antiderivative of $\sum _{ n=0 }^{ \infty }{ (n+1){ x }^{ 2n+2 } } $ I've proven that the radius of convergence of  $\sum _{ n=0 }^{ \infty  }{ (n+1){ x }^{ 2n+2 } } $ is $R=1$, and that it doesn't converge at the edges.
Now, I was told that this is the derivative of a function $f(x)$, which holds $f(0)=0$. 
My next step is to find this function in simple terms, and here I get lost. My attempt:
$f(x)=\sum _{ n=0 }^{ \infty  }{ \frac { n+1 }{ 2n+3 } { x }^{ 2n+3 } }  $
and this doesn't seem to help.
I'd like to use the fact that $|x|<1$ so I'll get a nice sum based on the sum of a geometric series but I have those irritating coefficients.
Any tips?
 A: First, consider
$$
g(w)=\sum_{n=0}^{\infty}(n+1)w^n.
$$
Integrating term-by-term, we find that the antiderivative $G(w)$ for $g(w)$ is
$$
G(w):=\int g(w)\,dw=C+\sum_{n=0}^{\infty}w^{n+1}
$$
where $C$ is an arbitrary constant. To make $g(0)=0$, we take $C=0$; then
$$
G(w)=\sum_{n=0}^{\infty}w^{n+1}=\sum_{n=1}^{\infty}w^n=\frac{w}{1-w}\qquad\text{if}\qquad\lvert w\rvert<1.
$$
(Here, we've used that this last is a geometric series with first term $w$ and common ratio $w$.) So, we find
$$
g(w)=G'(w)=\frac{1}{(1-w)^2},\qquad \lvert w\rvert<1.
$$
Now, how does this relate to the problem at hand? Note that
$$
\sum_{n=0}^{\infty}(n+1)x^{2n+2}=x^2\sum_{n=0}^{\infty}(n+1)(x^2)^n=x^2g(x^2)=\frac{x^2}{(1-x^2)^2}
$$
as long as $\lvert x^2\vert<1$, or equivalently $\lvert x\rvert <1$.
From here, you can finish your exercise by integrating this last function with respect to $x$, and choosing the constant of integration that makes its graph pass through $(0,0)$.
A: Hint: $$\frac{n+1}{2n+3} = \frac{1}{2}\left(1-\frac{1}{2n+3}\right)$$
