Find the sum of $\sum_{n=1}^\infty \frac{1}{n(n+2)}x^n$ $$S(x)=\sum_{n=1}^\infty \frac{1}{n(n+2)}x^n$$
The question is divided into three parts:
1. Determine its radius of convergence
2. By using the power series of $\frac{1}{1-x}$, show that for all x $\in$ ]-1,1[ , we get $\ln(1-x)=-\sum_{n}^\infty \frac{1}{n}x^n$
3. Find the value of the sum S(x)=$\sum_{n=1}^\infty \frac{1}{n(n+2)}x^n$
I was able to solve 1 and 2 but I found some difficulty in 3. I will write how I solved the first two questions and if possible I would like to know if I have made a mistake.*
1. Using the ratio test with $U_n=\frac{1}{n(n+2)}$
$$\frac{U_{n+1}}{U_n}=\frac{n(n+2)}{(n+1)(n+3)}=\frac{n^2+2n}{n^2+4n+3}$$
$$L=\lim\limits_{n \to \infty}(\frac{n^2+2n}{n^2+4n+3})=\lim\limits_{n\to \infty}(\frac{\frac{1}{n^2}(1+\frac{2}{n})}{\frac{1}{n^2}(1+\frac{4}{n}+\frac{3}{n^2})})=1$$
Radius of converge R=1/L $\rightarrow$ R = 1
2.
$$\frac{1}{1-x}=-\frac{dy}{dx}\ln(1-x)$$
and since it is known that $\sum_{n=0}^\infty x^n = \frac{1}{1-x}$ for $x \in]-1,1[$ (too lazy to write proof) we can take the integral of $x^n$ to get $\ln(1-x)=-\sum_{n}^\infty \frac{1}{n}x^n$
$$-\int_{0}^{x} x^n = -\frac {x^{n+1}}{n+1} \rightarrow -\sum_{n=0}^\infty\frac {x^{n+1}}{n+1}= -\sum_{n=1}^\infty\frac {x^n}{n}$$
3.
$$S(x)=\sum_{n=1}^\infty \frac{1}{n(n+2)}x^n$$
I took
$$U_n=\frac{1}{n(n+2)}=\frac{A}{n}+\frac{B}{n+2}$$
I got
$$U_n=\frac{1}{n(n+2)}=\frac{1}{2n}+\frac{-1}{2(n+2)}$$
$\therefore$
$$S(x)=\sum_{n=1}^\infty (\frac{1}{2n}+\frac{-1}{2(n+2)})x^n$$
First, I started to solve $\sum_{n=1}^\infty (\frac{1}{2n})x^n$, and by using the proof of question n.2 I got: $$\sum_{n=1}^\infty (\frac{1}{2n})x^n=\frac{1}{2}\sum_{n=1}^\infty (\frac{1}{n})x^n=\frac{1}{2}(-ln(1-x))=ln(\frac{1}{\sqrt{1-x}})$$
However, I was not able to solve the second part: $$\sum_{n=1}^{\infty}\frac{-1}{2(n+2)}x^n$$
How can I find the sum of the second part to find S(x)? I appreciate the help and I would be glad to know if I have made any mistakes too.
 A: Another approach is to start with the geometric series $\sum_{n=1} x^{n-1}=\frac{1}{1-x}$ (assuming you are in appropriate interval for convergence etc.).
\begin{align*}
\frac{1}{1-x} & = \sum_{n=1}x^{n-1}\\
\int \frac{1}{1-x} \, dx& =\sum_{n=1}\frac{x^n}{n}\\
-\ln(1-x)+c & = \sum_{n=1}\frac{x^n}{n}\\
-x\ln(1-x)+xc & = \sum_{n=1}\frac{x^{n+1}}{n}\\
\int -x\ln(1-x) \, dx+c\int x \, dx & = \sum_{n=1}\frac{x^{n+2}}{n(n+2)}\\
\int -x\ln(1-x) \, dx+c\int x \, dx & = x^2\left[\sum_{n=1}\frac{x^{n}}{n(n+2)}\right]\\
\end{align*}
A: Note that $$\sum \limits_{n=1}^{\infty} \frac{-1}{2(n+2)}x^n = \frac{1}{x^2}\sum \limits_{n=1}^{\infty} \frac{-1}{2(n+2)}x^{n+2} = -\frac{1}{x^2}(\frac{1}{2}x + \frac{1}{4}x^2 +\sum \limits_{n=1}^{\infty} \frac{1}{2n}x^n).$$
But you have already found $\sum \limits_{n=1}^{\infty} \frac{1}{2n}x^n$, so the rest is straightforward.
A: $$-\log(1-x)=\sum_{n=1}^{\infty}\frac{x^n}{n}$$
multiply by $x$
$$-x\log(1-x)=\sum_{n=1}^{\infty}\frac{x^{n+1}}{n}$$
$$\int_{0}^{x}-x\log(1-x)dx=\int_{0}^{x}\sum_{n=1}^{\infty}\frac{x^{n+1}}{n}dx$$
$$\frac{x(x+2)}{4}-\frac{(x^2-1)}{2}\log(1-x)=\sum_{n=1}^{\infty}\frac{x^{n+2}}{n(n+2)}$$
divide by $x^2$
$$\frac{x(x+2)}{4x^2}-\frac{(x^2-1)}{2x^2}\log(1-x)=\sum_{n=1}^{\infty}\frac{x^{n}}{n(n+2)}$$
