Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there a general way to find the sum

$$\sum_{n=1}^{\infty}\frac1{P(n)}$$

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?

share|improve this question
3  
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 $$ \eqalign{&\sum _{n=1}^{\infty } \left( 2\,{n}^{7}+{n}^{3}-5 \right) ^{-1}={ \frac {1}{2573576195}}\,\sum _{\alpha={\it RootOf} \left( 2\,{{\it \_Z }}^{7}+{{\it \_Z}}^{3}-5 \right) }\cr &\left( -2401000-8403500\,{\alpha}^{ 4}-73530913\,\alpha+6720\,{\alpha}^{2}+23520\,{\alpha}^{6}+205800\,{ \alpha}^{3}-576\,{\alpha}^{5} \right)\cr & \Psi \left( 1-\alpha \right)\cr} $$

share|improve this answer
    
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. –  J. M. 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 $2 x^7+14 x^6+42 x^5+70 x^4+71 x^3+45 x^2+17 x-2=0$. –  J. M. 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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.