Salut,
Punctul a):
[tex]a_n=\prod_{k=2}^n\left(1-\dfrac{1}{C_{k+1}^2}\right)=\prod_{k=2}^n\left[1-\dfrac{1}{\dfrac{(k+1)!}{2!\cdot(k+1-2)!}}\right]=\\=\prod_{k=2}^n\left[1-\dfrac{2!\cdot(k-1)!}{(k+1)!}\right]=\prod_{k=2}^n\left[1-\dfrac{2\cdot(k-1)!}{(k+1)\cdot k\cdot(k-1)!}\right]=\\=\prod_{k=2}^n\left[1-\dfrac{2}{(k+1)\cdot k}\right]=\prod_{k=2}^n\left[\dfrac{k^2+k-2}{(k+1)\cdot k}\right]=\\=\prod_{k=2}^n\left[\dfrac{k^2-1+k-1}{(k+1)\cdot k}\right]=\prod_{k=2}^n\left[\dfrac{(k-1)\cdot(k+1)+k-1}{(k+1)\cdot k}\right]=\\=\prod_{k=2}^n\left[\dfrac{(k-1)\cdot(k+2)}{(k+1)\cdot k}\right]=\left(\dfrac{1}{3}\cdot\dfrac{4}{2}\right)\cdot\left(\dfrac{2}{4}\cdot\dfrac{5}{3}\right)\cdot\left(\dfrac{3}{5}\cdot\dfrac{6}{4}\right)\cdot\ldots\cdot\\\cdot\left(\dfrac{n-2}{n}\cdot\dfrac{n+1}{n-1}\right)\cdot\left(\dfrac{n-1}{n+1}\cdot\dfrac{n+2}{n}\right)=\dfrac{1}{3}\cdot\dfrac{n+2}{n}=\dfrac{n+2}{3n}.[/tex]
Limita este deci 1/3.
Green eyes.
