Suppose $p(n)$ is a polynomial with rational coefficients and rational roots of degree at least $3$. If we know $$\sum_{n=1}^{\infty}\frac{1}{p(n)}\in\mathbb{Q}$$ are we able to infer that $$\sum_{n=1}^{\infty}\frac{n}{p(n)}\in\mathbb{Q}?$$

I've tried several approaches to proving (or disproving) this to include the following:

-Looking for counterexamples

-Generating functions


-Partial fraction decomposition

but nothing has yielded any positive or negative results. Any tips, terms, papers, methods, or generally topics that I could look into would also be welcome.

Edit: As noted by Carl Schildkraut below, if this is true, then we would automatically know that $\zeta(2k+1)$ was irrational. Since this seems to greatly increase the potential difficulty, I offer the following modification in order to simplify it:

Suppose $p(n)$ is a polynomial with rational coefficients, rational roots, $\deg(P)\geq 3$, and every root has order $1$. If we know $$\sum_{n=1}^{\infty}\frac{1}{p(n)}\in\mathbb{Q}$$ are we able to infer that $$\sum_{n=1}^{\infty}\frac{n}{p(n)}\in\mathbb{Q}?$$

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    $\begingroup$ What is the source of this problem? More particularly, why would one expect it to be true? $\endgroup$ Commented Sep 20, 2018 at 6:04
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    $\begingroup$ Also, one particular note is that, via taking $p(n)=n^{2k+1}$, this would imply the irrationality of $\zeta(2k+1)$ for all positive integers $k$ (as $\zeta(2k)$ is irrational). So, if it's true, it's most likely out of the reach of modern mathematics. $\endgroup$ Commented Sep 20, 2018 at 6:06
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    $\begingroup$ One more note (not sure if there's an easier way to see this): The case where $p(n)=n^2(n+1)$ gives the first sum as $\frac{\pi^2}{6}-1$ but the second as $1$, so the rationality of the second does not necessarily imply that of the first. $\endgroup$ Commented Sep 20, 2018 at 6:11
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    $\begingroup$ @CarlSchildkraut I wouldn't expect it to be true except that no matter what I try I am unable to find a counterexample. Also, I did see that it wasn't an if an only if statement which does seem to add to the difficulty. But dang, when you point out the thing about $\zeta(2k+1)$, it does seem to put this very far out of reach. I guess I shouldn't expect many answers. I'll edit an easier version. $\endgroup$
    – QC_QAOA
    Commented Sep 20, 2018 at 6:18
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    $\begingroup$ That makes sense. Perhaps this paper might be useful? $\endgroup$ Commented Sep 20, 2018 at 6:32

3 Answers 3


This is a partial answer. If $p(x)\in\mathbb{Q}[x]$ is a polynomial of degree at least $3$ such that all the roots of $p(x)$ are (pairwise distinct) integers, then there are two cases:

  • if one root of $p(x)$ is a positive integer, then the sum $\sum_{n=1}^\infty\,\frac{1}{p(n)}$ and $\sum_{n=1}^\infty\,\frac{n}{p(n)}$ are both undefined;
  • if every root of $p(x)$ is a nonnegative integer, then both $\sum_{n=1}^\infty\,\frac{1}{p(n)}$ and $\sum_{n=1}^\infty\,\frac{n}{p(n)}$ are rational numbers.

The first case is trivial, so I am dealing with the second case. Let $d\geq 3$ be the degree of $p(x)$. Then, there exist integers $k_1,k_2,\ldots,k_d$ and a nonzero rational number $r$ such that $0\leq k_1<k_2<\ldots<k_d$ for which $$p(x)=r(x+k_1)(x+k_2)\cdots (x+k_d)\,.$$ It follows that $$\frac{1}{p(x)}=\sum_{j=1}^d\,\frac{a_j}{x+k_j}\text{ and }\frac{x}{p(x)}=\sum_{j=1}^d\,\frac{b_j}{x+k_j}$$ for some rational numbers $a_1,a_2,\ldots,a_d$ and $b_1,b_2,\ldots,b_d$.

