Evaluate fraction of sum

Question

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 ?

Answer 1

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 $$ \frac{\mathrm{Cl}_2(2 \pi / 3)}{\mathrm{Cl}_2(\pi / 3)} = \frac{2}{3}.$$

Answer 2

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.