em cần gấp ạ.Giúp e với .
Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)
\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)
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\(\frac{1}{10^2}<\frac{1}{9\cdot10}=\frac19-\frac{1}{10}\)
Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{10^2}<1-\frac12+\frac12-\frac13+\cdots+\frac19-\frac{1}{10}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{10^2}<1-\frac{1}{10}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{10^2}<1\left(1\right)\)
Ta có: \(\frac{1}{2^2}>\frac{1}{2\cdot3}=\frac12-\frac13\)
\(\frac{1}{3^2}>\frac{1}{3\cdot4}=\frac13-\frac14\)
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\(\frac{1}{10^2}>\frac{1}{10\cdot11}=\frac{1}{10}-\frac{1}{11}\)
Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{10^2}>\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{10}-\frac{1}{11}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{10^2}>\frac12-\frac{1}{11}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{10^2}>\frac{11}{22}-\frac{2}{22}=\frac{9}{22}>\frac{8}{22}=\frac{4}{11}\) (2)
Từ (1),(2) suy ra \(\frac{4}{11}<\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{10^2}<1\)
em cần gấp ạ.Giúp em với.
em cần gấp ạ.Giúp em với!!!!!!!!!!!!!!!!!!!!!!
