Does anyone have a counterexample to the following conjecture?
Conjecture:
Suppose that the terms $a_{n}$ is a sequence of rational numbers and further that for all real numbers $r$, there are infinitely many $a_{n}$ with $\frac{a_{n+1}}{a_{n}} \neq r$. This means, in particular that the sequence is NOT geometric. Then, the series
$$\sum_{n=1}^{\infty} a_{n}$$
either diverges or converges to an irrational number.
If this statement conjecture were true, it would imply that $\zeta(2n+1)$ is irrational for all $n \in \mathbb{Z}^{+}$, which (I believe) is still an open problem.
I know there are some results that say that if $a_{n}$ is a sequence of integers that grows fast enough, then $\sum_{n=1} \frac{1}{a_{n}}$ is an irrational number, cf. results from Erdos and Strauss in the following paper
A derivative of an geometric power series answer your question. $$\sum_{n=0}^\infty x^n= \frac{1}{1-x} \Rightarrow \sum_{n=0}^\infty nx^{n-1}= \frac{1}{(1-x)^2}$$ in particular $$\sum_{n=0}^\infty \frac{n}{2^{n-1}}= \frac{1}{(1-\frac{1}{2})^2}=4$$ is a sequence of rational numbers converging to an rational number. Also taking $a_n=\frac{n}{2^{n-1}}$, we have $$\frac{a_{n+1}}{a_n}= \frac{1}{2} + \frac{1}{2n}$$ which is different from any real number infinitely many times.