Series

[Tolaso J Kos]

Evaluate the series:

$$\sum_{k=0}^{\infty} \frac{1}{k^2+k+1}$$

[r9m]

\begin{align*}\sum\limits_{k=1}^{\infty} \frac{1}{k^2+k+1} &= \frac{1}{1+2\alpha}\sum\limits_{k=1}^{\infty} \frac{1}{(k-\alpha)}+\frac{1}{(-k-1-\alpha)}\\&= \frac{1}{1+2\alpha}\left(\frac{1}{1+\alpha}+\frac{1}{\alpha}\right)-\frac{1}{1+2\alpha}\sum\limits_{k=-\infty}^{\infty} \frac{1}{k-\alpha}\\&= \frac{1}{\alpha^2+\alpha}+\frac{\pi\cot \pi \alpha}{1+2\alpha}\\&= \frac{\pi\cot \pi \alpha}{1+2\alpha} - 1\\&= -1 + \frac{\pi }{\sqrt 3}\tanh \frac{\pi \sqrt{3}}{2}\end{align*}

where, $\alpha$ and $-1-\alpha$ are the two roots of $x^2+x+1 = 0$ (imaginary cube roots of unity)

We used the partial fraction expansion of: $\displaystyle \pi\cot \pi z = \sum\limits_{k=-\infty}^{\infty} \frac{1}{z-k}$ ($z \notin \mathbb{Z}$)

[galactus]

Good job with digamma-related approach, R.

I will give it a go another way with residues. . Well, it still uses $\pi \cot(\pi z)$ like RD done.

Using the $$\sum_{-\infty}^{\infty}f(z)=-\text{sum of residues of} \cot(\pi z) \text{at the poles of f(z)}$$

$$\sum_{k=-\infty}^{\infty}f(k)=\sum_{k=-\infty}^{-1}f(k)+1+\sum_{k=1}^{\infty}f(k)$$

$$2\sum_{k=1}^{\infty}f(k)+1=-\text{sum of residues}$$

Consider: $$f(z)=\frac{\pi \cot(\pi z)}{z^{2}+z+1}$$

Use the usual square contour. The only poles are at

$$z=e^{\pm 2\pi i/3}, \;\ z=k$$

So, we may write:

$$\frac{\pi \cos(\pi z)}{(z-e^{2\pi i/3})(z-e^{4\pi i/3})}$$

The residue at $z=k$ is:

$$\lim_{z\to k}\frac{(z-k)\pi \cos(\pi z)}{\sin(\pi z)(z^{2}+z+1)}=\frac{1}{k^{2}+k+1}$$

The sum of the residues at $z=e^{\pm 2\pi i/3}$ is:

$$\lim_{z\to e^{2\pi i/3}}\frac{(z-e^{2\pi i/3})\pi \cot(\pi z)}{(z-e^{2\pi i/3})(z-e^{4\pi i/3})}=\frac{-\pi}{\sqrt{3}}\tanh\left(\frac{\pi\sqrt{3}}{2}\right)$$

same thing for $z=e^{4\pi i/3}$. Thus, summing them returns:

$$\frac{-2\pi}{\sqrt{3}}\tanh\left(\frac{\pi\sqrt{3}}{2}\right)$$

Thus, by the residue theorem we have:

$$\oint_{C}\frac{\pi \cot(\pi z)}{z^{2}+z+1}=\frac{-2\pi}{\sqrt{3}}\tanh\left(\frac{\pi\sqrt{3}}{2}\right)+2\sum_{k=1}^{\infty}\frac{1}{k^{2}+k+1}$$

Take the limit as $N\to \infty$, the integral on the left goes to 0 and we have:

$$\sum_{k=1}^{\infty}\frac{1}{k^{2}+k+1}=\frac{\pi}{\sqrt{3}}\tanh\left(\frac{\pi\sqrt{3}}{2}\right)$$

Subtract off the 1 because the sum begins with k=1 instead of k=0 and ultimately obtain:

$$\sum_{k=1}^{\infty}\frac{1}{k^{2}+k+1}=\frac{\pi}{\sqrt{3}}\tanh\left(\frac{\pi\sqrt{3}}{2}\right)-1$$