Is it true that if $\sum_{n=1}^\infty \frac{a_n}{b_n}$ is irrational then $\sum_{n=1}^\infty \frac{a_n}{b_n^2}$ is irrational.

Question

Assume that the: $$\sum_{n=1}^\infty \frac{a_n}{b_n}$$ is convergent and has an irrational sum, then if $$\sum_{n=1}^\infty \frac{a_n}{b_n^2}$$ is also convergent it should also be an irrational.

EDITED: According to comment the $a_n$ and $b_n$ are integer.

Answer 1

False anyway. It is enough to consider $$ a_n = (n+2)(3n+2),\qquad b_n = 2^n n(n+1). \tag{1}$$ Then $$ \sum_{n\geq 1}\frac{a_n}{b_n} = 2+6\log 2\not\in\mathbb{Q} \tag{2}$$ but $$ \sum_{n\geq 1}\frac{a_n}{b_n^2} = 1\in\mathbb{Q} \tag{3}$$ by creative telescoping: $\frac{a_n}{b_n^2}=g(n)-g(n+1)$ with $g(n)=\frac{4}{4^n n^2}$.

If details about $\sum_{n\geq 1}\frac{a_n}{b_n}=2+6\log 2$ or about $\log 2\not\in\mathbb{Q}$ are needed, please ask for them in the comments and I will provide them.

Answer 2

False.

$$a_n=2^{-n}$$ $$b_n=\sqrt2$$

False for rationals too:

$a_n$ is the coefficient $x^n$ in the Taylor series for $\sqrt{1+x}$ around $0$.

$$b_n=\frac{4^n}{3^n}$$

Sum of $a_n/b_n$ is $\frac{\sqrt7}2$

Sum of $a_n/b_n^2$ is $\frac54$.