Evaluar

Question

Evaluar %#% $ #%

Answer 1

$$\dfrac1n\sum{r=1}^n\dfrac r{\sqrt{n^2+r^2}} =\dfrac1n\sum{r=1}^n\dfrac{r/n}{\sqrt{1+(r/n)^2}}$$

Ahora uso $$\lim{n \to \infty} \frac1n\sum{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$ and set $\sqrt{1+x^2}=y$

¿Lo puede tomar desde aquí?

Answer 2

Es necesario aplicar el principio de suma de Riemann.

$$\begin{align} \ & \lim\limits{n \to \infty } \left(\frac{1}{{n\sqrt {{n^2} + 1} }} + \frac{2}{{n\sqrt {{n^2} + 4} }} + \frac{3}{{n\sqrt {{n^2} + 9} }} + \cdots + \frac{n}{{n\sqrt {{n^2} + {n^2}} }}\right)\ & = \lim\limits{n \to \infty }\sum{r=1}^n \frac{r}{n\sqrt{n^2+r^2}} \ & = \lim\limits{n \to \infty }\sum{r=1}^n \frac{\frac{r}{n}}{n\sqrt{1+(\frac{r}{n})^2}}\ & = \lim\limits{n \to \infty } \frac{1}{n} \sum_{r=1}^n \frac{\frac{r}{n}}{\sqrt{1+(\frac{r}{n})^2}} \end {Alinee el} $$

Ahora Supongamos $\frac1n=h$. Entonces $h\to 0$ $n\to\infty$.

Por lo tanto obtenemos $$ \begin{align} \ & \lim\limits{n \to \infty } \frac{1}{n} \sum{r=1}^n \frac{\frac{r}{n}}{\sqrt{1+(\frac{r}{n})^2}}\ & = \lim\limits{n \to \infty } h \sum{r=1}^n \frac{rh}{\sqrt{1+(rh)^2}}\ & = \int_0^1 \frac{x}{\sqrt{1+x^2}}dx \ & = \frac{1}{2}\int_0^1 \frac{d(1+x^2)}{\sqrt{1+x^2}}dx \ & = \frac{1}{2}\cdot 2\sqrt{1+x^2}|_0^1 \ & = \color{red}{\sqrt2-1} \end{Alinee el} $$