Let $A_0=0$ and $B_0=0$. For $j=1,2,\ldots,d$, set $$A_j=a_1+a_2+\ldots+a_j\in\mathbb{Q}\text{ and }B_j=b_1+b_2+\ldots+b_j\in\mathbb{Q}\,.$$ It can be easily seen that $A_d=\lim_{x\to\infty}\frac{x}{p(x)}=0$ and $B_d=\lim_{x\to\infty}\frac{x^2}{p(x)}=0$. We have $$\frac{1}{p(x)}=\sum_{j=1}^d\frac{A_j-A_{j-1}}{x+k_j}=\sum_{j=1}^{d-1}A_j\left(\frac{1}{x+k_j}-\frac{1}{x+k_{j+1}}\right)$$ and $$\frac{x}{p(x)}=\sum_{j=1}^d\frac{B_j-B_{j-1}}{x+k_j}=\sum_{j=1}^{d-1}B_j\left(\frac{1}{x+k_j}-\frac{1}{x+k_{j+1}}\right).$$ That is, $$\sum_{n=1}^\infty\frac{1}{p(n)}=\sum_{j=1}^{d-1}A_j\sum_{i=k_j+1}^{k_{j+1}}\frac{1}{i}\in\mathbb{Q}$$ and $$\sum_{n=1}^\infty\frac{n}{p(n)}=\sum_{j=1}^{d-1}B_j\sum_{i=k_j+1}^{k_{j+1}}\frac{1}{i}\in\mathbb{Q}.$$

Example: Let $d=3$ and $(k_1,k_2,k_3)=\left(0,3,7\right)$. Then, $(a_1,a_2,a_3)=\left(\frac1{21},-\frac1{12},\frac1{28}\right)$ and $(b_1,b_2,b_3)=\left(0,\frac14,-\frac1{4}\right)$. Hence, $(A_1,A_2)=\left(\frac1{21},-\frac1{28}\right)$ and $(B_1,B_2)=\left(0,\frac1{4}\right)$. We then get \begin{align}\sum_{n=1}^\infty\,\frac{1}{p(n)}&=A_1\left(\frac{1}{k_1+1}+\frac{1}{k_1+2}+\frac{1}{k_1+3}\right)+A_2\left(\frac{1}{k_2+1}+\frac{1}{k_2+2}+\frac{1}{k_2+3}+\frac{1}{k_2+4}\right)\\&=\frac1{21}\left(\frac{1}{1}+\frac{1}{2}+\frac13\right)-\frac1{28}\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac17\right)=\frac{2123}{35280}\end{align} and \begin{align}\sum_{n=1}^\infty\,\frac{n}{p(n)}&=B_1\left(\frac{1}{k_1+1}+\frac{1}{k_1+2}+\frac1{k_1+3}\right)+B_2\left(\frac{1}{k_2+1}+\frac{1}{k_2+2}+\frac{1}{k_2+3}+\frac1{k_2+4}\right)\\&=0\left(\frac{1}{1}+\frac{1}{2}+\frac13\right)+\frac1{4}\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac17\right)=\frac{319}{1680}.\end{align}

Remark: If $k_1,k_2,\ldots,k_d$ are all nonintegral rational, not necessarily nonnegative, numbers such that $k_i-k_j\in\mathbb{Z}$ for every pair $i,j=1,2,\ldots,d$, then the same proof works. That is, both $\sum_{n=1}^\infty\frac1{p(n)}$ and $\sum_{n=1}^\infty\frac{n}{p(n)}$ are rational numbers.

Example: Let $d=3$ and $(k_1,k_2,k_3)=\left(-\frac12,\frac32,\frac92\right)$. Then, $(a_1,a_2,a_3)=\left(\frac1{10},-\frac16,\frac1{15}\right)$ and $(b_1,b_2,b_3)=\left(\frac1{20},\frac14,-\frac3{10}\right)$. Hence, $(A_1,A_2)=\left(\frac1{10},-\frac1{15}\right)$ and $(B_1,B_2)=\left(\frac1{20},\frac3{10}\right)$. We then get \begin{align}\sum_{n=1}^\infty\,\frac{1}{p(n)}&=A_1\left(\frac{1}{k_1+1}+\frac{1}{k_1+2}\right)+A_2\left(\frac{1}{k_2+1}+\frac{1}{k_2+2}+\frac{1}{k_2+3}\right)\\&=\frac{1}{10}\left(\frac{1}{1/2}+\frac{1}{3/2}\right)-\frac1{15}\left(\frac{1}{5/2}+\frac{1}{7/2}+\frac{1}{9/2}\right)=\frac{974}{4725}\end{align} and \begin{align}\sum_{n=1}^\infty\,\frac{n}{p(n)}&=B_1\left(\frac{1}{k_1+1}+\frac{1}{k_1+2}\right)+B_2\left(\frac{1}{k_2+1}+\frac{1}{k_2+2}+\frac{1}{k_2+3}\right)\\&=\frac{1}{20}\left(\frac{1}{1/2}+\frac{1}{3/2}\right)+\frac3{10}\left(\frac{1}{5/2}+\frac{1}{7/2}+\frac{1}{9/2}\right)=\frac{71}{175}.\end{align}

  • $\begingroup$ I like the proof and I did something similar in my own partial answer (see comments above on original post). However, this method seems to have nothing to say about the general case unfortunately. $\endgroup$
    – QC_QAOA
    Commented Sep 20, 2018 at 14:02
  • $\begingroup$ Nope, unfortunately, it does not. And in all the cases I tried, when $p(x)$ has two roots whose difference is not in an integer, the sum $\sum_{n=1}^\infty\,\frac{1}{p(n)}$ is either undefined or irrational. So, I am guessing that the condition that the roots have integral differences is a necessary and sufficient condition for $\sum_{n=1}^\infty\,\frac{1}{p(n)}$ to be rational. However, I have no clue on how to prove this. $\endgroup$
    – user593746
    Commented Sep 20, 2018 at 14:03


I object a little bit to the "accepted" answer. It would require proving that $\sqrt{3} \pi - 3 \log(2)$ is not a rational number. I'm not sure this is so obvious.

