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.

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

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.

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


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

False for rationals too:

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


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

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

  • $\begingroup$ Sorry for mistake. The $a_n$ and $b_n$ are rational. $\endgroup$ – Gevorg Hmayakyan Jun 18 '17 at 15:56
  • $\begingroup$ @Gevorg Hmayakyan, Rational or İnteger? Or positive integer? Or negative integer?! $\endgroup$ – user453266 Jun 18 '17 at 16:04
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    $\begingroup$ 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. $\endgroup$ – Jack D'Aurizio Jun 18 '17 at 16:19
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    $\begingroup$ @Jack It answers the original, unedited question. That's when I got the votes. $\endgroup$ – Matt Samuel Jun 18 '17 at 16:21

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