Question

Salut, am nevoie de ajutor la ex. 322

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$\displaystyle \lim\limits_{n\to \infty}\dfrac{3}{n}\sum\limits_{k=1}^{n-1}\sqrt{\dfrac{n}{n+3k}} =\\ \\ = \lim\limits_{n\to \infty}\dfrac{3}{n}\sum\limits_{k=1}^{n}\sqrt{\dfrac{n}{n+3k}}=\\ \\ = \lim\limits_{n\to \infty}\dfrac{3}{n}\sum\limits_{k=1}^{n}\sqrt{\dfrac{1}{1+\dfrac{3k}{n}}}=\\ \\ = \lim\limits_{n\to \infty}\dfrac{4-1}{n}\sum\limits_{k=1}^{n}\left(1+\dfrac{(4-1)k}{n}\right)^{-\frac{1}{2}} =$

$\displaystyle = \int_{1}^4x^{-\frac{1}{2}}\, dx =\dfrac{x^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}\Bigg|_{1}^4 =2\sqrt{x}\Big|_{1}^4 = 2\cdot 2-2 = \boxed{2}$