Đáp án:
\[0\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
S = \frac{{2.4.6.8....\left( {2n} \right)}}{{3.5.7......\left( {2n + 1} \right)}} = \frac{2}{3}.\frac{4}{5}.\frac{6}{7}......\frac{{2n}}{{2n + 1}}\\
\frac{n}{{n + 1}} - \frac{{n + 1}}{{n + 2}} = \frac{{n\left( {n + 2} \right) - {{\left( {n + 1} \right)}^2}}}{{\left( {n + 1} \right)\left( {n + 2} \right)}} = \frac{{ - 1}}{{\left( {n + 1} \right)\left( {n + 2} \right)}} < 0,\,\,\,\forall n \in N\\
\Rightarrow \frac{n}{{n + 1}} < \frac{{n + 1}}{{n + 2}}\\
\Rightarrow \left\{ \begin{array}{l}
\frac{2}{3} < \frac{3}{4}\\
\frac{4}{5} < \frac{5}{6}\\
\frac{6}{7} < \frac{7}{8}\\
.....\\
\frac{{2n}}{{2n + 1}} < \frac{{2n + 1}}{{2n + 2}}
\end{array} \right.\\
\Rightarrow {S^2} = \frac{2}{3}.\frac{2}{3}.\frac{4}{5}.\frac{4}{5}.\frac{6}{7}.\frac{6}{7}......\frac{{2n}}{{2n + 1}}.\frac{{2n}}{{2n + 1}}\\
\Rightarrow {S^2} < \frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}.\frac{7}{8}......\frac{{2n}}{{2n + 1}}.\frac{{2n + 1}}{{2n + 2}}\\
\Rightarrow {S^2} < \frac{2}{{2n + 2}} = \frac{1}{{n + 1}}\\
\Rightarrow 0 < S < \frac{1}{{\sqrt {n + 1} }}\\
\lim \frac{1}{{\sqrt {n + 1} }} = 0 \Rightarrow \lim S = 0
\end{array}\)
