cho tôngr S=1/3+1/4+1/5+…+1/8+1/9. Chứng minh rằng 1 { "@context": "https://schema.org", "@type": "QAPage", "mainEntity": { "@type": "Question", "name": " cho tôngr S=1/3+1/4+1/5+...+1/8+1/9. Chứng minh rằng 1
0 bình luận về “cho tôngr S=1/3+1/4+1/5+…+1/8+1/9. Chứng minh rằng 1<S<2”
$\begin{array}{l}S=\dfrac13+\dfrac14+\dfrac15+\dfrac16+\dfrac17+\dfrac18+\dfrac19\\\quad=\left(\dfrac13+\dfrac14+\dfrac15\right)+\left(\dfrac16+\dfrac17+\dfrac18+\dfrac19\right)\\\quad>\left(\dfrac15+\dfrac15+\dfrac15\right)+\left(\dfrac19+\dfrac19+\dfrac19+\dfrac19\right)\\\quad=\dfrac15\cdot3+\dfrac19\cdot4\\\quad=\dfrac35+\dfrac49\\\quad=\dfrac{47}{45}>1\\\to S>1\quad(1)\\S=\dfrac13+\dfrac14+\dfrac15+\dfrac16+\dfrac17+\dfrac18+\dfrac19\\\quad=\left(\dfrac13+\dfrac14+\dfrac15\right)+\left(\dfrac16+\dfrac17+\dfrac18+\dfrac19\right)\\\quad<\left(\dfrac13+\dfrac13+\dfrac13\right)+\left(\dfrac16+\dfrac16+\dfrac16+\dfrac16\right)\\\quad=\dfrac13\cdot3+\dfrac16\cdot4\\\quad=1+\dfrac23\\\quad<1+1=2\\\to S<2\quad(2)\\\text{- Từ (1) và (2) $\to 1<S<2$} \end{array}$
$\begin{array}{l}S=\dfrac13+\dfrac14+\dfrac15+\dfrac16+\dfrac17+\dfrac18+\dfrac19\\\quad=\left(\dfrac13+\dfrac14+\dfrac15\right)+\left(\dfrac16+\dfrac17+\dfrac18+\dfrac19\right)\\\quad>\left(\dfrac15+\dfrac15+\dfrac15\right)+\left(\dfrac19+\dfrac19+\dfrac19+\dfrac19\right)\\\quad=\dfrac15\cdot3+\dfrac19\cdot4\\\quad=\dfrac35+\dfrac49\\\quad=\dfrac{47}{45}>1\\\to S>1\quad(1)\\S=\dfrac13+\dfrac14+\dfrac15+\dfrac16+\dfrac17+\dfrac18+\dfrac19\\\quad=\left(\dfrac13+\dfrac14+\dfrac15\right)+\left(\dfrac16+\dfrac17+\dfrac18+\dfrac19\right)\\\quad<\left(\dfrac13+\dfrac13+\dfrac13\right)+\left(\dfrac16+\dfrac16+\dfrac16+\dfrac16\right)\\\quad=\dfrac13\cdot3+\dfrac16\cdot4\\\quad=1+\dfrac23\\\quad<1+1=2\\\to S<2\quad(2)\\\text{- Từ (1) và (2) $\to 1<S<2$} \end{array}$