# If $\sum\limits_{n=1}^{\infty}\frac{1}{p(n)}\in\mathbb{Q}$, is $\sum\limits_{n=1}^{\infty}\frac{n}{p(n)}\in\mathbb{Q}$?

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

-Residues

-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 $$\sum_{n=1}^{\infty}\frac{n}{p(n)}\in\mathbb{Q}?$$

• What is the source of this problem? More particularly, why would one expect it to be true? Commented Sep 20, 2018 at 6:04
• 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. Commented Sep 20, 2018 at 6:06
• 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. Commented Sep 20, 2018 at 6:11
• @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. Commented Sep 20, 2018 at 6:18
• That makes sense. Perhaps this paper might be useful? Commented Sep 20, 2018 at 6:32

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}

• 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. Commented Sep 20, 2018 at 14:02
• 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.
– user593746
Commented Sep 20, 2018 at 14:03

$$\newcommand\Q{\mathbf{Q}}$$

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.

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