<s>[INST] Find the explicit form of $ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n(n+2)}x^{n-1} $. Find the explicit form of
$$
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n(n+2)}x^{n-1}.
$$
Let $S(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n(n+2)}x^{n-1}$. It has radius of convergence $1$.
Let $S_1(x)=xS(x)$. Then $S_1'(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(n+2)}x^{n-1}$ for $|x|<1$.
Let $S_2(x)=x^3S_1'(x)$. Then $S_2'(x)=\sum_{n=1}^{\infty}(-x)^{n-1}=\frac{x^2}{1+x}$.
By integration, I obtained $S_1'(x)=\frac{1}{2x}-\frac{1}{x^2}+\frac{\ln (x+1)}{x^3}$. Then how to obtain $S(x)$? Or there is other method to do this problem?
 [/INST] $$\log(1+x)=\sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n}~~~(1)$$
Multipluy by $x$ on both sides and integrate w.r.t. $x$
$$\int x \log(1+x) dx= \sum_{n=1}^{\infty} (-1)^{n-1}\int \frac{x^{n+1}}{n} dx= \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^{n+2}}{n(n+2)} .$$
$$\implies S(x)=\sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^{n-1}}{n(n+2)}= \frac{1}{x^3} \int x \log(1+x) ~dx.$$
Finally by carrying out integration by parts, we get the required sum
$$S(x)=\frac{1}{4x^3}[2x-x^2-2\log(1+x)+2x^2 \log(1+x)]$$
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