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?
 A: 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*}
A: 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}
$
A: And here's my attempt
to generalize this
as much as possible.
Let
$f(x)
=\sum_{n=0}^{\infty} (-1)^{n}\sum_{k=1}^m \dfrac{a_kx^{c_kn}}{n+b_k} 
$
where the $b_k$
are positive integers
and the
$a_k$ and $c_k$
 are real.
Then
$\begin{array}\\
f(x)
&=\sum_{n=0}^{\infty} (-1)^{n}\sum_{k=1}^m \dfrac{a_kx^{c_kn}}{n+b_k} \\
&=\sum_{k=1}^m \sum_{n=0}^{\infty} (-1)^{n}\dfrac{a_kx^{c_kn}}{n+b_k} \\
&=\sum_{k=1}^m a_k\sum_{n=b_k}^{\infty} (-1)^{n-b_k}\dfrac{x^{c_k(n-b_k)}}{n} \\
&=\sum_{k=1}^m a_kx^{-b_kc_k}\sum_{n=b_k}^{\infty} (-1)^{n-b_k}\dfrac{x^{c_kn}}{n} \\
&=\sum_{k=1}^m a_kx^{-b_kc_k}(-1)^{b_k+1}\sum_{n=b_k}^{\infty} (-1)^{n+1}\dfrac{x^{c_kn}}{n} \\
&=\sum_{k=1}^m a_kx^{-b_kc_k}(-1)^{b_k+1}(\sum_{n=1}^{\infty} (-1)^{n+1}\dfrac{x^{c_kn}}{n}-\sum_{n=1}^{b_k-1} (-1)^{n+1}\dfrac{x^{c_kn}}{n}) \\
&=\sum_{k=1}^m a_kx^{-b_kc_k}(-1)^{b_k+1}(\ln(1+x^{c_k})-\sum_{n=1}^{b_k-1} (-1)^{n+1}\dfrac{x^{c_kn}}{n}) \\
&=\sum_{k=1}^m a_kx^{-b_kc_k}(-1)^{b_k+1}\ln(1+x^{c_k})-\sum_{k=1}^m a_kx^{-b_kc_k}(-1)^{b_k+1}\sum_{n=1}^{b_k-1} (-1)^{n+1}\dfrac{x^{c_kn}}{n} \\
&=\sum_{k=1}^m a_kx^{-b_kc_k}(-1)^{b_k+1}\ln(1+x^{c_k})-\sum_{k=1}^m a_k\sum_{n=1}^{b_k-1} (-1)^{n+b_k}\dfrac{x^{c_k(n-b_k)}}{n} \\
&=\sum_{k=1}^m a_kx^{-b_kc_k}(-1)^{b_k+1}\ln(1+x^{c_k})-\sum_{k=1}^m a_k\sum_{n=1}^{b_k-1} (-1)^{n}\dfrac{x^{-c_kn}}{b_k-n} \\
\end{array}
$
Similarly,
let
$f(x)
=\sum_{n=0}^{\infty} \sum_{k=1}^m \dfrac{a_kx^{c_kn}}{n+b_k} 
$
where the $b_k$
are positive integers
and the
$a_k$ and $c_k$
 are real.
Here
I use
$-\ln(1-x)
=\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}x^n}{n}
$.
Then
$\begin{array}\\
f(x)
&=\sum_{n=0}^{\infty} \sum_{k=1}^m \dfrac{a_kx^{c_kn}}{n+b_k} \\
&=\sum_{k=1}^m \sum_{n=0}^{\infty}\dfrac{a_kx^{c_kn}}{n+b_k} \\
&=\sum_{k=1}^m a_k\sum_{n=b_k}^{\infty} \dfrac{x^{c_k(n-b_k)}}{n} \\
&=\sum_{k=1}^m a_kx^{-b_kc_k}\sum_{n=b_k}^{\infty} \dfrac{x^{c_kn}}{n} \\
&=\sum_{k=1}^m a_kx^{-b_kc_k}\sum_{n=b_k}^{\infty} \dfrac{x^{c_kn}}{n} \\
&=\sum_{k=1}^m a_kx^{-b_kc_k}(\sum_{n=1}^{\infty} \dfrac{x^{c_kn}}{n}-\sum_{n=1}^{b_k-1} \dfrac{x^{c_kn}}{n}) \\
&=-\sum_{k=1}^m a_kx^{-b_kc_k}(\ln(1-x^{c_k})-\sum_{n=1}^{b_k-1} \dfrac{x^{c_kn}}{n}) \\
&=-\sum_{k=1}^m a_kx^{-b_kc_k}\ln(1-x^{c_k})-\sum_{k=1}^m a_kx^{-b_kc_k}\sum_{n=1}^{b_k-1} \dfrac{x^{c_kn}}{n} \\
&=-\sum_{k=1}^m a_kx^{-b_kc_k}\ln(1-x^{c_k})-\sum_{k=1}^m a_k\sum_{n=1}^{b_k-1} \dfrac{x^{c_k(n-b_k)}}{n} \\
&=-\sum_{k=1}^m a_kx^{-b_kc_k}\ln(1-x^{c_k})-\sum_{k=1}^m a_k\sum_{n=1}^{b_k-1} \dfrac{x^{-c_kn}}{b_k-n} \\
\end{array}
$
A: $$\sum_{n=0}^{\infty} \dfrac{(-1)^{n}x^{2n}}{(n+1)(n+2)}=\sum_{n=0}^\infty(-1)^nx^{2n}\int_0^1\int_0^1 y^nz^{n+1}dydz$$
$$=\int_0^1\int_0^1z\sum_{n=0}^\infty(-x^2yz)^ndydz$$
$$=\int_0^1\int_0^1\frac{z}{1+x^2yz}dydz$$
$$=\int_0^1z\left(\int_0^1\frac{dy}{1+x^2yz}\right)dz$$
$$=\int_0^1z\left(\frac{\ln(1+x^2z}{x^2z}\right)dz$$
$$=\frac1{x^2}\int_0^1\ln(1+x^2z)dz$$
$$=\frac1{x^2}\cdot\frac{(1+x^2)\ln(1+x^2)-x^2}{x^2}$$
