Prove that for all $x\in [0,1],$ the sequence of fractional parts $(\{\sqrt{n}\})_{n\geq 1} $ contains a subsequence with the limit $x$.
Limit of a subsequence of fractioned part
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real-analysis
sequences-and-series
2 Answers
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In order to have $\{\sqrt{n}\}\approx x $ it is enough to have $\sqrt{n}\approx M+x$ with $M\in\mathbb{N}$ or $n \approx M^2+2M x +x^2$. So it is enough to consider a diverging sequence of (large) natural numbers $M_1,M_2,M_3,\ldots$ and define $n_k$ as the closest integer to $M_k^2+2M_k x+x^2$. The sequence of fractional parts $\{\sqrt{n_1}\},\{\sqrt{n_2}\},\{\sqrt{n_3}\},\ldots$ will converge to $x$ by the concavity of the square root function.
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Observation. Let the sequence $a_n=\{\sqrt{n}\}$, $n\in\mathbb N$, and observe that
$$
0=a_{k^2}< a_{k^2+1}<\ldots