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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 ?

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  • $\begingroup$ 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 $\sum a_n t^{n} \log t$. $\endgroup$
    – Sangchul Lee
    Feb 5, 2012 at 7:08
  • $\begingroup$ Oh I see. That is really useful $\endgroup$
    – Geralt of Rivia
    Feb 5, 2012 at 8:56

2 Answers 2

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

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  • $\begingroup$ Does anyone have a good reference for connections between the Clausen functions and the polygamma functions? Wolfram alpha evalulates the numerator and denominator in terms of the polygammas. $\endgroup$
    – graveolensa
    Feb 5, 2012 at 8:05
  • $\begingroup$ wow a very elegant solution! thank you $\endgroup$
    – Geralt of Rivia
    Feb 5, 2012 at 8:50
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    $\begingroup$ @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 $$\psi_1 (x) = \sum_{n=0}^{\infty} \frac{1}{(n+x)^2}.$$ $\endgroup$
    – Sangchul Lee
    Feb 5, 2012 at 9:16
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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.

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