Is there possibly a closed form for $$\sum_{m=1}^{\infty} \left(\sum_{k=1}^{6} \dfrac{1}{m-\alpha^k}\right)^2$$ where $\alpha=e^{2\pi i/7}$ ?
I'm having problems evaluating the first sum, let alone the second. I tried some algebraic manipulation, but I could only find that $\displaystyle\sum_{k=1}^{6} \dfrac{1}{1-\alpha^k}=3$ using the fact that $\dfrac{1}{1-\alpha^k}+\dfrac{1}{1-\alpha^{7-k}}=1$
Hints and answers appreciated.
On
Let us reorder our sum by summing over $m$ first, we have:
$$\sum_{m=1}^\infty \left(\sum_{j=1}^6 \frac{1}{m-\alpha^j}\right)^2 = \sum_{j=1}^6 \left( \sum_{m=1}^\infty\frac{1}{(m-\alpha^j)^2}\right) + \sum_{j=1}^6\sum_{k=1,\ne j}^6\left(\sum_{m=1}^\infty\frac{1}{(m-\alpha^j)(m-\alpha^k)}\right) $$ Recall the infinite product expansion of Gamma function,
$$\frac{1}{\Gamma(z)} = z e^{\gamma z} \prod_{m=1}^\infty\left(1+\frac{z}{m}\right)e^{-\frac{z}{m}}$$
Taking logarithm and differentiate twice and using the relation $\Gamma(z+1) = z\Gamma(z)$, we find:
$$-\psi(1+z) = \gamma + \sum_{m=1}^\infty \left(\frac{1}{m+z} - \frac{1}{m}\right) \quad\implies\quad \psi'(1+z) = \sum_{m=1}^\infty \frac{1}{(m+z)^2} \tag{*1}$$
where $\psi(z) = \frac{\Gamma'(z)}{\Gamma(z)}$ is the digamma function. From this, we find
$$\begin{align} \sum_{m=1}^\infty \frac{1}{(m-\alpha^j)^2} &= \psi'(1-\alpha^j)\\ \sum_{m=1}^\infty \frac{1}{(m-\alpha^j)(m-\alpha^k)} &= \frac{1}{\alpha^j-\alpha^k}\sum_{m=1}^\infty \left(\frac{1}{m-\alpha^j} - \frac{1}{m-\alpha^k}\right)\\ &= - \frac{\psi(1-\alpha^j)-\psi(1-\alpha^k)}{\alpha^j-\alpha^k} \end{align} $$ and our sum is
$$\sum_{j=1}^6 \psi'(1-\alpha^j) - \sum_{j=1}^6\sum_{k=1,\ne j}^6\frac{\psi(1-\alpha^j)-\psi(1-\alpha^k)}{\alpha^j - \alpha^k}\\ \approx 28.3895279881439339857786846204067953172346... $$
The six numbers $m-\alpha^k$ are the roots of $f(X) = \frac{(m-X)^7 - 1}{m-X-1} = X^6 + a_5 X^5 + \cdots + a_1 X + a_0$, so the sum of their reciprocals is $-\frac{a_1}{a_0}$.
Letting $X=0$, we have $a_0 = \frac{m^7-1}{m-1}$. Taking a derivative, we have $a_1 = -\frac{6m^7 -7m^6+1}{(m-1)^2}$, therefore:
$$s_m = \sum_{k=1}^6 \frac{1}{m-\alpha^6} = \frac{(6m^7 - 7m^6 + 1)}{(m-1)(m^7-1)} = \frac{6m^5 + 5m^4 + 4m^3 + 3m^2 + 2m + 1}{m^6 + m^5 + m^4 + m^3 + m^2 + m + 1}$$ $$= \frac{7m^6}{m^7-1} - \frac{1}{m-1}$$
(the last form comes from taking the logarithmic derivative, which ccorn suggested)
You can check that this gives $3$ for $m=1$. You can also quickly verify that $s_m \leq 21/m$, so $\sum_{m=1}^\infty s_m^2$ converges, and is less than $\frac{147}{2}\pi^2$. This is a very crude bound; after a few initial terms we should have $s_m \approx 6/m$, so the sum is closer to $6\pi^2$.
As for the possible existence of a closed form... well, good luck with that.
Addendum: I think this should not even be attempted until there is a solution for the same problem with third roots of unity, as this already seems quite hard, yet the corresponding problem for second roots of unity is quite tractable. In other words, is there a closed form for the following sum?:
$$\sum_{m=1}^\infty \left( \frac{2m+1}{m^2 + m + 1} \right)^2 = 1 + \sum_{m=2}^\infty \left( \frac{3m^2}{m^3-1} - \frac{1}{m-1} \right)^2$$
There are closed forms involving the logarithmic derivative of the gamma function... but I generally don't accept something as a closed form if it's both conceptually more difficult and harder to evaluate than the original formula.
Another observation:
$$\frac{(2m+1)^2}{(m^2 + m + 1)^2} = 4\cdot \frac{1}{m^2 + m + 1} - 3\cdot \frac{1}{(m^2 + m + 1)^2}$$
The second sum actually does have a closed form, though it's a doozy:
$$\sum_{m=0}^\infty \frac{1}{m^2 + m + 1} = \frac{2\pi}{\sqrt{3}} \left( \coth(\pi \sqrt{3}) -\frac{1}{2}\coth(\frac{\pi\sqrt{3}}{2}) \right)$$
(this is derived from the identity $\coth x = \sum_{m=-\infty}^\infty \frac{x}{\pi^2m^2 + x^2}$, which comes from Fourier something something)
What this means is that there is a closed form for $\sum_{m=1}^\infty \frac{(2m+1)^2}{(m^2 + m + 1)^2}$ if and only if there is a closed form for $\sum_{m=1}^\infty \frac{1}{(m^2 + m + 1)^2}$. This seems like a useful transformation, since the latter looks more standard and has probably been studied.
This seems to come down to finding a closed form for the real part of $\sum_{m=1}^\infty \frac{1}{(m-\omega)^2}$, where $\omega$ is a third root of unity. This looks promising, but there might still be a negative answer.