Đặt :
\(S=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+=..+\dfrac{100}{3^{100}}\)
\(\Leftrightarrow3S=1+\dfrac{2}{3}+\dfrac{3}{3^2}+\dfrac{3}{3^3}+--...+\dfrac{100}{3^{99}}\)
\(\Leftrightarrow3S-S=\left(1+\dfrac{2}{3}+\dfrac{3}{3^2}+--.+\dfrac{100}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+--..+\dfrac{100}{3^{99}}\right)\)
\(\Leftrightarrow2S=1+\dfrac{1}{3}+\dfrac{1}{3^2}+--...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
Đặt :
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+--+\dfrac{1}{3^{99}}\)
\(\Leftrightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+--.+\dfrac{1}{3^{99}}\)
\(\Leftrightarrow3A-A=\left(1+\dfrac{1}{3}+--+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+-+\dfrac{1}{3^{99}}\right)\)
\(\Leftrightarrow2A=1-\dfrac{1}{3^{99}}\)
\(\Leftrightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}\)
\(\Leftrightarrow S=1+\dfrac{1}{2}-\dfrac{1}{2.3^{99}}+\dfrac{100}{3^{100}}< 1+\dfrac{1}{2}=\dfrac{3}{2}< \dfrac{3}{4}\)
\(\Leftrightarrow S< \dfrac{3}{4}\left(đpcm\right)\)
