# 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.

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.

• Are the $a_n,b_n$ supposed to be integers? – Hagen von Eitzen Jun 18 '17 at 15:48

## 2 Answers

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.

• Wow. I thought the statement was wrong, but I doubt I'd come up with a counter-example in a reasonable stretch of time. – user436658 Jun 18 '17 at 16:30

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 $\frac54$.

• Sorry for mistake. The $a_n$ and $b_n$ are rational. – Gevorg Hmayakyan Jun 18 '17 at 15:56
• @Gevorg Hmayakyan, Rational or İnteger? Or positive integer? Or negative integer?! – user453266 Jun 18 '17 at 16:04
• I wonder how this question got seven upvotes without actually answering OP's question about $a_n, b_n\in\mathbb{Z}$. But I won't downvote: the above lines are trivial but they are correct. – Jack D'Aurizio Jun 18 '17 at 16:19
• @Jack It answers the original, unedited question. That's when I got the votes. – Matt Samuel Jun 18 '17 at 16:21