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

\({u_n} = \dfrac{n}{2}\)

\({u_n} = \dfrac{2}{n}\)

\({u_n} = n\)

\({u_n} = \sqrt n \)

$\dfrac{1}{5}.$

$ - \dfrac{4}{5}.$     

$\dfrac{4}{5}.$

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

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

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

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

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

$1.$

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

$\dfrac{4}{5}.$

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

\({u_n} = \dfrac{1}{{\sqrt n }}\)                  

\({u_n} = \dfrac{1}{{\sqrt[3]{n}}}\)

\({u_n} = \dfrac{{\sqrt[3]{n}}}{2}\)

\({u_n} = 0\)

$0.$    

$ - \dfrac{1}{4}.$      

$\dfrac{3}{4}.$

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

\(\lim {u_n} = 0\)                   

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

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

\(\lim {u_n} < 0\)