Find the closed form for this series I found this interesting series from the , it is from an old math books. It is as followed:
$\dfrac{1}{2}-\dfrac{x^2}{6}+\dfrac{x^4}{12}-\dfrac{x^6}{20}+\dfrac{x^8}{30}-...$
I notice that one can rewrite this series as followed:
$\dfrac{1}{2}-\dfrac{x^2}{2\cdot 3}+\dfrac{x^4}{3\cdot 4}-\dfrac{x^6}{4\cdot 5}+\dfrac{x^8}{5\cdot 6}-...$
So the general formula for this series is 
$$\sum_{n=0}^{\infty} \dfrac{(-1)^{n}x^{2n}}{(n+1)(n+2)}$$
Is there a closed form for this series?

Start with a geometric series
\begin{eqnarray*}
\sum_{n=0}^{\infty} (-1)^n y^n = \frac{1}{1+y}.
\end{eqnarray*}
Integerate 
\begin{eqnarray*}
\sum_{n=0}^{\infty} \frac{(-1)^n y^{n+1}}{n+1}  = \ln(1+y).
\end{eqnarray*}
Integerate again
\begin{eqnarray*}
\sum_{n=0}^{\infty} \frac{(-1)^n y^{n+2}}{(n+1)(n+2)}  = (1+y)\ln(1+y)-y.
\end{eqnarray*}
Now divide by $y^2$ and let $y=x^2$ and we have
\begin{eqnarray*}
\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(n+1)(n+2)}  = \frac{(1+x^2)\ln(1+x^2)-x^2}{x^4}.
\end{eqnarray*}
