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 $n$.
 A: Let's set $A = 4(x+4)^2$. You want to find 
$$\sum\limits_{n=1}^{\infty} \frac{1}{n}A^n = \sum\limits_{n=1}^{\infty} \int_{0}^A t^{n-1}\,dt = \int_{0}^A \sum\limits_{n=1}^{\infty} t^{n-1} \,dt = \int_0^A \frac{1}{1-t}\,dt = \cdots$$
A: I'd try the following:
$$|x|<1\implies \frac1{1-x}=\sum_{n=0}^\infty x^n\stackrel{\text{diff.}}\implies\frac1{(1-x)^2}=\sum_{n=1}nx^{n-1}$$
Check now that $\;|4(x+4)^2|<1\;$ for $\;|x+4|<\frac14\;$ and you get $\;S(x)\;$ after integrating (or integrating directly)
A: A quick trick to compute $S(x)$ is the following. Use 
$$
\frac{1}{n}=\int_0^\infty ds\ e^{-sn}
$$
to write
$$
S(x)=\int_0^\infty ds \sum_{n=1}^\infty (4e^{-s} (x+4)^2)^n=\int_0^\infty ds \frac{4e^{-s} (x+4)^2}{1-4e^{-s} (x+4)^2}=-\ln \left(-4 x^2-32 x-63\right)\ ,
$$
where one uses the geometric series and the simple substitution $e^{-s}=z$.
A: Let $t=4(x+4)^2$ and the series becomes
$$\sum_{k=1}^\infty\frac{t^n}n.$$
You can recognize the Taylor development of $-\ln(1-t)$, or derive the series to get
$$\sum_{k=1}^\infty t^{n-1}=\frac1{1-t}.$$
A: The approach in the OP can work.  We need only make a simple substitution to facilitate.  To that end, we proceed.
Let $S_N(x)$ denote the partial sums of the series $S(x)=\sum_{n=1}^\infty \frac{4^n(x+2)^{2n}}{n}$. Then, letting $y=(2x+8)^2$, with $|y|<1$,  we have
$$\begin{align}
S_N(x)&=\sum_{n=1}^N \frac{4^n(x+2)^{2n}}{n}\\\\
&=\sum_{n=1}^N \frac{y^n}{n}\\\\
&=\sum_{n=1}^N \int_0^y z^{n-1}\,dz\\\\
&=\int_0^y \frac{1-z^N}{1-z}\,dz\\\\
&\to -\log|1-y|\,\,\text{as}\,\,N\to \infty \tag 1\\\\
&=-\log\left|1-(2x+8)^2\right|\\\\
&=-\log|2x+9|-\log|2x+7|
\end{align}$$
where the justification for interchanging the limit with the integral in $(1)$ is provided by the Dominated Convergence Theorems since $\left|\frac{1-z^N}{1-z}\right|\le \frac{2}{|1-z|}$ for $|z|<1$.
A: 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.
