Question

Salut, ma puteti ajuta la problema 284...?

Answer 1

$\sum\limits_{k=1}^{n} \dfrac{6}{k(k+1)(k+3)} =\sum\limits_{k=1}^{n} \dfrac{6}{k(k+1)}\cdot \dfrac{1}{k+3} = 6\sum\limits_{k=1}^{n} \dfrac{1}{k(k+1)(k+3)} = \\ \\ = 6\sum\limits_{k=1}^{n} \Big(\dfrac{1}{k}-\dfrac{1}{k+1}\Big)\cdot \dfrac{1}{k+3}} = 6\sum\limits_{k=1}^{n} \Big(\dfrac{1}{k(k+3)}-\dfrac{1}{(k+1)(k+3)}\Big)=$

$=6\sum\limits_{k=1}^{n} \Big(\dfrac{1}{3}\cdot \dfrac{k+3-k}{k(k+3)}-\dfrac{1}{2}\cdot \dfrac{k+3-(k+1)}{(k+1)(k+3)}\Big)=\\ \\ =6\sum\limits_{k=1}^{n} \Big[\dfrac{1}{3}\cdot \Big(\dfrac{1}{k}-\dfrac{1}{k+3}\Big)-\dfrac{1}{2}\cdot \Big(\dfrac{1}{k+1}-\dfrac{1}{k+3}\Big)\Big]= \\ \\ = \sum\limits_{k=1}^{n} \Big[2\cdot \Big(\dfrac{1}{k}-\dfrac{1}{k+3}\Big)-3\cdot \Big(\dfrac{1}{k+1}-\dfrac{1}{k+3}\Big)\Big]$

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

$\Rightarrow \lim\limits_{n\to \infty}\sum\limits_{k=1}^{n} \dfrac{6}{k(k+1)(k+3)} = 2+1+\dfrac{2}{3}-\dfrac{3}{2}-1 =\dfrac{12+4-9}{6} = \boxed{\dfrac{7}{6}}$