\(A=\dfrac{1}{3^1}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\)
\(A=\dfrac{1}{3}.\left(1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\right)\)
\(\Rightarrow3A=3.\dfrac{1}{3}.\left(1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\right)\)
\(\Rightarrow3A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\)
\(\Rightarrow3A-A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...\dfrac{1}{3^{2022}}-\left(\dfrac{1}{3^1}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\right)\)
\(\Rightarrow2A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...\dfrac{1}{3^{2022}}-\dfrac{1}{3^1}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-...\dfrac{1}{3^{2022}}-\dfrac{1}{3^{2023}}\)
\(\Rightarrow2A=1-\dfrac{1}{3^{2023}}\)
\(\Rightarrow A=\dfrac{1}{2}\left(1-\dfrac{1}{3^{2023}}\right)\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{1}{3^{2023}}< \dfrac{1}{2}\)
\(B=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{12}=\dfrac{4+3+1}{12}=\dfrac{8}{12}=\dfrac{2}{3}\)
mà \(\dfrac{2}{3}>\dfrac{1}{2}\) \(\left(\dfrac{2}{3}=\dfrac{4}{6}>\dfrac{1}{2}=\dfrac{3}{6}\right)\)
\(\Rightarrow A< B\)
A = \(\dfrac{1}{3}\)+ \(\dfrac{1}{3^2}\)+ \(\dfrac{1}{3^3}\)+............+\(\dfrac{1}{3^{2023}}\)
3A = 1+ \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+...+\(\dfrac{1}{3^{2022}}\)
3A - A = 1 - \(\dfrac{1}{3^{2023}}\)
2A = 1 - \(\dfrac{1}{3^{2023}}\) < 1
B = \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\)+ \(\dfrac{1}{12}\)
B = \(\dfrac{4}{12}\) + \(\dfrac{3}{12}\) + \(\dfrac{1}{12}\)
B = \(\dfrac{8}{12}\)
B = \(\dfrac{2}{3}\) ⇒ 2B = \(\dfrac{4}{3}\) > 1
2A < 2B ⇒ A < B
