Cho cấp số nhân lùi vô hạn \(\left( {{u_n}} \right)\) công bội $q$. Đặt \(S = {u_1} + {u_2} + ... + {u_n} + ...\) thì:

$\dfrac{1}{5}.$

$ - \dfrac{4}{5}.$     

$\dfrac{4}{5}.$

$ - \dfrac{1}{5}.$

\({u_n} = \dfrac{{{{\left( {2017 - n} \right)}^{2018}}}}{{n{{\left( {2018 - n} \right)}^{2017}}}}\).

\({u_n} = n\left( {\sqrt {{n^2} + 2018}  - \sqrt {{n^2} + 2016} } \right)\).

\(\left\{ \begin{array}{l}{u_1} = 2017\\{u_{n + 1}} = \dfrac{1}{2}\left( {{u_n} + 1} \right),\,\,\,\,n = 1,2,3...\end{array} \right.\).

\({u_n} = \dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + ... + \dfrac{1}{{n\left( {n + 1} \right)}}\).

\(\lim \dfrac{{3{u_n} - 1}}{{{u_n} + 1}} = 3\)

\(\lim \dfrac{{3{u_n} - 1}}{{{u_n} + 1}} =  - 1\)

\(\lim \dfrac{{3{u_n} - 1}}{{{u_n} + 1}} = 2\)

\(\lim \dfrac{{3{u_n} - 1}}{{{u_n} + 1}} = 1\)

$1.$

$ - \dfrac{1}{4}.$

$\dfrac{4}{5}.$

$ - \dfrac{3}{4}.$

\(\lim n\sqrt {{u_n}}  = \sqrt 2 .\)

\(\lim n\sqrt {{u_n}}  = \dfrac{1}{{\sqrt 3 }}.\)

\(\lim n\sqrt {{u_n}}  = \sqrt 3 .\)     

\(\lim n\sqrt {{u_n}}  = \dfrac{1}{{\sqrt 2 }}.\)

\(\lim {u_n} = 0\)                   

\(\lim {u_n} > \lim {v_n}\)

\(\lim {u_n} < \lim {v_n}\)

\(\lim {u_n} < 0\) 

$0.$     

$1.$    

$\dfrac{3}{5}.$         

$ + \infty .$