D=\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
=>3D=1+\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
=>3D-D=(1+\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\))-(\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\))
=>2D=1-\(\dfrac{1}{3^{100}}< 1\)
=>D<\(\dfrac{1}{2}\)
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