Đáp án:

$S<\dfrac15$.

Giải thích các bước giải:

$S=\dfrac13-\dfrac2{3^2}+\dfrac3{3^3}-\dfrac4{3^4}\ +\,.\!.\!.+\ \dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\\\Rightarrow 3S=3.\!\;\!\!\left(\dfrac13-\dfrac2{3^2}+\dfrac3{3^3}-\dfrac4{3^4}\ +\,.\!.\!.+\ \dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\right)\\\Rightarrow 3S=1-\dfrac23+\dfrac3{3^2}-\dfrac4{3^3}\ +\,.\!.\!.+\ \dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\\\Rightarrow 3S+S=\left(1-\dfrac23+\dfrac3{3^2}-\dfrac4{3^3}\ +\,.\!.\!.+\ \dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\right)+\left(\dfrac13-\dfrac2{3^2}+\dfrac3{3^3}-\dfrac4{3^4}\ +\,.\!.\!.+\ \dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\right)\\\Rightarrow 4S=1-\dfrac13+\dfrac1{3^2}-\dfrac1{3^3}\ +\,.\!.\!.+\ \dfrac1{3^{99}}-\dfrac{100}{3^{100}}\\\Rightarrow 12S=3.\!\;\!\!\left(1-\dfrac13+\dfrac1{3^2}-\dfrac1{3^3}\ +\,.\!.\!.+\ \dfrac1{3^{99}}-\dfrac{100}{3^{100}}\right)\\\Rightarrow 12S=3-1+\dfrac13-\dfrac1{3^2}\ +\,.\!.\!.+\ \dfrac1{3^{98}}-\dfrac{100}{3^{99}}\\\Rightarrow 12S+4S=\left(3-1+\dfrac13-\dfrac1{3^2}\ +\,.\!.\!.+\ \dfrac1{3^{98}}-\dfrac{100}{3^{99}}\right)+\left(1-\dfrac13+\dfrac1{3^2}-\dfrac1{3^3}\ +\,.\!.\!.+\ \dfrac1{3^{99}}-\dfrac{100}{3^{100}}\right)\\\Rightarrow 16S=3-\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\\\Rightarrow S=\dfrac{3-\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}}{16}\\\Rightarrow S=\dfrac3{16}-\dfrac{\dfrac{99}{3^{99}}+\dfrac{100}{3^{100}}}{16}<\dfrac3{15}=\dfrac15\\\Rightarrow S<\dfrac15$

Vậy $S<\dfrac15$.