<s>[INST] 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 .
You may refer to this and this articles.
 [/INST] 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.
</s>