Cho dãy số $({u_n})$ với ${u_n} = \dfrac{1}{{1.3}} + \dfrac{1}{{3.5}} + ... + \dfrac{1}{{\left( {2n - 1} \right).\left( {2n + 1} \right)}}$

Khi đó $\lim {u_n}$ bằng?

$1.$    

$\dfrac{2}{3}.$

$- 1.$

$- \dfrac{1}{3}.$

$\lim \sqrt[3]{{{u_n}}} = L$

$\lim \sqrt {{u_n}}  = L$

$\lim \sqrt {{u_n}}  = \sqrt L$         

$\lim \sqrt[3]{{{u_n}}} = \sqrt[3]{L}$

$1.$

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

$\dfrac{4}{5}.$

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

$\lim \left( {{u_n}.{v_n}} \right) = 0$          

$\lim \left( {{u_n}.{v_n}} \right) =  + \infty$          

$\lim \left( {{u_n}.{v_n}} \right) =  - \infty$

$\lim \left( {{u_n}.{v_n}} \right) = 1$

$0.$     

$1.$    

$\dfrac{3}{5}.$         

$+ \infty .$

${u_n} = \dfrac{n}{2}$

${u_n} = \dfrac{2}{n}$

${u_n} = n$

${u_n} = \sqrt n$

$\lim n =  + \infty$    

$\lim \sqrt n  =  + \infty$      

$\lim \sqrt[3]{n} =  + \infty$

$\lim \dfrac{1}{n} =  + \infty$