Question

Ma ajuta cineva cu aceasta limita, va rog ?

Answer 1

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

$\displaystyle= \dfrac{1}{2}+\lim\limits_{n\to \infty}\dfrac{1}{n}\sum\limits_{k=1}^n\left(1+\sqrt{\dfrac{k}{n}\Big(\dfrac{k}{n}+2\Big)}\right)=\\ \\ = \dfrac{1}{2}+\int_{0}^1 \left(1+\sqrt{x(x+2)}\right)\, dx = \\ \\ = \dfrac{1}{2}+1-0+\int_{0}^1\dfrac{(x+1)^2-1}{\sqrt{(x+1)^2-1}}\, dx = \\ \\ = \dfrac{3}{2}+\int_{1}^{2}\dfrac{x^2-1}{\sqrt{x^2-1}}\, dx = \\ \\ = \dfrac{3}{2}+\int_{1}^2x(\sqrt{x^2-1})'\, dx - \int_{1}^2 \dfrac{1}{\sqrt{x^2-1}}\, dx =$

$\displaystyle =\dfrac{3}{2}+2\sqrt 3 - 0-\int_{1}^2\sqrt{x^2-1}\, dx -\ln\Big|\sqrt{x^2-1}+x\Big|\Bigg|_{1}^2 = \\ \\ = \dfrac{3}{2}+2\sqrt 3 - \int_{1}^2\sqrt{x^2-1}\, dx -\ln(2+\sqrt 3)\\ \\ \\l = \dfrac{3}{2}+\int_{1}^2\sqrt{x^2-1}\, dx\\ l =\dfrac{3}{2}+2\sqrt 3 - \int_{1}^2\sqrt{x^2-1}\, dx -\ln(2+\sqrt 3)\\ \\\\ \Rightarrow 2l = 3+2\sqrt 3 - \ln(2+\sqrt 3)\\ \\ \\ \Rightarrow \boxed{l = \dfrac{3+2\sqrt 3-\ln(2+\sqrt 3)}{2}}$