$$\sum_{n=1}^\infty \frac{1}{n^4+n^2+1}$$
I tried breaking into factors but it is not telescoping.
$$\frac {1}{(n^2+n+1)(n^2-n+1)} = \frac {1}{2n} \left(\frac {1}{n^2-n+1} - \frac {1}{n^2+n+1}\right)$$
Asked
Active
Viewed 94 times
4
-
1but what about the $\frac {1}{2n}$ – Soham Apr 18 '15 at 15:44
-
Couple of suggestions. When using partial fractions you should have the degree of the numerator to be one less than that of the denominator. Once you've got that worked out try completing the square of each denominator. – Tim Raczkowski Apr 18 '15 at 15:48
-
@TimRaczkowski but how can I apply it here? The numerator has degree 0. How can increase the degree? – Soham Apr 18 '15 at 15:51
-
Well once you find the right coefficients, they would cancel correctly if you performed the addition. – Tim Raczkowski Apr 18 '15 at 15:53
-
in other words, you should be solving for $\frac{1}{n^4+n^2+1} = \frac{An+B}{n^2+n+1} + \frac{Cn+D}{n^2-n+1}$ – Rolf Hoyer Apr 18 '15 at 15:55
-
See this – science Apr 18 '15 at 16:04
1 Answers
2
Let $\omega=\exp\frac{2\pi i}{3}$. Then:
$$\sum_{n\geq 1}\frac{1}{n^4+n^2+1}=\frac{1}{i\sqrt{3}}\left(\sum_{n\geq 1}\frac{1}{n^2-\omega}-\sum_{n\geq 1}\frac{1}{n^2-\omega^2}\right)=\color{red}{-\frac{1}{2}+\frac{\pi\sqrt{3}}{6}\tanh\frac{\pi\sqrt{3}}{2}}$$ since we can deal with series like the ones appearing in the middle term through the logarithmic derivative of the Weierstrass product for the hyperbolic sine function. In particular, the identity: $$ \sum_{n\geq 0}\frac{1}{n^2+z^2}=\frac{1+\pi z\coth(\pi z)}{2z^2}$$ is well-known.
Jack D'Aurizio
- 353,855
-
-
@LucyferZedd: do you consider Fourier/Laplace/Mellin transforms as elementary? – Jack D'Aurizio Apr 18 '15 at 16:04
-
1