Giải thích các bước giải:
\(\begin{array}{l}
13,\\
\frac{1}{{n\left( {n + 1} \right)}} = \frac{{\left( {n + 1} \right) - n}}{{n\left( {n + 1} \right)}} = \frac{1}{n} - \frac{1}{{n + 1}}\\
{u_n} = \frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + .... + \frac{1}{{n\left( {n + 1} \right)}}\\
= \left( {1 - \frac{1}{2}} \right) + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + ..... + \left( {\frac{1}{n} - \frac{1}{{n + 1}}} \right)\\
= 1 - \frac{1}{{n + 1}}\\
\lim \frac{1}{{n + 1}} = 0 \Rightarrow \lim {u_n} = \lim \left( {1 - \frac{1}{{n + 1}}} \right) = 1\\
14,\\
\frac{1}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}} = \frac{1}{2}.\frac{2}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}} = \frac{1}{2}.\frac{{\left( {2n + 1} \right) - \left( {2n - 1} \right)}}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}} = \frac{1}{2}\left[ {\frac{1}{{2n - 1}} - \frac{1}{{2n + 1}}} \right]\\
{u_n} = \frac{1}{{1.3}} + \frac{1}{{3.5}} + \frac{1}{{5.7}} + .... + \frac{1}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}}\\
= \frac{1}{2}.\left[ {\frac{2}{{1.3}} + \frac{2}{{3.5}} + \frac{2}{{5.7}} + ..... + \frac{2}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}}} \right]\\
= \frac{1}{2}.\left[ {1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + .... + \frac{1}{{2n - 1}} - \frac{1}{{2n + 1}}} \right]\\
= \frac{1}{2}.\left( {1 - \frac{1}{{2n + 1}}} \right)\\
\lim \frac{1}{{2n + 1}} = 0 \Rightarrow \lim {u_n} = \lim \left[ {\frac{1}{2}\left( {1 - \frac{1}{{2n + 1}}} \right)} \right] = \frac{1}{2}.1 = \frac{1}{2}\\
15,\\
1 - \frac{1}{{{n^2}}} = \frac{{{n^2} - 1}}{{{n^2}}} = \frac{{\left( {n - 1} \right)\left( {n + 1} \right)}}{{{n^2}}}\\
{u_n} = \left( {1 - \frac{1}{{{2^2}}}} \right)\left( {1 - \frac{1}{{{3^2}}}} \right).....\left( {1 - \frac{1}{{{n^2}}}} \right)\\
= \frac{{\left( {{2^2} - 1} \right)\left( {{3^2} - 1} \right)......\left( {{n^2} - 1} \right)}}{{{2^2}{{.3}^2}....{n^2}}}\\
= \frac{{\left( {1.3} \right).\left( {2.4} \right)......\left[ {\left( {n - 1} \right)\left( {n + 1} \right)} \right]}}{{\left( {2.3.4.....n} \right)\left( {2.3.4....n} \right)}}\\
= \frac{{\left( {1.2.3.....\left( {n - 1} \right)} \right).\left( {3.4.5.6....\left( {n + 1} \right)} \right)}}{{\left( {2.3.4.....n} \right)\left( {2.3.4.....n} \right)}}\\
= \frac{{1.2.3.....\left( {n - 1} \right)}}{{2.3.4....n}}.\frac{{3.4.5.....\left( {n + 1} \right)}}{{2.3.4....n}}\\
= \frac{1}{n}.\frac{{n + 1}}{2}\\
= \frac{{n + 1}}{{2n}}\\
\lim {u_n} = \lim \frac{{n + 1}}{{2n}} = \lim \frac{{1 + \frac{1}{n}}}{2} = \frac{1}{2}
\end{array}\)
