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
cho tôngr S=1/3+1/4+1/5+…+1/8+1/9. Chứng minh rằng 1
By Jade
By Jade
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
+ )S = \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + … + \dfrac{1}{8} + \dfrac{1}{9}\\
> \dfrac{1}{3} + \left( {\dfrac{1}{9} + \dfrac{1}{9} + … + \dfrac{1}{9} + \dfrac{1}{9}} \right)\\
= \dfrac{1}{3} + \dfrac{6}{9}\\
= 1\\
+ )S = \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + … + \dfrac{1}{8} + \dfrac{1}{9}\\
< \dfrac{1}{3} + \left( {\dfrac{1}{4} + \dfrac{1}{4} + … + \dfrac{1}{4} + \dfrac{1}{4}} \right)\\
= \dfrac{1}{3} + \dfrac{6}{4}\\
= \dfrac{1}{3} + \dfrac{3}{2}\\
= \dfrac{{11}}{6}\\
< 2
\end{array}$
Như vậy $1<S<2$