Closed form of $\sum\limits_{n=1}^\infty \frac{4^n(x+4)^{2n}}n$ 
Let
  $$S(x) = \sum_{n=1}^\infty \frac{4^n(x+4)^{2n}}n$$
  1. Find the radius of convergence.
  2. Calculate $S(x)$.
  3. Find $S^{(n)}(x)$ without computing the derivatives of $S(x)$.

From the root test I find $R = 1/4$. It's the second point that troubles me. This is my attempt:
$$\begin{align}
S(x) &= \sum\limits_{n=1}^\infty \frac{4^n(x+4)^{2n}}n =\\
&= \sum\limits_{n=1}^\infty 2^{2n+1}\int_{-4}^x (t + 4)^{2n-1}\mathrm dt =\\
&= \int_{-4}^x \sum\limits_{n=1}^\infty 2^{2n+1}(t + 4)^{2n-1}\mathrm dt =\\
&= 4\int_{-4}^x \sum\limits_{n=1}^\infty 2^{2n-1}(t + 4)^{2n-1}\mathrm dt =\\
&=\ ???
\end{align}$$
I don't know how to continue from there. I know that I should transform the inner sum into a known Taylor expansion or a geometric series, but I don't see how I could do that.
As for the last point, we have that
$$S(x) = \sum_{n=1}^\infty \frac{4^n(x+4)^{2n}}n = \sum_{n = 0}^\infty \frac{S^{(n)}(x)}{n!}(x + 4)^n,$$
as per the Taylor series definition. However, I don't know how to reconcile the indices and the two powers $2n$ and $n$.

Let's put $\alpha:=4(x+4)^2$; we want to find
$$
\sum_{n=1}^{+\infty}\frac{\alpha^n}{n}
$$
We want to compute the following sum:
$$
\sum_{n=1}^{+\infty}\frac{1}{nz^n},\;\;\;\; z\in\mathbb C\;.
$$
We immediately see that $|z|>1$, in order to have absolute convergence.
We recall first two results:
$\bullet\;\;$First:
$$
\log(1+z)=\sum_{n=1}^{+\infty}(-1)^{n+1}\frac{z^n}{n},\;\;\;\forall |z|<1
$$
$\bullet\;\;$Second:
$$
\prod_{n=0}^{+\infty}\left(1+z^{2^{n}}\right)=
\sum_{n=0}^{+\infty}z^{n}=\frac{1}{1-z},\;\;\;\forall |z|<1
$$
The last one can be proved, showing by induction that $\prod_{k=0}^{N}\left(1+z^{2^{k}}\right)=\sum_{k=0}^{2^{N+1}-1}z^{k}$.
Ok:
\begin{align*}
\sum_{n=1}^{+\infty}\frac{1}{nz^n}=&
\sum_{n=1}^{+\infty}\frac{1}{n}\left(\frac{1}{z}\right)^n\\
=&\underbrace{\sum_{k=0}^{+\infty}\frac{1}{2k+1}\left(\frac{1}{z}\right)^{2k+1}-
\sum_{k=1}^{+\infty}\frac{1}{2k}\left(\frac{1}{z}\right)^{2k}}_{\log\left(1+\frac{1}{z}\right)}+
2\sum_{k=1}^{+\infty}\frac{1}{2k}\left(\frac{1}{z}\right)^{2k}\\
=&\log\left(1+\frac{1}{z}\right)+
\sum_{k=1}^{+\infty}\frac{1}{k}\left(\frac{1}{z^2}\right)^{k}\\
=&\log\left(1+\frac{1}{z}\right)+
\log\left(1+\frac{1}{z^2}\right)+\cdots\\
=&\sum_{n=0}^{+\infty}\log\left(1+\frac{1}{z^{2^n}}\right)\\
=&\log\left(\prod_{n=0}^{+\infty}\left(1+\left(\frac{1}{z}\right)^{2^n}\right)\right)\\
=&\log\left(\frac{z}{z-1}\right)
\end{align*}
just put now $z=1/\alpha$ and conclude.
