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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!

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    $\begingroup$ The denominator is just geometric series, $$ 1 + \ldots + n^a = \frac{n^{a+1}-1}{n-1} $$ so you are trying to find $$ \sum_{n=1}^\infty \frac{n-1}{n^{a+1}-1}, $$ just be careful with the first term ($n=1$) $\endgroup$ – gt6989b Oct 13 '15 at 2:41
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    $\begingroup$ If $a$ is odd you should be able to evaluate$$\sum_{n=-\infty}^\infty\frac1{1+n+\cdots+n^a}$$by integrating $(\pi\cot\pi z)/(1+z+\cdots+z^a)$ around a square contour, but this is not quite what you want. $\endgroup$ – David Oct 13 '15 at 3:11
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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 $\zeta_k=e^{{2\pi i k}/({a+1})}$.

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