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Thread: Lim inf and sup of fractional part

  1. #1
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    Lim inf and sup of fractional part

    How to proove that franctional part of sqrt(n) has the limes inferior (lower limit) equal to 0 and limes superior (upper limit) equal to 1?

    Thanks in advance!
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  2. #2
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    Re: Lim inf and sup of fractional part

    For the limit inferior, show that there are an unlimited number of $\sqrt{n_i}$ with a fractional part equal to zero.

    For the limit superior, consider the values of the $\sqrt{n_i-1}$.
    Thanks from romsek and divergent
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  3. #3
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    Re: Lim inf and sup of fractional part

    One approach for the limit superior is to recognise that for each $n_i=k_i^2$, we have $$\frac{n_i-1}{\sqrt{n_i}} = \frac{k_i^2 - 1}{k_i} = k_i - \frac1{k_i} \lt \sqrt{n_i-1}$$
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  4. #4
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    Re: Lim inf and sup of fractional part

    Here's my take on the problem. I'm probably just dense, but I didn't understand Archie's solution.

    Thanks from Plato
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  5. #5
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    Re: Lim inf and sup of fractional part

    Using $[x]$ to mean the fractional part of $x$...

    As you point out, $\underline{\lim} [ \sqrt{n} ] = 0$ because for every $n$ there is a $k \gt n$ such that $k = i^2$ (for some integer $i$) and thus $$[ \sqrt{k} ] = [ \sqrt{i^2} ] = [ i ] = 0$$

    For $\overline{\lim} [ \sqrt{n} ]$ we can consider the subsequence formed only of the elements $[ \sqrt{n} ]$ where $n=k^2-1$ for some integer $k$. For each element in this sequence (for large enough $k$), $$\left[ \frac{k^2 - 1}{k} \right] \lt [ \sqrt{n} ] \lt 1$$
    Then $$\lim_{k \to \infty} \left[ \frac{k^2 - 1}{k} \right] = \lim_{k \to \infty} \left(1 - \frac1k\right) < \lim_{k \to \infty} [ \sqrt{n} ] \lt 1$$
    and by the squeeze theorem $ \displaystyle \lim_{k \to \infty} [ \sqrt{n} ] = 1$
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