When is the series $\sum_{n=1}^\infty \frac1{a n^2 + b n + c}$ rational?

Question

Let $a,b,c$ be integers such that $a\neq 0$ and $$ a n^2 + b n + c \neq 0 $$ for all positive integers $n$.

(a) Prove that if there exists a positive integer $k$ such that $$ b^2 - 4ac = k^2 a^2, $$ then $$ \sum_{n=1}^\infty \frac{1}{a n^2 + b n + c} $$ is rational.

(b)* Is the converse of (a) true?

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(a)

Since $b^2 - 4ac = a^2k^2$, the roots of $aX^2 + bX + c=0$ are

$$ \alpha = \frac{-b + a\,k}{2a}, \quad \beta = \frac{-b - a\,k}{2a}, \quad \alpha - \beta = k. $$

Hence

$$ \frac1{a n^2 + b n + c} = \frac1{a(n-\alpha)(n-\beta)} = \frac1{a(\alpha-\beta)} \Bigl(\frac1{n-\alpha} - \frac1{n-\beta}\Bigr) = \frac1{a k}\Bigl(\frac1{n-\alpha} - \frac1{n-\beta}\Bigr). $$

Telescoping

$$ U_N = \sum_{n=1}^N\Bigl(\frac1{n-\alpha}-\frac1{n-\beta}\Bigr) = \sum_{n=1}^N\Bigl(\frac1{n-\alpha}-\frac1{n-\alpha+k}\Bigr). $$

Writing out these terms from $n=1$ up to $n=N$ shows everything cancels except

$$ U_N = \Bigl(\frac1{1-\alpha} + \frac1{2-\alpha} + \cdots + \frac1{k-\alpha}\Bigr) \;-\; \Bigl(\frac1{N+1-\alpha} + \cdots + \frac1{N+k-\alpha}\Bigr). $$

As $N\to\infty$ the second bracket vanishes, so

$$ \sum_{n=1}^\infty\Bigl(\frac1{n-\alpha}-\frac1{n-\beta}\Bigr) = \sum_{j=1}^k \frac1{j-\alpha}. $$

Therefore

$$ \sum_{n=1}^\infty \frac1{a n^2 + b n + c} = \frac1{a k}\sum_{j=1}^k \frac1{j-\alpha} \;\in\;\mathbb Q. $$

For b (Partial attempt)

Suppose $b=0$ and moreover $\sqrt{-\,c/a}$ is a rational number that is not a positive integer. Call that rational number “q” and set

$$ p = \sqrt{a}, \quad q = \sqrt{-\,c/a}, \quad \frac{q}{p}\notin\{1,2,3,\dots\}. $$

Then

$$ a n^2 + c = p^2n^2 - q^2, \quad S = \sum_{n=1}^\infty \frac1{p^2n^2 - q^2}. $$

A standard Fourier‐series expansion of the function $x\mapsto\cos\bigl(\tfrac qp\,x\bigr)$ on $[-π,π]$ shows

$$ S = \frac1{2q^2} \;-\;\frac{π}{2\,p\,q}\,\tan\!\Bigl(\tfrac{π\,q}{p}\Bigr). $$

If $S$ were rational, then $\tan\bigl(\tfrac{πq}{p}\bigr)$ would have to equal $π$ times a rational number. But by expanding $\bigl(\cos(π\lambda)+i\sin(π\lambda)\bigr)^p$ (De Moivre’s formula) one obtains a nontrivial polynomial relation

$$ \sum_{k=0}^{\lfloor(p-1)/2\rfloor} \binom p{2k+1}(-1)^k\bigl(π\lambda\bigr)^{2k+1} = 0, $$

contradicting the fact that $π$ is transcendental. Thus in this special case rationality of the sum forces an impossibility giving partial evidence toward the full converse.