Evaluate $\sum_{n=0}^\infty \frac1{n^3+1}$ if it can express in terms of elementary functions .
Asked
Active
Viewed 180 times
3
-
3@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. – hmakholm left over Monica Feb 27 '12 at 13:49
-
1.6865 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. – Manishearth Feb 27 '12 at 13:52
1 Answers
6
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.
Sasha
- 70,631
-
1The 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 $\frac{e^{\pi i /k}}{k}\left(\gamma+\psi(-e^{\pi i /k})\right)+ \frac{e^{-\pi i /k}}{k}\left(\gamma+\psi(-e^{-\pi i /k})\right)$? – graveolensa Feb 27 '12 at 16:16
-