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\(B=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2014}}\)
\(5B=5\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2014}}\right)\)
\(5B=1+\frac{1}{5}+...+\frac{1}{5^{2013}}\)
\(5B-B=\left(1+\frac{1}{5}+...+\frac{1}{5^{2013}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2014}}\right)\)
\(4B=1-\frac{1}{5^{2014}}\Rightarrow B=\frac{1-\frac{1}{5^{2014}}}{4}\)
Ta có: \(1-\frac{1}{5^{2014}}< 1\Rightarrow\frac{1-\frac{1}{5^{2014}}}{4}< \frac{1}{4}\)
\(\Rightarrow B< \frac{1}{4}\)(Đpcm)



TỪ ĐỀ BÀI => 5A=1+1/5+1/5^2+......+1/5^2013
CÓ 4A=5A-A
=>4A=(1+1/5+1/5^2+.....+1/5^2013)-(1/5+1/5^2+1/5^3+....+1/5^2014)
=>4A= 1- 1/5^2014
=>A= (1-1/5^2014)/4 ;CÓ 1-1/5^2014 <1
=>A<1/4


Đặt \(A=\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{2014^3}< B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2013.2014.2015}\)
Mà \(2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2013.2014.2015}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2013.2014}-\frac{1}{2014.2015}\)
\(=\frac{1}{2}-\frac{1}{2014.2015}< \frac{1}{2}\)
\(\Rightarrow B< \frac{1}{4}\)
Vậy \(A< \frac{1}{4}\)

\(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.......+\dfrac{1}{10^2}\)
\(D< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{9.10}\)
\(D< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{9}-\dfrac{1}{10}\)
\(D< 1-\dfrac{1}{10}\Leftrightarrow D< 1\left(đpcm\right)\)

\(A=\frac{1}{5}+\frac{1}{5^2}+........+\frac{1}{5^{2014}}\)
\(\Rightarrow5A=1+\frac{1}{5}+...........+\frac{1}{5^{2013}}\)
\(\Rightarrow5A-A=1+...........+\frac{1}{5^{2013}}-\frac{1}{5}+...........+\frac{1}{5^{2014}}\)
\(\Rightarrow4A=1-\frac{1}{5^{2014}}\)
\(\Rightarrow4A< 1\Rightarrow A< \frac{1}{4}\)
=> 5A = 1 + 1/5 +...+1/5^2013
=>4A= 1- 1/5^2014
=> 4A< 1 => A < 1/4