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A limit on Dirichlet function

Let f:\mathbb{R} \rightarrow \mathbb{R} be the Dirichlet function;

    \[f(x) = \left\{\begin{matrix} 1 & , & x\in \mathbb{Q} \\ 0& , & x\in \mathbb{R} \setminus \mathbb{Q} \end{matrix}\right.\]

Evaluate the limit

    \[\ell = \lim_{n \rightarrow +\infty} \frac{1}{n} \left ( f \left ( 1 \right ) + f \left ( \sqrt{2} \right ) + f \left ( \sqrt{3} \right )+ \cdots + f \left ( \sqrt{n} \right ) \right )\]

We simply note that

    \[0 \leq \frac{f \left ( 1 \right ) + f \left ( \sqrt{2} \right ) + f \left ( \sqrt{3} \right )+ \cdots + f \left ( \sqrt{n} \right )}{n} = \frac{\left \lfloor n \right \rfloor}{n} \leq \frac{1}{\sqrt{n}}\]

and the limit follows to be 0. The reason why

    \[f \left ( 1 \right ) + f \left ( \sqrt{2} \right ) + f \left ( \sqrt{3} \right )+ \cdots + f \left ( \sqrt{n} \right ) = \left \lfloor n \right \rfloor\]

is because \sqrt{m} is rational if-f m is a perfect square.

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