Số lượng số của S là :
\(\left(20-11\right):1+1=10\)( số )
Ta có :
\(\frac{1}{11}>\frac{1}{20};\frac{1}{12}>\frac{1}{20};...;\frac{1}{19}>\frac{1}{20};\frac{1}{20}=\frac{1}{20}\)
\(\Rightarrow\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)
\(\Rightarrow S>\frac{1}{20}.10\)
\(\Rightarrow S>\frac{1}{2}\)
Vậy \(S>\frac{1}{2}\)
Ta có:
1/11 + 1/12 + 1/13 + ....................... + 1/ 20 > 1/20 +1/20 +1/ 20 +1/20 +1/20 +1/20 +1/20 +1/ 20 +1/20 +1/20 = 1/2
=> S > 1/2
Vậy S > 1/2
\(\frac{1}{11}>\frac{1}{20};\frac{1}{12}>\frac{1}{20};............;\frac{1}{20}=\frac{1}{20}\) nên cái tổng ấy sẽ > 10 lần 1/20 =1/2
Ta có: \(\frac{1}{11}>\frac{1}{20}\)
\(\frac{1}{12}>\frac{1}{20}\)
\(\frac{1}{13}>\frac{1}{20}\)
\(........\)
\(\frac{1}{19}>\frac{1}{20}\)
suy ra: \(S=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}+\frac{1}{20}=\frac{1}{2}\)
Vậy \(S>\frac{1}{2}\)
\(\frac{1}{11}>\frac{1}{20};\frac{1}{12}>\frac{1}{20};\frac{1}{13}>\frac{1}{20};\frac{1}{14}>\frac{1}{20};\frac{1}{15}>\frac{1}{20}\)
\(\frac{1}{16}>\frac{1}{20};\frac{1}{17}>\frac{1}{20};\frac{1}{18}>\frac{1}{20};\frac{1}{19}>\frac{1}{20};\frac{1}{20}=\frac{1}{20}\)
cộng vế với vế ta đc:
\(S=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}\)
> \(\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}=\frac{1}{20}\cdot10=\frac{1}{2}\)
vậy \(S>\frac{1}{2}\)
