Find $\sum_{n=1}^∞ \frac{n^{2}}{n^{8}+n^{4}+1}$

Question

Find $\sum_{n=1}^∞ \frac{n^{2}}{n^{8}+n^{4}+1}$

This series is equivalent to $\frac{1}{2} \sum_{n=1}^∞ \frac{1}{n^{4} - n^{2} +1} - \frac{1}{n^{4} + n^{2} +1}$

Them I'm stuck. Any hint$?$

Answer 1

You properly wrote $$\sum_{n=1}^\infty \frac{n^{2}}{n^{8}+n^{4}+1}=\frac 12\left(\sum_{n=1}^\infty\frac{1}{n^{4} - n^{2} +1}-\sum_{n=1}^\infty\frac{1}{n^{4} + n^{2} +1} \right)$$

Now, write $$\frac{1}{n^{4} - n^{2} +1}=\frac{1}{(n^2-a)(n^2-b)}=\frac 1 {a-b}\left(\frac 1 {n^2-a}-\frac 1 {n^2-b}\right)$$ $$\frac{1}{n^{4} + n^{2} +1}=\frac{1}{(n^2-c)(n^2-d)}=\frac 1 {c-d}\left(\frac 1 {n^2-c}-\frac 1 {n^2-d}\right)$$ which means that you face a series of $$\sum_{n=1}^\infty\frac{1}{n^2-k}=\frac{1-(\pi \sqrt{k}) \cot \left(\pi \sqrt{k}\right)}{2 k}$$

For the first summation, you have $a=\frac{1}{2}-\frac{i \sqrt{3}}{2}$ and $b=\frac{1}{2}+\frac{i \sqrt{3}}{2}$. For the second summation $c=-\frac{1}{2}-\frac{i \sqrt{3}}{2}$ and $d=-\frac{1}{2}+\frac{i \sqrt{3}}{2}$. That is to say $$\sqrt a=\frac{\sqrt{3}}{2}-\frac{i}{2} \qquad \sqrt b=\frac{\sqrt{3}}{2}+\frac{i}{2}\qquad \sqrt c=\frac{1}{2}-\frac{i \sqrt{3}}{2}\qquad \sqrt d=\frac{1}{2}+\frac{i \sqrt{3}}{2}$$

All of the above makes $$\sum_{n=1}^\infty\frac{1}{n^{4} - n^{2} +1}=\frac{1}{6} \left(\frac{\pi \left(\sqrt{3} \sin \left(\sqrt{3} \pi \right)+3 \sinh (\pi )\right)}{\cosh (\pi )-\cos \left(\sqrt{3} \pi \right)}-3\right)$$ $$\sum_{n=1}^\infty\frac{1}{n^{4} + n^{2} +1}=\frac{1}{6} \left(\sqrt{3} \pi \tanh \left(\frac{\sqrt{3} \pi }{2}\right)-3\right)$$ $$\color{blue}{\sum_{n=1}^\infty \frac{n^{2}}{n^{8}+n^{4}+1}=\frac{\pi \sqrt 3}{12} \left(\frac{\sin \left(\sqrt{3} \pi \right)+\sqrt{3} \sinh (\pi )+\tanh \left(\frac{\sqrt{3} \pi }{2}\right) \left(\cos \left(\sqrt{3} \pi \right)-\cosh (\pi )\right) } { \cosh (\pi )-\cos \left(\sqrt{3} \pi \right)}\right)}$$