Question

Am nevoie de ajutor cu ex 9

Answer 1

$\displaystyle \prod_{k=2}^n\Big(1-\dfrac{1}{k^2}\Big) = \prod_{k=2}^n\dfrac{k^2-1}{k^2}= \prod_{k=2}^n\dfrac{(k-1)(k+1)}{k\cdot k}=\\ \ =\prod_{k=2}^n\Big(\dfrac{k-1}{k}\cdot\dfrac{k+1}{k}\Big)=\prod_{k=2}^n\dfrac{k-1}{k}\cdot\prod_{k=2}^n\dfrac{k+1}{k} =$

$= \Big(\dfrac{1}{2}\cdot \dfrac{2}{3}\cdot \dfrac{3}{4}\cdot ...\cdot \dfrac{n-2}{n-1}\cdot \dfrac{n-1}{n}\Big)\cdot \Big(\dfrac{3}{2}\cdot \dfrac{4}{3}\cdot \dfrac{5}{4}\cdot ...\cdot \dfrac{n}{n-1}\cdot \dfrac{n+1}{n}\Big)=$

$\displaystyle =\Big(\dfrac{1}{1}\cdot \dfrac{1}{n}\Big)\cdot \Big(\dfrac{1}{2}\cdot \dfrac{n+1}{1}\Big) =\dfrac{1}{2}\cdot \dfrac{n+1}{n}\\ \\\\ \Rightarrow \lim\limits_{n\to \infty}\prod_{k=2}^n\Big(1-\dfrac{1}{k^2}\Big)=\lim\limits_{n\to \infty }\Big( \dfrac{1}{2}\cdot \dfrac{n+1}{n}\Big) =\boxed{\dfrac{1}{2}}$