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Evaluate $\sum_{n=0}^\infty \frac1{n^3+1}$ if it can express in terms of elementary functions .

You may refer to this and this articles.

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Posting in the imperative is extremely rude. Also: – kahen Feb 27 '12 at 13:41
@kahen: The imperative is just a grammatical form. We use it in proofs all the time; nothing rude about that. What is rude is the OP just posting an exercise text without adding anything of his own. This is independent of the grammar used in the exercise text. – Henning Makholm Feb 27 '12 at 13:49
1.6865 I have no clue what they did there, I dunno what digamma functions are. It seems that its a sort of derivative of the gamma function, but I fail to see how they got there. Might help you. – Manishearth Feb 27 '12 at 13:52
up vote 6 down vote accepted

Use $n^3+1 = (n+1)(n^2-n+1) = (n+1)(n-\omega)(n - \bar{\omega})$, where $\omega = \mathrm{e}^{i \pi/3}$. Then $$ \frac{1}{n^3+1} = \frac{1}{3} \frac{1}{n+1} - \frac{1}{3} \frac{\omega}{n -\omega} - \frac{1}{3} \frac{\bar{\omega}}{n -\bar{\omega}} = \frac{1}{3} \frac{\omega +\bar{\omega}}{n+1} - \frac{1}{3} \frac{\omega}{n -\omega} - \frac{1}{3} \frac{\bar{\omega}}{n -\bar{\omega}} $$ Therefore $$ \sum_{n=0}^m \frac{1}{n^3+1} = \frac{\omega}{3} \sum_{n=0}^m \left(\frac{1}{n+1} - \frac{1}{n-\omega} \right) + \frac{\bar{\omega}}{3} \sum_{n=0}^m \left(\frac{1}{n+1} - \frac{1}{n-\bar{\omega}} \right) $$ Thus $$ \begin{eqnarray} \sum_{n=0}^\infty \frac{1}{n^3+1} &=& \frac{\omega}{3} \sum_{n=0}^\infty \left(\frac{1}{n+1} - \frac{1}{n-\omega} \right) + \frac{\bar{\omega}}{3} \sum_{n=0}^\infty \left(\frac{1}{n+1} - \frac{1}{n-\bar{\omega}} \right) \\ &=& \frac{\omega}{3} \left( \gamma + \psi(-\omega) \right) + \frac{\bar{\omega}}{3} \left( \gamma + \psi(-\bar{\omega}) \right) \end{eqnarray} $$ where $\psi(x)$ is the digamma function, and $\gamma$ is the Euler-Mascheroni constant.

Thus the sum is not elementary.

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The sum has a non-elementary expression but it does not really mean that it cannot be expressed elementary... – lhf Feb 27 '12 at 14:47
Is this generally true -- if you change the denominator to $n^{k}+1$, is it true that the sum is $\frac{e^{\pi i /k}}{k}\left(\gamma+\psi(-e^{\pi i /k})\right)+ \frac{e^{-\pi i /k}}{k}\left(\gamma+\psi(-e^{-\pi i /k})\right)$? – deoxygerbe Feb 27 '12 at 16:16
@deoxygerbe No, we should sum over roots of $n^k + 1$. – Sasha Feb 27 '12 at 17:22

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