Closed form for $\sum_{n=1}^{\infty} \frac{1}{1+n+n^2+\cdots+n^a}$ Does anyone know if there happens to exist a closed form solution for this sum:
$$\sum_{n=1}^{\infty} \frac{1}{1+n+n^2+\cdots+n^a}$$
For low values of $a$, Wolfram Alpha gives a closed form in terms of the polygamma function function of order $0$ (derivative of the logarithm of the gamma function), so I am wondering if there is a general closed form in terms of $a$.
Thanks!
 A: Just note that
$$1+n+\ldots+n^a=\prod_{k=1}^{a}\left(n-\zeta_k\right),$$
where $\zeta_k$'s are $(a+1)$th roots of unity excluding $1$, and decompose the general term of the series into partial fractions:
$$\frac{1}{1+n+\ldots+n^a}=\sum_{k=1}^a \frac{\alpha_k}{n-\zeta_k}.\tag{1}$$
It is not difficult to show that $\displaystyle \alpha_k=\frac{\zeta_k\left(\zeta_k-1\right)}{a+1}$.
Consider the finite sum
$$\mathcal{S}_N:=\sum_{n=1}^N\frac{1}{1+n+\ldots+n^a}=\sum_{k=1}^a\alpha_k\sum_{n=1}^N\frac{1}{n-\zeta_k}=\sum_{k=1}^a\alpha_k\Bigl[\psi\left(1-\zeta_k+N\right)-\psi\left(1-\zeta_k\right)\Bigr].$$
Next, let us use two observations: 


*

*the asymptotic behavior of $\psi(z)$ as $z\rightarrow +\infty$ $\psi\left(1+z\right)=\ln z+O\left(z^{-1}\right)$, 

*$\sum_{k=1}^a\alpha_k=0$ (compute the coefficient of $n^{a-1}$ in the numerator of the right hand side of (1))
They imply that the value of the sum of interest is given by
$$\mathcal{S}_{\infty}=-\sum_{k=1}^a\alpha_k\psi\left(1-\zeta_k\right)=\frac1{a+1}\sum_{k=1}^a\zeta_k\left(1-\zeta_k\right)\psi\left(1-\zeta_k\right),$$
with $\zeta_k=e^{{2\pi i k}/({a+1})}$.
