Cho

- Hướng dẫn giải

$\begin{array}{l}S=\frac{9}{2.5}+\frac{9}{5.8}+\frac{9}{8.11}+\mathrm{...}+\frac{9}{29.32}\\ =3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\mathrm{...}+\frac{1}{29}-\frac{1}{32}\right)\\ =3\left(\frac{1}{2}-\frac{1}{32}\right)=\frac{45}{32}>1\\ \Rightarrow S>1\end{array}$