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Does there exist integer sequences $a_n$ and $b_n$ such that $\sum_{n=1}^\infty (a_n\sqrt{2}+b_n)$ is absolutely convergent and $a_n \sqrt{2}+b_n>0 \ \ \forall \ n\geq N$, for some $N\in\mathbb{N}$.

I tried for a few integer sequences but could not get an example. Any help would be highly appreciated. Thank you.

Max
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    Yes. $\sqrt{2}$ can be approximated arbitrarily well by fractions of the form $\frac{a_n}{b_n}$, so $a_n \sqrt{2} + b_n$ can be arbitrarily small. Now you just make it small enough that the series becomes absolutely convergent. I will post this as an answer later. – Smiley1000 Jan 25 '25 at 19:02
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    For an explicit example, think of what happens when you set $a_n\sqrt2+b_n=(\sqrt2-1)^n$. – Christian E. Ramirez Jan 25 '25 at 19:11

2 Answers2

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Note that any power of $a\sqrt{2} + b$ will only contain multiples of 1 and multiples of $\sqrt{2}$ and can be written as $c\sqrt{2} + d$. Hence, if we can find $0 < a\sqrt{2} + b < 1$. Then we can let $a_n\sqrt{2} + b_n = (a\sqrt{2} + b)^n$ and we have a geometric series with ratio less than 1. Hence the series will converge absolutely.

Nic
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It is well-known that the fractional part of positive integer multiples of an irrational number is dense in $[0,1).$

Since $\sqrt{2}$ is irrational, it follows that, for every $n\in\mathbb{N},\ \exists\ a_n\in\mathbb{N}$ such that $0<\{a_n\sqrt{2}\}<\frac{1}{n^2}.$

Then let $b_n = -\left\lfloor a_n\sqrt{2}\right\rfloor,$ and the result is immediate.

Adam Rubinson
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