Tính giới hạn cho sau: \(\lim \left[ {\dfrac{1}{{1.4}} + \dfrac{1}{{2.5}} + ... + \dfrac{1}{{n(n + 3)}}} \right]\)

* Hướng dẫn giải

\(\lim \left[ {\dfrac{1}{{1.4}} + \dfrac{1}{{2.5}} + ... + \dfrac{1}{{n(n + 3)}}} \right]\)

Ta có:

 \(\begin{array}{l}\dfrac{1}{{1.4}} + \dfrac{1}{{2.5}} + ... + \dfrac{1}{{n(n + 3)}}\\ = \dfrac{1}{3}\left( {\dfrac{3}{{1.4}} + \dfrac{3}{{2.5}} + ... + \dfrac{3}{{n(n + 3)}}} \right)\\ = \dfrac{1}{3}\left( {1 - \dfrac{1}{4} + \dfrac{1}{2} - \dfrac{1}{5} + ... + \dfrac{1}{n} - \dfrac{1}{{n + 3}}} \right)\\ = \dfrac{1}{3}\left[ {\left( {1 + \dfrac{1}{2} + ... + \dfrac{1}{n}} \right) - \left( {\dfrac{1}{4} + \dfrac{1}{5} + ... + \dfrac{1}{{n + 3}}} \right)} \right]\\ = \dfrac{1}{3}\left( {1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{{n + 1}} + \dfrac{1}{{n + 2}} + \dfrac{1}{{n + 3}}} \right)\\ \Rightarrow \lim \left( {\dfrac{1}{{1.4}} + \dfrac{1}{{2.5}} + ... + \dfrac{1}{{n(n + 3)}}} \right)\\ = \lim \dfrac{1}{3}\left( {1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{{n + 1}} + \dfrac{1}{{n + 2}} + \dfrac{1}{{n + 3}}} \right)\\ = \dfrac{1}{3}\left( {1 + \dfrac{1}{2} + \dfrac{1}{3}} \right) = \dfrac{{11}}{{18}}\end{array}\)