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Let $q$ be a rational number with $0\lt q\leq\dfrac{\pi^2}6-1$. Then show that there exists a set $S\subset \{2,3,4,\dotsc\}$, such that

$$q=\sum_{n\in S}\frac1{n^2}$$

I have no clue about it. Could anyone help me? Thanks a lot.

p.s. I encountered it when surfing the Internet. I only know the problem was from a math student who got perfect score on the IMO

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What is $\sum_{n=2}^\infty \frac1{n^2}$? – LutzL Dec 30 '13 at 18:35
@ziang chen Even though you are not that new...Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level.Also, many find the use of imperative ("Show") to be rude when asking for help; please consider rewriting your post. – Matthew Conroy Dec 30 '13 at 21:42
You seem to posting a large number of unmotivated questions with no evidence of any efforts of your own to solve them, and most phrased as imperatives ("show", "prove", etc.). Are you just copying all the problems you have found in a book, or do you actually have any interest in the solutions that people are posting – Old John Dec 30 '13 at 22:03
@ziangchen So, have you tried anything? Have you thought about it? Where did you find it? – Matthew Conroy Dec 30 '13 at 22:04
@ziangchen I don't know what that means. – Matthew Conroy Dec 30 '13 at 23:06
up vote 3 down vote accepted

This problem if i am not mistaken was proposed by P. Erdos and proved by R.L. Graham (and maybe Sierpsinski ) in this article

You can find much more than you want there.
I hope that this helps.

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Very true (+1), although the result by Graham seems to cover finite sums of this type, whereas the OP here does not specify finite, so there might be a simpler proof in this case. – Old John Dec 30 '13 at 23:08
I suppose that the "powerfull" mathematician that ziang chen is referring to is simply P. Erdos – Konstantinos Gaitanas Dec 30 '13 at 23:10
Yes, I have a suspicion that (s)he is going through some publication about Erdős and posting all the questions here for us to solve. Certainly does not seem to display much interest in any of the answers posted. – Old John Dec 30 '13 at 23:13
@OldJohn yes i have just read another question which seems "Erdosian" – Konstantinos Gaitanas Dec 30 '13 at 23:15
@panoramix Thank you very much! I will read the paper carefully – ziang chen Dec 30 '13 at 23:21

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