|
|
$\lim(\frac{1}{1.3}+\frac{1}{2.4}+...+\frac{1}{n(n+2)})=\dfrac{1}{2}\lim\left ( \dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+2} \right )$
$=\dfrac{1}{2}\lim\left ( 1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\right )-\dfrac{1}{2}\lim\left (\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{n+2} \right )$
$=\dfrac{1}{2}\lim\left ( 1+\dfrac{1}{2}-\dfrac{1}{n+1}-\dfrac{1}{n+2}\right )=\dfrac{1}{2}\left ( 1+\dfrac{1}{2}\right )=\dfrac{3}{4}$.
|