$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^n\left[\sqrt{4i/n}\right]$

Question

$$ \lim_{n \to \infty} \sum_{i = 1}^{n}{1 \over n} \,\left\lfloor\,\sqrt{\,{4i \over n}\,}\,\right\rfloor\quad \mbox{where}\ \left\lfloor\,x\,\right\rfloor\ \mbox{is the greatest integer function.} $$ Enfrenté el problema de esta manera $$ \lim_{n \to \infty} \sum_{i = 1}^{n}{1 \over n} \,\left\lfloor\,\sqrt{\,{4i \over n}\,}\,\right\rfloor = \lim_{n \to \infty}\frac{1}{n}\int_{n/4}^{n}\mathrm{d}x = \lim_{n \to \infty}\left(\frac{1}{n}\times\frac{3n}{4}\right) = \frac{3}{4} $$
Me sentí como un malote haciendo esto, sólo para descubrir que la respuesta es en realidad $3$ . ¿En qué me he equivocado? ¿Cómo puedo corregirlo?

Answer 1

La integral de Riemann es $$\lim_{n \to \infty} \frac{b-a}{n} \sum_{k=1}^n f\left(a + \frac{b-a}{n} k\right) = \int_{x=a}^b f(x) \, dx,$$ y con $a = 0$ , $b = 1$ y $f(x) = \lfloor 2 \sqrt{x} \rfloor$ obtenemos $$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \lfloor 2 \sqrt{k/n} \rfloor = \int_{x=0}^1 \lfloor 2 \sqrt{x} \rfloor \, dx.$$ Desde $2 \sqrt{x} = 1$ cuando $1/4 \le x < 1$ y $0$ si $0 \le x < 1/4$ esto se convierte en $$\int_{x=1/4}^1 \, dx = \frac{3}{4}.$$ No es posible que la suma dada sea igual a $3$ ya que está acotado trivialmente por encima de $$\frac{1}{n} \sum_{k=1}^n \left\lfloor 2\sqrt{\frac{k}{n}} \right\rfloor < \frac{1}{n} \sum_{k=1}^n 2 \sqrt{\frac{k}{n}} < \frac{1}{n} \sum_{k=1}^n 2 = \frac{2n}{n} = 2.$$

Answer 2

Es una suma de Riemann.

El límite es $$\int_0^1\lfloor 2\sqrt {x}\rfloor dx $$

$$=\int_0^\frac {1}{4}\lfloor 2\sqrt {x}\rfloor dx +\int_{\frac {1}{4}}^1\lfloor 2 \sqrt {x}\rfloor dx$$ $$=(\frac {1}{4}-0).0+(1-\frac {1}{4}).1=\frac {3}{4} $$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

A ' enfoque directo ' que significa "  calcular explícitamente la "suma inicial  " !!!.

$$ \mbox{Note that}\qquad\left\{\begin{array}{l} \ds{\color{#f00}{0} \leq \root{4i \over n} < 1 \implies \bbox[#ffe,10px]{\ds{0 \leq i < {n \over 4}}}} \\[2mm] \ds{\color{#f00}{1} \leq \root{4i \over n} < 2 \implies \bbox[#ffe,10px]{\ds{{n \over 4} \leq i < n}}} \\[2mm] \ds{\color{#f00}{2} \leq \root{4i \over n} < 3 \implies \bbox[#ffe,10px]{\ds{n \leq i < {9 \over 4}\,n}}} \end{array}\right. $$

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\begin{align} \lim_{n \to \infty}\sum_{i = 1}^{n}{1 \over n} \,\left\lfloor\root{4i \over n}\right\rfloor & = \lim_{n \to \infty}\pars{% {1 \over n}\sum_{0\ \leq\ i\ <\ n/4}\color{#f00}{0}\ +\ {1 \over n}\sum_{n/4\ \leq\ i\ <\ n }\color{#f00}{1} + {1 \over n}\,\color{#f00}{2}} \\[5mm] & = \lim_{n \to \infty}\braces{\vphantom{\huge A}% {1 \over n} \bracks{4 \not\mid n}\pars{n - \left\lfloor\,{n \over 4}\right\rfloor} + {1 \over n} \bracks{4 \mid n}\pars{n - \left\lfloor\,{n \over 4}\right\rfloor + 1} + {2 \over n}} \\[5mm] & = \lim_{n \to \infty}\braces{\vphantom{\huge A}% {n - \left\lfloor\,n/4\,\right\rfloor \over n} + {\bracks{4 \mid n} \over n} + {2 \over n}} \\[5mm] & = \lim_{n \to \infty}\braces{\vphantom{\huge A}% {n - \bracks{n/4 - \braces{n/4}} \over n} + {\bracks{4 \mid n} + 2 \over n}} \\[5mm] & = \lim_{n \to \infty}\braces{\vphantom{\huge A}% \color{#f00}{3 \over 4} + {\braces{n/4} + \bracks{4 \mid n} + 2 \over n}} = \bbx{\large{3 \over 4}} \end{align}