+1)2}={1\over2}.
$$
The radius of convergence is the reciprocal of the above limit: $1/(1/2)=2$.
This tells you that:
$\ \ \ \ $ the series converges whenever $|x-2|<2$
and
$\ \ \ \ $ the series diverges whenever
$|x-2|>2$.
Since $|x-2|<2$ if and only if $0
So we almost have the answer. But, we have not said anything about the endpoints of this interval ($x=0$ and $x=4$).
The series may or may not converge at the points $x=0$ and $x=4$. You need to check what happens when $x=0$ and $x=4$ separately.
This is a matter of replacing $x$ by the appropriate value in the series and seeing what you get:
When $x=0$, the series becomes:
$$
\sum_{n=1}^\infty {(-1)^{n+1}\over n 2^n}( 0-2)^n
=\sum_{n=1}^\infty {(-1)^{2n+1}\over n }
=\sum_{n=1}^\infty {-1\over n }
$$
which diverges. The series does not converge for $x=0$.
When $x=4$, the series becomes:
$$
\sum_{n=1}^\infty {(-1)^{n+1}\over n 2^n}(4- 2)^n=
\sum_{n=1}^\infty {(-1)^{n+1}\over n 2^n}( 2)^n
=\sum_{n=1}^\infty {(-1)^{n+1}\over n }
$$
which is a convergent alternating series. The series converges for $x=4$.
Putting everything together, the interval of convergence for the series is $(0,4]$; that is the series converges if and only if $x$ is in the interval $(0,4]$.