Giá trị của giới hạn \(\lim...

A

- Hướng dẫn giải

Ta có 

\(\begin{aligned} \frac{1}{1.4}+\frac{1}{2.5}+\ldots \ldots+\frac{1}{n(n+3)} &=\frac{1}{3}\left[1-\frac{1}{4}+\frac{1}{2}-\frac{1}{5}+\frac{1}{3}-\frac{1}{6}+\cdots+\frac{1}{n}-\frac{1}{n+3}\right] \\ &=\frac{1}{3}\left[\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)-\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\cdots+\frac{1}{n+3}\right)\right] \\ &=\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}\right) \\ &=\frac{1}{3}\left(\frac{11}{6}-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}\right) \end{aligned}\)

Do đó

\(\lim \left(\frac{1}{1.4}+\frac{1}{2.5}+\ldots . .+\frac{1}{n(n+3)}\right)=\lim \frac{1}{3}\left(\frac{11}{6}-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}\right)=\frac{11}{18}\)