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I had this task long long time ago in a calculus class. I remember it had a very elegant solution using nice property of $\text{Spec}(\sqrt{2})$, where $$ \text{Spec}(\alpha)=\{\lfloor\alpha n\rfloor\,:\,n=1,2,3,\ldots\}. $$ I wasn't able to recreate the solution so I asked this question here.

Thanks to zwim I was able to recall the original solution. See my answer below.

kubus
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  • And what have you tried ? – user577215664 Jul 22 '20 at 19:16
  • There is one solution here that seems ok : https://artofproblemsolving.com/community/c7h39398p245740 – zwim Jul 22 '20 at 19:16
  • You would be better to write the solution out as an independent answer, which you can then accept. This would mark the question as "answered". – user1729 Jul 24 '20 at 16:24
  • @user1729 It would be, but the question was closed, and I was requested to show my effort and why it is interesting. So I made an effort and I solved it and wrote it in the question statement. Should I move the solution to an answer? Will not my question be blocked again then? – kubus Jul 24 '20 at 16:52
  • I guess there are two answers to your comment. Firstly, if you make clear in your question that you solved it yourself after asking it, and you thought people might be interested in the solution then I don't see it getting closed again. Secondly, the close reason was really more complicated than "what have you tried". The link was too this thread, and only the first point is about showing what you've tried. – user1729 Jul 24 '20 at 17:02
  • Personally, I think it would be find (and better) if you were to expand on the background, so the bit starting "I had this task long long time ago...". For example, what does "double star" mean, and what do you mean by "the sequence $\lfloor n\sqrt{2}\rfloor$ described in Concrete Mathematics" (if you have the book to hand then you could give a page reference, but anyway it isn't currently clear that this is a book...). – user1729 Jul 24 '20 at 17:05
  • Thanks. I didn't get these subtle issues what background is enough. I thought "Concrete Mathematics" is well-known book. I just wanted to share this interesting problem and a solution, but I couldn't move forward without a hint (which I luckily get from the community in the form of comment). I will try to improve the question using your suggestions and move the solution to an answer when I will have some free time. – kubus Jul 24 '20 at 17:16

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Thanks to zwim I was able to recreate original solution which was beautiful. I just wanted recall that solution because it uses nice property from

Graham, Ronald L.; Knuth, Donald E.; Patashnik, Oren, Concrete mathematics: a foundation for computer science., Amsterdam: Addison-Wesley Publishing Group. xiii, 657 p. (1994). ZBL0836.00001, Section 3.2, Page 77.

Define a multiset of integers $\text{Spec}(\alpha)$: $$ \text{Spec}(\alpha)=\{\lfloor\alpha n\rfloor\,:\,n=1,2,3,\ldots\}. $$ Define a function $f(x)$ for positive real $x$: $$ f(x)=\sum_{0<n<x}(-1)^{\lfloor n\sqrt{2}\rfloor}. $$ There is a beautiful proof in "Concrete Mathematics" that $\text{Spec}(\sqrt{2})$ and $\text{Spec}(2+\sqrt{2})$ form a disjoint partition of the set of positive integers, thus $$ \sum_{0<n\sqrt{2}<x}(-1)^{\lfloor n\sqrt{2}\rfloor} +\sum_{0<n(2+\sqrt{2})<x}(-1)^{\lfloor n(2+\sqrt{2})\rfloor} = \sum_{0<n< \lceil x\rceil}(-1)^n\in\{-1,0\}. $$ Since $(-1)^{\lfloor n(2+\sqrt{2})\rfloor}=(-1)^{\lfloor n\sqrt{2}\rfloor}$ we get $$ f\left(\frac{x}{\sqrt{2}}\right)+f\left(\frac{x}{2+\sqrt{2}}\right)\in\{-1,0\}, $$ which leads to inequality: $$ -1\le f((1+\sqrt{2})x)+f(x)\le 0 $$ for all positive real $x$. Now, by induction over $k=0,1,2,\ldots$ one can prove such claim: $$ 0<x<(1+\sqrt{2})^k \implies |f(x)|\le k. $$ As mentioned in https://artofproblemsolving.com/community/c7h39398p245740 (link provided by zwim) using the Abel transform we conclude that our series converges if and only if the series $\sum_{n\geqslant 1}\frac {f(n+1)}{n(n+1)}$ converges and, in case of convergence, they converge to the same sum. It is enough to show that $|f(x)|=O(\log x)$ - which we did.

kubus
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