Ta có : \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+....+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{1}{20}.10=\frac{1}{2}\) ( 10 số hạng 1/20)
\(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+....+\frac{1}{29}>\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{1}{30}.10=\frac{1}{3}\) ( 10 số hạng 1/30 )
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\(\frac{1}{90}+\frac{1}{91}+...+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{1}{100}.10=\frac{1}{10}\). Và: \(\frac{1}{100}=\frac{1}{100}\)
Nên: \(C=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}+\frac{1}{100}>1\) (đpcm)
Ta có:
\(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)
\(\frac{1}{20}+\frac{1}{21}+...+\frac{1}{19}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)
\(\frac{1}{30}+\frac{1}{31}+...+\frac{1}{39}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}+\frac{10}{40}+\frac{1}{4}\)
\(=>\frac{1}{10}+\frac{1}{11}+...+\frac{1}{39}>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}>1\)
Vậy \(C>1\)
