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I am looking at a certain sequence, and consequently I am wondering if anyone knows if there happens to exist a closed form solution for this sum:

$$\sum_{n=1}^{\infty} \frac{1}{1+n+n^2+\cdots+n^a}$$

or, an alternate form:

$$\sum_{n=1}^{\infty} \frac{n-1}{n^{a+1}-1}$$

For low values of $a$, Wolfram Alpha gives a closed form in terms of the polygamma function function of order $0$, so I am wondering if there is a general closed form in terms of $a$.

I asked this question on MSE, but have not received answers yet. I realize I did ask it recently there, but I also wanted to ask it here just to see if anyone had seen this sum before.

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    $\begingroup$ Have you tried the old trick of contour integration of $\int \cot\pi z/P(z)\, dz$ over a large square, where $P(z)=(z^{a+1}-1)/(z-1)$ is your polynomial? This should work fine for $\deg P\ge 3$. $\endgroup$
    – Christian Remling
    Commented Oct 13, 2015 at 7:05

2 Answers 2

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We use expansion $$\frac{z-1}{z^{a+1}-1}=\frac1{a+1}\sum_{w^{a+1}=1} \frac{w(w-1)}{z-w}.$$ Now use Abel-Poisson regularization of your sum, multiplying $n$-th term by $t^n$ for $t<1$: $$ \sum \frac{n-1}{n^{a+1}-1}=\frac1{a+1}\lim_{t\rightarrow 1-0} \sum_{w^{a+1}=1} w(w-1)\sum_n \frac{t^n}{n-w}. $$ We have for $w\ne 1$ $$\sum \frac{t^n}{n-w}=\int_0^1 \sum_{n=1}^{\infty} t^nz^{n-w-1}dz=t\int_0^1 \frac{z^{-w}dz}{1-tz}.$$ Hence our sum equals $$ \frac1{a+1}\int_0^1 \frac{\sum_{w^{a+1}=1} w(w-1)z^{-w}}{1-z}dz. $$ This may be further rewritten in several ways. For example, we may replace $(1-z)^{-1}$ to $(1-z)^{\varepsilon-1}$ and then tend $\varepsilon$ to $+0$. For each single $w$ we get $$\int z^{-w}(1-z)^{\varepsilon-1}dz=B(1-w,\varepsilon)=\frac{\Gamma(1-w)\Gamma(\varepsilon)}{\Gamma(1-w+\varepsilon)}=\Gamma(\varepsilon)(1-\varepsilon \psi(1-w)+O(\varepsilon^2)).$$ Multiply this by $w(w-1)$ and sum up. Singularity $\Gamma(\varepsilon)$ goes out as expected, we get a sum like $$ \frac1{a+1}\sum_{w^{n+1}=1,w\ne 1} w(1-w)\psi(1-w).$$ Or just use the formula (proved by the same way) $\int_0^1 \frac{1-z^{-w}}{1-z}=\gamma+\psi(1-w).$

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This is not an answer but a conjecture. It based on a first examples examples calculated by Mathematica. $$S(a)=\frac1{a+1}-\sum_{i=1}^a\frac{\psi(-\alpha_i)}{f_a'(\alpha_i)},$$ where $f_0(x)=1$, $f_1(x)=x+3$, $f_2(x)=x^2+5x+7$,... $$f_{n+1}(x)=(x+2)f_n(x)+1\qquad(n\ge 0),$$ (see A193844) and $\alpha_i$ are the roots of $f_a$.

UPD: From Fedor's formula $f_{n}(x)=\frac{(x+2)^{n+1}-1}{x+1}$ follows that for $\alpha_k=w_k-2$ we have $$f_n'(\alpha_k)=\frac{n+1}{w_k(w_k-1)}.$$

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    $\begingroup$ That is, $f_{n}(x)=\frac{(x+2)^{n+1}-1}{x+1}$, i.e. roots are $w-2$, where $w^{n+1}=1$. $\endgroup$
    – Fedor Petrov
    Commented Oct 13, 2015 at 8:51
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    $\begingroup$ It looks like my answer above modified by using functional equation $\psi(x+1)=\psi(x)+1/x$. $\endgroup$
    – Fedor Petrov
    Commented Oct 13, 2015 at 9:25

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