Question

Nu prea ma descurc la genul asta de exercitiiimi poate explica cineva?

Answer 1

$\lim\limits_{n\to \infty} \dfrac{1\cdot 2+2\cdot 3+...+n(n+1)}{n(n+1)(n+2)} = \\ \\ = \lim\limits_{n\to \infty} \dfrac{\sum\limits_{k=1}^{n}\Big[k(k+1)\Big]}{n(n+1)(n+2)} =\lim\limits_{n\to \infty} \dfrac{\sum\limits_{k=1}^{n}(k^2+k)}{n(n+1)(n+2)} =$

$=\lim\limits_{n\to \infty} \dfrac{\sum\limits_{k=1}^{n}k^2+\sum\limits_{k=1}^{n}k}{n(n+1)(n+2)}=\lim\limits_{n\to \infty} \dfrac{\dfrac{n(n+1)(2n+1)}{6}+\dfrac{n(n+1)}{2}}{n(n+1)(n+2)}=\\ \\=\lim\limits_{n\to \infty} \dfrac{\dfrac{(2n+1)}{6}+\dfrac{1}{2}}{(n+2)}\lim\limits_{n\to \infty} \dfrac{2n+4}{6(n+2)}=\lim\limits_{n\to\infty}\dfrac{2(n+2)}{6(n+2)}=\dfrac{2}{6}=\dfrac{1}{3}$