First of all the task was to find the radius of convergence $\rho$ of:
$$\sum\limits_{k=1}^\infty \left ( \frac{(-1)^{k+1}}{2k}\right)(2x-4)^{2k}$$
I am pretty sure $\rho=\frac{1}{2}$ is correct.
Now we had to look for the derivative of the power series which I think should be:
$$\sum\limits_{k=1}^\infty (-1)^{k+1}2^{2k}(x-2)^{2k-1}$$
And finally we should find the closed form of:
$$f:(2-\rho,2+\rho) \longrightarrow \mathbb{R}:x \mapsto \sum\limits_{k=1}^\infty (-1)^{k+1}2^{2k}(x-2)^{2k-1}$$
$\sum\limits_{k=0}^\infty (-1)^k2^{2k+2}(x-2)^{2k+1}=4(x-2)\sum\limits_{k=0}^\infty(-1)^k2^{2k}(x-2)^{2k}=4(x-2)\sum\limits_{k=0}^\infty \left ( -4(x-2)^2 \right )^k$
$$=4(x-2)\left( \frac{1}{1+4(x-2)^2} \right ) = \frac{4(x-2)}{1+4(x-2)^2}$$
This should be justified, since $x \in (1.5,2.5)$
Is this correct? Would be great if someone could look over it :)
Now the sum of the antiderivative of the series I found the closef form already:
$$\int \frac{4(x-2)}{4(x-2)^2+1}dx$$
$u=x-2 \Longrightarrow dx=du$
Substituting:
$$\int \frac{4u}{4u^2+1}du$$
$v=4u^2+1 \Longrightarrow du=\frac{dv}{8u}$
Substituting:
$$\int \frac{4u}{v}\frac{1}{8u}dv=\int \frac{1}{2v}dv=\frac{1}{2}\int \frac{1}{v}dv=\frac{1}{2}\left [ \ln(|v|) \right ]$$
Resubstituting:
$$\frac{1}{2}\left [ \ln(|4u^2+1|) \right ]$$
Resubstituting:
$$\frac{1}{2}\left [ \ln(|4(x-2)^2+1|) \right ]=\frac{1}{2}\left [ \ln(4(x-2)^2+1) \right ]=\int\sum\limits_{k=1}^\infty(-1)^{k+1}2^{2k}(x-2)^{2k-1}dx=\sum\limits_{k=1}^\infty\frac{(-1)^{k+1}2^{2k}}{2k}(x-2)^{2k}$$
So yes as @jijijojo promised me the closed form of the second series was $\frac{1}{2}\left [ \ln(4(x-2)^2+1) \right ]$ :)