Question

Salut, ma puteti ajuta la problema 963...?

Answer 1

$\displaystyle \lim\limits_{n\to \infty}\sum\limits_{k=1}^n\left(\dfrac{1}{n+\sqrt{k^2+1}}\right) =\lim\limits_{n\to \infty}\sum\limits_{k=1}^n\left(\dfrac{1}{n}\cdot \dfrac{1}{1+\dfrac{\sqrt{k^2+1}}{n}}\right) = \\ \\ =\lim\limits_{n\to \infty}\sum\limits_{k=1}^n\left(\dfrac{1}{n}\cdot \dfrac{1}{1+\sqrt{\dfrac{k^2+1}{n^2}}}\right) = \lim\limits_{n\to \infty}\sum\limits_{k=1}^n\left(\dfrac{1}{n}\cdot \dfrac{1}{1+\sqrt{\dfrac{k^2}{n^2}+\dfrac{1}{n^2}}}\right)=\\ \\ \dfrac{1}{n^2}\to 0$

$\displaystyle=\lim\limits_{n\to \infty}\sum\limits_{k=1}^n\left(\dfrac{1}{n}\cdot \dfrac{1}{1+\sqrt{\dfrac{k^2}{n^2}}}\right) = \lim\limits_{n\to \infty}\sum\limits_{k=1}^n\left(\dfrac{1}{n}\cdot \dfrac{1}{1+\dfrac{k}{n}}}\right)= \\ \\ f(x) = \dfrac{1}{1+x} \\ \\ =\lim\limits_{n\to \infty}\sum\limits_{k=1}^n\left(\dfrac{1}{n}\cdot f\Big(\dfrac{k}{n}\Big)}\right) = \displaystyle\int_{0}^1f(x)\,dx = \\ \\ = \int_{0}^1 \dfrac{1}{1+x}\, dx = \ln(1+x)\Big|_0^1 = \ln 2-\ln 1 = \boxed{\ln 2}$