Evaluate $\sum_{n=0}^\infty \frac1{n^3+1}$ if it can express in terms of elementary functions .
Evaluate $\sum_{n=0}^\infty \frac1{n^3+1}$ if it can express in terms of elementary functions.
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01.6865 http://www.wolframalpha.com/input/?i=sum+1%2F%28n%5E3%2B1%29 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. – 2012-02-27
1 Answers
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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0@deoxygerbe No, we should sum over roots of $n^k + 1$. – 2012-02-27