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\(Ta\) \(có\) \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{256}\)
\(Vì\) \(1>\frac{1}{256},\frac{1}{2}>\frac{1}{256},....,\frac{1}{255}>\frac{1}{256},\frac{1}{256}=\frac{1}{256}\)
\(\Rightarrow1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{256}>\frac{1}{256}+\frac{1}{256}+...+\frac{1}{256}\)
\(=\frac{1}{256}.256=1\)\(< 5\)
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Ta có :
\(\dfrac{1}{2}>\dfrac{1}{5}\)
\(\dfrac{1}{3}>\dfrac{1}{5}\)
\(\dfrac{1}{4}>\dfrac{1}{5}\)
\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}>\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}\)
\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}>\dfrac{4}{5}\)
\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}\)
\(=\dfrac{7}{6}+\dfrac{1}{4}+\dfrac{1}{5}\)
\(=\dfrac{17}{12}+\dfrac{1}{5}\)
\(=\dfrac{97}{60}\)
\(\dfrac{4}{5}=\dfrac{4.12}{5.12}=\dfrac{48}{60}\)
Mà \(\dfrac{97}{60}>\dfrac{48}{60}\)
\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}>\dfrac{4}{5}\left(đpcm\right)\).
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Ta có: 1/3^2<1/2.3;1/4^2<1/3.4;........
=>1/3^2+1/4^2+1/5^2+......+1/100^2
< 1/2.3+1/3.4+1/4.5+.....+1/99.100
=1/2-1/3+1/3-1/4+1/4-1/5+...+1/99-1/100
=1/2-1/100
=49/100
Mà 49/100<1/2
Nên 1/3^2+1/4^2+1/5^2+......+1/100^2<1/2
Đ ú n g nha......................