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By definition, a point $x$ is a cluster point of a sequence $\{x_n\}_{n=1}^\infty$ if $\forall\epsilon>0$, there exist infinitely many $n$'s such that $|x_n-x|<\epsilon$. Now I'd like to find all cluster points of the sequence $x_n=\sqrt{n\pi}-\lfloor\sqrt{n\pi}\rfloor$ defined by the floor function. Only problem is, I don't know how to get rid of the disturbing floor function.

After computing the first few terms of the sequence, I found that only the decimal part of each $\sqrt{n\pi}$ is retained. Therefore, it seems very plausible that our sequence has the interval $[0,1]$ as its collection of cluster points. But the proof remains unknown to me. Does anyone have an idea? Thank you.

Boar
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    In fact I think if $x_n$ is increasing and diverges to $\infty$ but $x_{n+1}-x_n$ is decreasing and converges to $0$ then the fractional parts ${x_n}$ will be dense in the interval $[0,1]$. – anon Oct 31 '21 at 09:15
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    https://math.stackexchange.com/questions/843763/for-x-in-mathbb-r-setminus-mathbb-q-the-set-nx-lfloor-nx-rfloor-n-in-ma – N.Quy Oct 31 '21 at 09:31
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    N.Quy's answer applies because $x_{n^2}$ is a subsequence of yours. – Empy2 Oct 31 '21 at 09:43

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