Đáp án:
$\begin{array}{l}
A = \dfrac{1}{{1.3}} + \dfrac{1}{{3.5}} + ... + \dfrac{1}{{\left( {2n - 1} \right).\left( {2n + 1} \right)}}\\
= \dfrac{1}{2}.\left( {\dfrac{2}{{1.3}} + \dfrac{2}{{3.5}} + ... + \dfrac{2}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}}} \right)\\
= \dfrac{1}{2}.\left( {\dfrac{{3 - 1}}{{1.3}} + \dfrac{{5 - 3}}{{3.5}} + ... + \dfrac{{2n + 1 - \left( {2n - 1} \right)}}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}}} \right)\\
= \dfrac{1}{2}.\left( {1 - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{5} + ... + \dfrac{1}{{2n - 1}} - \dfrac{1}{{2n + 1}}} \right)\\
= \dfrac{1}{2}\left( {1 - \dfrac{1}{{2n + 1}}} \right)\\
= \dfrac{1}{2} - \dfrac{1}{{2\left( {2n + 1} \right)}} < \dfrac{1}{2}
\end{array}$
Vậy A<1/2
