Question

Cho A= \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+.......+\dfrac{1}{3^{2018}}+\dfrac{1}{3^{2019}}\) . Chứng minh rằng: A<\(\dfrac{1}{2}\)

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Answer 1

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2018}}+\dfrac{1}{3^{2019}}\)

\(3A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{2017}}+\dfrac{1}{3^{2018}}\)

\(3A-A=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{2018}}\right)-\left(\dfrac{1}{3}+...+\dfrac{1}{3^{2019}}\right)\)

\(2A=1-\dfrac{1}{3^{2019}}\)

\(A=\dfrac{1}{2}-\dfrac{1}{2\cdot3^{2019}}< \dfrac{1}{2}\) (DPCM)

Answer 2

\(3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2017}}+\dfrac{1}{3^{2018}}\)

\(\Rightarrow3A-A=1-\dfrac{1}{3^{2019}}\)

\(\Rightarrow2A=1-\dfrac{1}{3^{2019}}\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{2019}}\)

\(\Rightarrow A< \dfrac{1}{2}\)