Evaluate fraction of sum So i have to evaluate this sum:
$\displaystyle \frac{1-2^{-2}+4^{-2}-5^{-2}+7^{-2}-8^{-2}+10^{-2}-11^{-2}+\cdots}{1+2^{-2}-4^{-2}-5^{-2}+7^{-2}+8^{-2}-10^{-2}-11^{-2}+\cdots}$
it has the form :
$\displaystyle \frac{\sum^{\infty}_0 [(3n+1)^{-2}-(3n+2)^{-2}]}{\sum^{\infty}_0 (-1)^n[(3n+1)^{-2}+(3n+2)^{-2}]}$
My current attempt : Trying to convert this into power series 
$\displaystyle a(n) = (3n+1)^{-2}-(3n+2)^{-2} ~~~~~~ b(n) = (3n+1)^{-2}+(3n+2)^{-2}$
Can a(n) and b(n) be the definite integral of certain polynomial function f(x) ?  
Maybe there is a better direction. Can someone give me a hint ?
 A: Let
$$\begin{align*}
\mathrm{Li}_2(z) & = \sum_{n=1}^{\infty} \frac{z^n}{n^2}, \\
\mathrm{Cl}_2(\theta) & = \sum_{n=1}^{\infty} \frac{\sin n\theta}{n^2} = \Im \left[ \mathrm{Li}_2(e^{i\theta}) \right].
\end{align*}$$
be dilogarithm and Clausen function, respectively. Then it is easy to see that the expression in question reduces to
$$ \frac{\mathrm{Cl}_2(2 \pi / 3)}{\mathrm{Cl}_2(\pi / 3)}.$$
By comparing the power series of both sides, we obtain
$$ \mathrm{Li}_2(z) + \mathrm{Li}_2(-z) = \frac{1}{2}\mathrm{Li}_2(z^2).$$
Now plugging $z = e^{\pi i / 3}$ gives
$$\mathrm{Li}_2(e^{2\pi i/3}) = 2 \mathrm{Li}_2(e^{\pi i/3}) + 2 \mathrm{Li}_2(-e^{\pi i/3}).$$
Now taking imaginary part, we obtain
$$\mathrm{Cl}_2(2 \pi / 3) = 2 \mathrm{Cl}_2(\pi / 3) - 2 \mathrm{Cl}_2(2 \pi / 3),$$
since $-e^{\pi i/3} = e^{-2\pi i/3}$. Therefore we have
$$ \frac{\mathrm{Cl}_2(2 \pi / 3)}{\mathrm{Cl}_2(\pi / 3)} = \frac{2}{3}.$$
A: First, let $A = 1-\frac{1}{2^2} + \frac{1}{4^2}-\frac{1}{5^2}+... $ and $B = 1 +\frac{1}{2^2}-\frac{1}{4^2}-\frac{1}{5^2}+...  $
Notice that the both A and B converges absolutely (by comparison test). Thus, limit law applies: $B-A=2\times\frac{1}{2^2}-2\times\frac{1}{4^2}-2\times\frac{1}{8^2}-2\times\frac{1}{10^2}+... =\frac{A}{2} $ and $\frac{A}{B}=\frac{2}{3}$.
Lol i was so naive back then. 
