Đặt :
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+................+\dfrac{1}{2^n}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.........+\dfrac{1}{2^{n-1}}\)
\(\Rightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+..........+\dfrac{1}{2^{n-1}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+............+\dfrac{1}{2^n}\right)\)
\(\Rightarrow A=1-\dfrac{1}{2^n}< 1\)
\(\Rightarrow A< 1\rightarrowđpcm\)
Vậy \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...............+\dfrac{1}{2^n}< 1\rightarrowđpcm\)
Ta thấy :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
................
\(\dfrac{1}{2^n}< \dfrac{1}{n.\left(n-1\right)}\)
\(\)- > \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2^n}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}\)= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)= \(1-\dfrac{1}{n}< 1\left(ĐPCM\right)\)
A=12+122+123+124+................+12nA=12+122+123+124+................+12n
⇒2A=1+12+122+123+.........+12n−1⇒2A=1+12+122+123+.........+12n−1
⇒2A−A=(1+12+122+..........+12n−1)−(12+122+............+12n)⇒2A−A=(1+12+122+..........+12n−1)−(12+122+............+12n)
⇒A=1−12n<1⇒A=1−12n<1
⇒A<1→đpcm⇒A<1→đpcm
Vậy 12+122+123+...............+12n<1→đpcm