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, we see that the series converges on the interval $(0,4)$. Outside the interval $[0,4]$, the series diverges.
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 $(0,4]$.