Question

[Đào Lâm Nguyên]

S= 1/3 - 2/3^2 + 3/3^3 - 4/3^4 +..... + 99/3^99 - 100/3^100. So sánh S và 1/2

Answer 1

[Nguyễn Gia Khánh]

\(S=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)

\(3S=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)

\(3S+S=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)

\(3\times4S=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{97}}-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)

\(S=\dfrac{3}{12}-\dfrac{1}{12}+\dfrac{1}{3\times12}-\dfrac{1}{3^2\times12}+...-\dfrac{100}{3^{99}\times12}\)

\(\Rightarrow S< \dfrac{3}{12}=\dfrac{1}{4}< \dfrac{1}{2}\)