$s_n=\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}\right)$

Question

\begin{align*}S_n=\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}\right)\end{align*}

¿Cómo calcular el % de límite $s_n$?

\begin{align*}\lim_{n\to \infty } \, S_n\end{align*}

Answer 1

Considere la curva $y=\frac{1}{\sqrt{x}}$. Tenemos $$\int_1^{n+1}\frac{1}{\sqrt{x}}\,dx\lt 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}\lt \int_0^n \frac{1}{\sqrt{x}}\,dx.$ $ evaluar los integrales. Tenemos $$2\sqrt{n+1}-2\lt 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}\lt 2\sqrt{n}.$ $ divida todo por $\sqrt{n}$ y utilice el exprimir a la conclusión de que nuestro límite es $2$.

Answer 2

Una respuesta utilizando el teorema de Stolz-Cesàro: $$\lim_{n\to\infty} \frac{ \sum_{k=1}^n 1/\sqrt{k} }{\sqrt{n}} = \lim_{n\to\infty} \frac{1/\sqrt{n} }{\sqrt{n} - \sqrt{n-1}} = \lim_{n\to\infty} \frac{\sqrt{n}+\sqrt{n-1} }{\sqrt{n}}=2.$ $

Answer 3

Desde $$\frac{1}{\sqrt{k}} \ge \frac{2}{\sqrt{k}+\sqrt{k+1}} = 2(\sqrt{k+1}-\sqrt{k}) \ge \frac{1}{\sqrt{k+1}}$ $

Nos encontramos. $$\begin{align} & \sum_{k=1}^{n} \frac{1}{\sqrt{k}} \ge 2\sum_{k=1}^{n}(\sqrt{k+1}-\sqrt{k}) = 2(\sqrt{n+1}-1) \ge 2\sqrt{n} - 2\\ \text{and}\quad & \sum_{k=1}^{n} \frac{1}{\sqrt{k}} = 1 + \sum_{k=1}^{n-1} \frac{1}{\sqrt{k+1}} \le 1 + 2\sum_{k=1}^{n-1}(\sqrt{k+1}-\sqrt{k}) = 2\sqrt{n} - 1 \end {Alinee el} $$ como resultado, $$ 2 - \frac{2}{\sqrt{n}} \le S_n \le 2 - \frac{1}{\sqrt{n}} \quad\implies\quad \lim_{n\to\infty} S_n = 2$ $

Answer 4

\begin{align} {1 \over \sqrt{n}}\,\sum_{k = 1}^{n}{1 \over \sqrt{k\,}} &= {1 \over \sqrt{n}}\,\sum_{k = 1}^{n}{1 \over \sqrt{n\xi_{k}\,}}\,n\Delta\xi = \sum_{k = 1}^{n}{1 \over \sqrt{\xi_{k}\,}}\,\Delta\xi \sim \int_{1/n}^{1}{{\rm d}\xi \over \xi^{1/2}} = \left.\vphantom{\LARGE A}\;2\xi^{1/2}\right\vert_{1/n}^{1} \\[3mm]&= 2\left(1 - {1 \over \sqrt{n\,}}\right) = 2 - {2 \over \sqrt{n\,}} \to \color{#ff0000}{\Large 2} \quad\mbox{when}\quad n \to \infty \end {Alinee el}