Here is an alternate solution which is less random:

Let~$b > a$ be distinct non-zero rational numbers such that $2a$ and $2b$ are integers but $a$ and $b$ are not. Note that $a+b$ and $a-b$ will both be integers. Let $$R= \displaystyle{\frac{a+b}{a-b}},$$ and let $$f(x) = \frac{1}{a b(a+b)} \cdot x(x+a)(x+b)(x+a+b).$$ Note that $$ \sum_{n=1}^{\infty} \frac{1}{f(n)} = \sum_{n=1}^{\infty} \frac{ab(a+b)}{n(n+a)(n+b)(n+a+b)} = \sum_{n=1}^{\infty} \frac{1}{n} - \frac{1}{n+a+b} + \frac{R}{n+a} - \frac{R}{n+b} $$ $$= \sum_{n=1}^{b + a } \frac{1}{n} + R \sum_{n=1}^{b-a} \frac{1}{n+a} \in \Q + \Q = \Q$$ is a telescoping sum and thus rational. On the other hand, if $$F(a,b) = \frac{1}{R} \sum_{n=1}^{\infty} \frac{n}{f(n)},$$ then $$F(a,b) = \frac{1}{R} \sum_{n=1}^{\infty} \frac{n ab(a+b)}{n(n+a)(n+b)(n+a+b)} = \sum_{n=1}^{\infty} \frac{b}{n+b} - \frac{a}{n+a} - \frac{b-a}{n+a+b}$$ $$= \sum_{n=1}^{\infty} \frac{a}{n+b} + \frac{b-a}{n+b} - \frac{a}{n+a} - \frac{b-a}{n+a+b}$$ $$= \sum_{n=1}^{\infty} \frac{a}{n+b} - \frac{a}{n+a} + \frac{b-a}{n+b} - \frac{b-a}{n+a+b}$$ $$= \left(\sum_{n=1}^{b-a} - \frac{a}{n+a} \right) + (b-a) \sum_{n=1}^{\infty} \frac{1}{n+b} - \frac{1}{n+a+b}$$ $$= \left(\sum_{n=1}^{b-a} - \frac{a}{n+a} \right) + (b-a)\left( \sum_{n=0}^{\infty} \frac{1}{n+1/2} - \frac{1}{n+1} \right) - (b-a) \sum_{n=0}^{b-1/2} \frac{1}{n+1/2} + (b-a) \sum_{n=0}^{a+b-1} \frac{1}{n+1}$$ $$ \in \Q + 2(b-a) \log 2 + \Q + \Q = \Q + 2(b-a) \log 2.$$

This is similar to the other solution, except showing that $\log(2)$ is irrational is a direct consequence of the transcendence of $e$.


Answering my own question as I believe I have finally found a counterexample:

$$\sum_{n=1}^{\infty}\frac{1}{\left(n+\frac{1}{3}\right) \left(n+\frac{5}{6}\right) \left(n+\frac{11}{6}\right) \left(n+\frac{7}{3}\right)}=\frac{9}{154}$$

$$\text{but }\sum_{n=1}^{\infty}\frac{n}{\left(n+\frac{1}{3}\right) \left(n+\frac{5}{6}\right) \left(n+\frac{11}{6}\right) \left(n+\frac{7}{3}\right)}=\frac{1}{90} \left(-261+80 \sqrt{3} \pi -240 \log (2)\right).$$

The reason I said believe is that it is still an open problem whether $\{\pi,\log(r),s\}$ for $r,s\in\mathbb{Q}$ are algebraically independent, but most would agree that they probably are.

I did manage to prove that the conjecture was true if $p(n)$ was degree two or three, but it seems that for degree four and above it is not true. Based on how I constructed this (combining two degree two polynomials in a certain way) it might be the case that counterexamples only exist for polynomials of even degree larger than four (or four might be the only degree with a counterexample), but further study is needed.

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    $\begingroup$ Do you have a proof that $\sqrt 3\pi - 3\log(2)$ is irrational? $\endgroup$
    – Alex Ortiz
    Commented Feb 22, 2019 at 21:21

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