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Is there a general way to find the sum


where $P(x)$ is a polynomial of degree $k\geq 2$, with coefficients $a_0,a_1,\dots,a_k$? (possibly restricted to integers)

What is $\sum\limits_{n=1}^{\infty}\frac1{2n^7+n^3-5}$?

And how to find $\sum\limits_{n=1}^{\infty}\frac{Q(n)}{P(n)}$ for two polynomials?

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Considering that people already don't know how to express the sums $\sum \frac{1}{n^{2k+1}}$ in terms of other constants... –  Qiaochu Yuan Nov 17 '11 at 18:51
Thats strange, how come $\sum1/(n^7+n^3)$ can be expressed in terms of the gamma function? –  100101011011010011010110101110 Nov 17 '11 at 18:55
Is the degree of $P(n)$ supposed to vary with the index variable or what? –  Henning Makholm Nov 17 '11 at 19:03

2 Answers 2

up vote 2 down vote accepted

Of course we must assume $P(n) \ne 0$ for all positive integers $n$. Start with the partial fraction decomposition to express $1/P(n)$ (or $Q(n)/P(n)$ where the degree of $P$ is at least 2 more than the degree of $Q$) as a linear combination of $1/(n - \alpha)^j$ where $\alpha$ are the roots of $P$. Then your sum can be expressed in terms of $\Psi(1-\alpha)$ and its derivatives, where $\Psi$ is the digamma function. For example, according to Maple $$ \eqalign{&\sum _{n=1}^{\infty } \left( 2\,{n}^{7}+{n}^{3}-5 \right) ^{-1}={ \frac {1}{2573576195}}\,\sum _{\alpha={\it RootOf} \left( 2\,{{\it \_Z }}^{7}+{{\it \_Z}}^{3}-5 \right) }\cr &\left( -2401000-8403500\,{\alpha}^{ 4}-73530913\,\alpha+6720\,{\alpha}^{2}+23520\,{\alpha}^{6}+205800\,{ \alpha}^{3}-576\,{\alpha}^{5} \right)\cr & \Psi \left( 1-\alpha \right)\cr} $$

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Abramowitz and Stegun display a method for summing rational series in terms of polygamma functions here. If convenient, one might be able to convert the polygamma functions to (generalized) harmonic numbers. –  Guess who it is. Nov 18 '11 at 5:57
For comparison: Mathematica says that $$\sum_{n=1}^{\infty}\frac1{2n^7+n^3-5}=-\sum_k \frac{\psi(-x_k)}{14 x_k^6+84 x_k^5+210 x_k^4+280 x_k^3+213 x_k^2+90 x_k+17}$$, where the $x_k$ are the roots of $2 x^7+14 x^6+42 x^5+70 x^4+71 x^3+45 x^2+17 x-2=0$. –  Guess who it is. Nov 18 '11 at 6:05

When $P(x)$ is a polynomial in $x^2$, that is, has only even-degree terms, an integration-by-residues trick succeeds, as follows. $$ 2\sum_{n=1}^\infty {1\over P(n)} \;=\; \sum_{n\not=0} {1\over P(n)} \;=\; \sum_{n\not=0} {\rm Res}_{z=n}{\cot(\pi z)\over P(n)} $$ For degree $P$ at least $2$, a contour integral of $\cot(\pi z)/P(z)$ over a large circle goes to $0$. Such an integral includes not only non-zero integers $n$, but also $0$ and all zeros of $P(z)$. Thus, $$ 2\sum_{n=1}^\infty {1\over P(n)} \;=\; -{\rm Res}_{z=0}{\cot(\pi z)\over P(z)} - \sum_{{\rm zeros}\;w\not=0\,{\rm of}\,P} {\rm Res}_{z=w} {\cot(\pi z)\over P(z)} $$ This also applies to ratios $Q/P$ with all monomials of the same parity.

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