Is it true: $\sum_{n=1}^\infty \frac{a_n}{b_n}$&$\sum_{n=1}^\infty \frac{b_n}{c_n}$ are irrational=>$\sum_{n=1}^\infty \frac{a_n}{c_n}$ is irrational

Question

Assume that the: $$\sum_{n=1}^\infty \frac{a_n}{b_n}$$ and $$\sum_{n=1}^\infty \frac{b_n}{c_n}$$ are convergent and irrational, then if $$\sum_{n=1}^\infty \frac{a_n}{c_n}$$ is convergent should it also be irrational then for the integer $a_n$, $b_n$ and $c_n$?

I assume this is false, but can not find any counterexample.

Answer 1

Let $a_n = n+1, b_n = (-1)^n n(n+1), c_n = n(n+1)^2$

Then $$\sum_{n=1}^{\infty} \frac{a_n}{b_n} = -\ln 2$$ $$\sum_{n=1}^{\infty} \frac{b_n}{c_n} = \ln 2 -1$$ $$\sum_{n=1}^{\infty} \frac{a_n}{c_n} = 1$$

Answer 2

Another example with all terms positive is

$$ a_n = 2n+1, \qquad b_n = n(n+1)(2n-1), \qquad c_n = n(n+1)(n+2)(n+3)(2n+1). $$

Then we can prove that

$$ \sum_{n=1}^{\infty} \frac{a_n}{b_n} = \frac{1}{3} + \frac{8}{3}\log 2, \qquad \sum_{n=1}^{\infty} \frac{b_n}{c_n} = \frac{9}{15} - \frac{8}{15}\log 2, \qquad \sum_{n=1}^{\infty} \frac{a_n}{c_n} = \frac{1}{18}. $$