\(A=\dfrac{5}{4}+\dfrac{5}{4^2}+\dfrac{5}{4^3}+...+\dfrac{5}{4^{99}}\\ 4A=5+\dfrac{5}{4}+\dfrac{5}{4^2}+...+\dfrac{5}{4^{98}}\\ 4A-A=\left(5+\dfrac{5}{4}+\dfrac{5}{4^2}+...+\dfrac{5}{4^{98}}\right)-\left(\dfrac{5}{4}+\dfrac{5}{4^2}+\dfrac{5}{4^3}+...+\dfrac{5}{4^{99}}\right)\\ 3A=5-\dfrac{5}{4^{99}}\\ A=\left(5-\dfrac{5}{4^{99}}\right):3\\ A=\dfrac{5}{3}-\dfrac{5}{4^{99}}:3\\ A=\dfrac{5}{3}-\dfrac{5}{4^{99}\cdot3}< \dfrac{5}{3}\)
Vậy \(A< \dfrac{5}{3}\)
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