Đc lém Min đúng lúc tui đang định đăng câu ó

\(Ta\)  \(có\)  \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{256}\)

                   \(Vì\)  \(1>\frac{1}{256},\frac{1}{2}>\frac{1}{256},....,\frac{1}{255}>\frac{1}{256},\frac{1}{256}=\frac{1}{256}\)

                 \(\Rightarrow1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{256}>\frac{1}{256}+\frac{1}{256}+...+\frac{1}{256}\)

                  \(=\frac{1}{256}.256=1\)\(< 5\)

Đặt \(A=\frac15+\frac{2}{5^2}+\cdots+\frac{2016}{5^{2016}}\)

=>\(5A=1+\frac25+\cdots+\frac{2016}{5^{2015}}\)

=>\(5A-A=1+\frac25+\cdots+\frac{2016}{5^{2015}}-\frac15-\frac{2}{5^2}-\cdots-\frac{2016}{5^{2016}}\)

=>\(4A=1+\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{2015}}-\frac{2016}{5^{2016}}\)

Đặt \(B=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{2015}}\)

=>\(5B=1+\frac15+\cdots+\frac{1}{5^{2014}}\)

=>\(5B-B=1+\frac15+\cdots+\frac{1}{5^{2014}}-\frac15-\frac{1}{5^2}-\cdots-\frac{1}{5^{2015}}\)

=>\(4B=1-\frac{1}{5^{2015}}=\frac{5^{2015}-1}{5^{2015}}\)

=>\(B=\frac{5^{2015}-1}{4\cdot5^{2015}}\)

TA có: \(4A=1+\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{2015}}-\frac{2016}{5^{2016}}\)

\(=1+\frac{5^{2015}-1}{4\cdot5^{2015}}-\frac{2016}{5^{2016}}=1+\frac{5^{2016}-5-8064}{4\cdot5^{2016}}=1+\frac14-\frac{8069}{4\cdot5^{2016}}\)

=>\(4A<1+\frac14=\frac54\)

=>\(A<\frac{5}{16}\)

mà \(\frac{5}{16}<\frac{5}{15}=\frac13\)

nên \(A<\frac13\) (1)

Ta có: \(4A=1+\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{2015}}-\frac{2016}{5^{2016}}\)

=>\(20A=5+1+\frac15+\cdots+\frac{1}{5^{2014}}-\frac{2016}{5^{2015}}\)

=>\(20A-4A=5+1+\frac15+\cdots+\frac{1}{5^{2014}}-\frac{2016}{5^{2015}}-1-\frac15-\frac{1}{5^2}-\cdots-\frac{1}{5^{2015}}-\frac{2016}{5^{2016}}\)

=>\(16A=5-\frac{2017}{5^{2015}}-\frac{2016}{5^{2016}}>5\)

=>\(A>\frac{5}{16}\)

=>\(A>\frac{4}{16}=\frac14\) (2)

Từ (1),(2) suy ra 1/4<A<1/3

thiếu đề :(

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}< \frac{1}{2}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}< \frac{1}{2}\)

\(=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\left(đpcm\right)\)

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 \(\dfrac{97}{60}>\dfrac{48}{60}\)

\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}>\dfrac{4}{5}\left(đpcm\right)\).

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......................