So sánh tổng sau với ` 1`. `S = 1/3 + 1/7 + 1/13 + 1/21+...+ 1/91 + 1/111`

Đáp án+Giải thích các bước giải:

$\\S=\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{1}{13}+...+\dfrac{1}{91}+\dfrac{1}{111}$ $\\\text{Ta có:}$$\\\dfrac{1}{3}<\dfrac{1}{2}\\\dfrac{1}{7}<\dfrac{1}{6}\\\dfrac{1}{13}<\dfrac{1}{12}\\.........\\\dfrac{1}{91}<\dfrac{1}{90}\\\dfrac{1}{111}<\dfrac{1}{110}$ $\\=>\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{1}{13}+...+\dfrac{1}{91}+\dfrac{1}{111}<\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}+\dfrac{1}{110}=\dfrac{1}{2.1}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}+\dfrac{1}{10.11}\\=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\\=1-\dfrac{1}{11}\\=\dfrac{10}{11}<1\\=>S<1$