<s>[INST] 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?
 [/INST] I get
$(x^{-2}+x^{-4})\ln(1+x^2)-x^{-2}
$.
I use
$\ln(1+x)
=\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}x^n}{n}
$.
$\begin{array}\\
f(x)
&=\sum_{n=0}^{\infty} \dfrac{(-1)^{n}x^{2n}}{(n+1)(n+2)}\\
&=\sum_{n=0}^{\infty} (-1)^{n}x^{2n}(\dfrac1{n+1}-\dfrac1{n+2})\\
&=\sum_{n=0}^{\infty} (-1)^{n}x^{2n}\dfrac1{n+1}-\sum_{n=0}^{\infty} (-1)^{n}x^{2n}\dfrac1{n+2}\\
&=\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}x^{2n-2}}{n}-\sum_{n=2}^{\infty} \dfrac{(-1)^{n+2}x^{2n-4}}{n}\\
&=x^{-2}\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}x^{2n}}{n}-x^{-4}\sum_{n=2}^{\infty} \dfrac{(-1)^{n}x^{2n}}{n}\\
&=x^{-2}\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}x^{2n}}{n}+x^{-4}\sum_{n=2}^{\infty} \dfrac{(-1)^{n+1}x^{2n}}{n}\\
&=x^{-2}\ln(1+x^2)+x^{-4}(-x^2+\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}x^{2n}}{n})\\
&=x^{-2}\ln(1+x^2)+x^{-4}(-x^2+\ln(1+x^2))\\
&=(x^{-2}+x^{-4})\ln(1+x^2)-x^{-2}\\
\end{array}
$
</s>