# 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 ?

• Of course such choice is possible, but in the expense of coefficient or exponent. That is, either the coefficient or the exponent becomes complicated. Rather, if we surrender and consider 'nearly polynomial' functions, we can find much more useful ones. For example, the integral of $t^{n-1} \log t$ from 0 to 1 gives $n^{-2}$, which allows us to relate the given sums as an integral of the function of the form $\sum a_n t^{n} \log t$. – Sangchul Lee Feb 5 '12 at 7:08
• Oh I see. That is really useful – Geralt of Rivia Feb 5 '12 at 8:56

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 $$\frac{\mathrm{Cl}_2(2 \pi / 3)}{\mathrm{Cl}_2(\pi / 3)} = \frac{2}{3}.$$
• @deoxygerbe : If $\theta$ is a rational multiple of $\pi$, then there is a certain relation between Clausen function $\mathrm{Cl}_2(\theta)$ and trigamma function $\psi_1$, by noticing that $$\psi_1 (x) = \sum_{n=0}^{\infty} \frac{1}{(n+x)^2}.$$ – Sangchul Lee Feb 5 '12 at 9:16
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 $\frac{A}{B}=\frac{2}{3}$.