¿Cuál es el límite de esta secuencia?

Question

$\lim_{n \to \infty}\frac{1}{n}\bigg(\sqrt{\frac{1}{n}}+\sqrt{\frac{2}{n}}+\cdots + 1 \bigg)$

Mi intento:

$\lim_{n \to \infty}\frac{1}{n}\bigg(\sqrt{\frac{1}{n}}+\sqrt{\frac{2}{n}}+\cdots + 1 \bigg)=\lim_{n \to \infty}\bigg(\frac{\sqrt{1}}{\sqrt{n^3}}+\frac{\sqrt{2}}{\sqrt{n^3}}+\cdots + \frac{\sqrt{n}}{\sqrt{n^3}} \bigg)=0+\cdots+0=0$

Answer 1

Una pista: Dejemos que $f(x) := \sqrt{x}$ . Entonces $$\lim_{n \to \infty} \sum_{k=1}^n f\left(\dfrac{k}{n}\right) \dfrac1{n} = \int_0^1 f(x)dx.$$

Answer 2

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$$ \int_{0}^{n}x^{1/2}\,\dd x < \sum_{k = 1}^{n}\sqrt{\vphantom{\large A}k\,} < \int_{1}^{n + 1}x^{1/2}\,\dd x \quad\imp\quad {2 \over 3}\,n^{3/2} < \sum_{k = 1}^{n}\sqrt{\vphantom{\large A}k\,} < {2 \over 3}\bracks{\pars{n + 1}^{3/2} - 1} $$

$$ {2 \over 3}\quad <\quad {1 \over n}\sum_{k = 1}^{n}\sqrt{\vphantom{\large A}k \over n\,}\quad <\quad {2 \over 3}\bracks{\pars{1 + {1 \over n}}^{3/2} - {1 \over n^{3/2}}} $$

$$ \vphantom{\Huge A} $$

$${\large% \lim_{n \to \infty} \pars{{1 \over n}\sum_{k = 1}^{n}\sqrt{\vphantom{\large A}k \over n\,}\,\,} = {2 \over 3}} $$

Answer 3

Por ese mismo argumento, el límite de $$\frac1n(\underbrace{1+\cdots+1}_{\text{$ n $ summands}})$$ es cero. ¿Lo es?

Answer 4

Su límite parece ser 2/3, pero no sé cómo demostrarlo, excepto usando la pista de Njguliyev