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How do I evaluate $$\lim_{n\to\infty}\left(\sqrt{n^2-n+1}-\left\lfloor\sqrt{n^2 - n+1}\right\rfloor\right),n\in\Bbb N$$

Attempt:

I thought of using Squeeze theorem but that could not help.

Secondly, we know that $x- \lfloor x\rfloor=\{x\}$ where $\{\}$ denotes the fractional part function. But I am not sure how to actually evaluate limits involving the fractional part function.

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    $\begingroup$ Hint: $n - 1 < \sqrt{n^2 - n + 1} < n$. $\endgroup$ Oct 2, 2018 at 13:34
  • $\begingroup$ "But I am not sure how to actually evaluate limits involving the fractional part function." One common way is to switch out $\{x\}$ with $x-[x]$, and then... Oh, wait. $\endgroup$
    – Arthur
    Oct 2, 2018 at 13:39

1 Answer 1

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Since $n - 1 < \sqrt{n^2 - n + 1} < n$, then\begin{align*} &\mathrel{\phantom{=}}{} \sqrt{n^2 - n + 1} - [\sqrt{n^2 - n + 1}] = \sqrt{n^2 - n + 1} - (n - 1)\\ &= \frac{n}{\sqrt{n^2 - n + 1} + (n - 1)} → \frac{1}{2}. \quad (n → ∞) \end{align*}

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  • $\begingroup$ I dont see why the first line is true. $\endgroup$ Oct 2, 2018 at 13:57
  • $\begingroup$ @Neymar Squaring them and you could see. $\endgroup$ Oct 2, 2018 at 13:58
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    $\begingroup$ (n-1)^2 < n^2-n+1<n^2 how come this true ? $\endgroup$ Oct 2, 2018 at 13:59
  • $\begingroup$ also, why the second is equality but you has inequality in the first line. so confusing $\endgroup$ Oct 2, 2018 at 13:59
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    $\begingroup$ As $n\to\infty$, we can assume that $n>1$, then $(n-1)^2=n^2-2n+1<n^2-n+1<n^2$. Take the square root and we arrive at the first inequality. If we take $n\in\mathbb Z$, then we can use the definition of the floor function to write $[\sqrt{n^2-n+1}]=n-1$. $\endgroup$
    – user170231
    Oct 2, 2018 at 14:16

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