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How to find $$\sum_{k \text{ odd}} \frac{2(k^2-1)}{k^4+k^2+1}$$ Here we find $\displaystyle\sum_{k=1}^{\infty} \frac{2(k^2-1)}{k^4+k^2+1}=1$ and we know that $\displaystyle\sum_{k \,\text{odd}} + \sum_{k \,\text{even}}=1.$ Can we use this information to find sum? Or, maybe we can find it on other way?

We can say that our sum is equal to $$\color{red}{2}\sum_{k=1}^{\infty}\frac{(2k-1)^2-1}{(2k-1)^4+(2k-1)^2+1}=\sum_{k=1}^{\infty} \left(\frac{1-4k}{4k^2-2k+1}+\frac{4k-3}{4k^2-6k+3}\right),$$ but I don't think we can see something from that.

EDIT: Actually, Wolfram find that $$2\sum_{k=1}^{\infty}\frac{(2k-1)^2-1}{(2k-1)^4+(2k-1)^2+1}=\pi \text{sech}\left(\frac{\sqrt{3} \pi}{2}\right).$$ It's pretty nice closed form.

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I write here so everyone can see. I finally solve it.

First, note that $$\sum_{k \text{ odd}} \frac{2(k^2-1)}{k^4+k^2+1}=\sum_{k=1}^{\infty}(-1)^{k-1}\frac{2k-1}{k^2-k+1},$$ because $$\sum_{k=1}^{\infty}(-1)^{k-1}\frac{2k-1}{k^2-k+1}=\sum_{k\, \text{odd}}\left(\frac{2k-1}{k^2-k+1}-\frac{2(k+1)-1}{(k+1)^2-(k+1)+1}\right).$$

Now, use formula $$\frac{1}{\cos \pi z}=\frac{4}{\pi}\sum_{k=0}^{\infty} (-1)^k \frac{2k+1}{(2k+1)^2-(2z)^2}=\frac{4}{\pi}\sum_{k=1}^{\infty} (-1)^{k-1} \frac{2k-1}{(2k-1)^2-(2z)^2}$$ and set $z=i\cdot \alpha.$ We find $$\text{sech}(\pi \alpha)=\frac{4}{\pi}\sum_{k=1}^{\infty}(-1)^{k-1}\frac{2k-1}{(2k-1)^2+4\alpha^2}$$ and then setting $\displaystyle\alpha = \frac{\sqrt{3}}{2}$ we find $$\text{sech}\left(\frac{\sqrt{3}\pi}{2}\right)=\frac{4}{\pi}\sum_{k=1}^{\infty}(-1)^{k-1}\frac{2k-1}{(2k-1)^2+3}=\frac{1}{\pi}\sum_{k=1}^{\infty}(-1)^{k-1}\frac{2k-1}{k^2-k+1}$$ or $$\sum_{k \text{ odd}} \frac{2(k^2-1)}{k^4+k^2+1}=\sum_{k=1}^{\infty}(-1)^{k-1}\frac{2k-1}{k^2-k+1}=\pi \text{sech}\left(\frac{\sqrt{3}\pi}{2}\right).$$


Added: Formula $$\frac{1}{\cos \pi z}=\frac{4}{\pi}\sum_{k=0}^{\infty} (-1)^k \frac{2k+1}{(2k+1)^2-(2z)^2}$$ is derived from (same link as above) $$\pi \tan \pi z = \sum_{k=0}^{\infty} \frac{8z}{(2k+1)^2 - 4z^2}, \qquad (2z \neq \pm 1, \pm 3, \dots).$$ From $\dfrac{1}{\sin z}=\cos z + \tan \dfrac{z}{2}$ we find, further, that $$\frac{\pi}{\sin \pi z}=\frac{1}{z}-\frac{2z}{z^2-1^2}+\frac{2z}{z^2-2^2}\pm\cdots, \qquad (z \neq 0, \pm 1, \pm 2, \dots),$$ and finally, replacing $z$ here by $\dfrac{1}{2}-z,$ $$\frac{\pi}{4 \cos \pi z}=\frac{1}{1^2-(2z)^2}-\frac{3}{3^2-(2z)^2}+\frac{5}{5^2-(2z)^3}\pm \cdots, \qquad (2z \neq \pm 1, \pm 3, \dots).$$

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  • $\begingroup$ Just beautiful! (+1) $\endgroup$ – user 1591719 Feb 20 '13 at 21:22
  • $\begingroup$ Thanks. I have to admit that I cheated a bit, because I start from Wolfram solution and then find answer. $\endgroup$ – Cortizol Feb 20 '13 at 21:27
  • $\begingroup$ It doesn't matter. The solution is very clever. $\endgroup$ – user 1591719 Feb 20 '13 at 21:27
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Maple gets $${\frac {2}{229}}\,\sum _{r={\it RootOf} \left( 16\,{{\it \_Z}}^{4}-32 \,{{\it \_Z}}^{3}+24\,{{\it \_Z}}^{2}-6\,{\it \_Z}+1 \right) } \left( 56\,{r}^{3}-45\,{r}^{2}-23\,r+21 \right) \Psi \left( 1-r \right)$$

The quartic is irreducible with two pairs of complex-conjugate roots. I don't know if there's any reason to expect the values of $\Psi(1-r)$ to have any special relations (other than $\overline{\Psi(1-r)} = \Psi(1-\overline{r})$).

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  • $\begingroup$ I don't know if there is nice solution. It's interesting that if we , for example, put that sum start from $k=1$ Wolfram can't find closed form, but if we set $k=0$ Wolfram find nice closed form. I'm talking about wolframalpha.com/input/… and wolframalpha.com/input/… $\endgroup$ – Cortizol Feb 19 '13 at 19:56
  • $\begingroup$ for which sum Maple gets that result? $\endgroup$ – Cortizol Feb 20 '13 at 20:14

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