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Lời giải:
\(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(A+3A=1+\frac{1-2}{3}+\frac{-2+3}{3^2}+\frac{3-4}{3^3}+\frac{-4+5}{3^4}+...+\frac{99-100}{3^{99}}-\frac{100}{3^{100}}\)
\(4A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-.....+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(4A=(1-\frac{1}{3})+(\frac{1}{3^2}-\frac{1}{3^3})+...+(\frac{1}{3^{98}}-\frac{1}{3^{99}})-\frac{100}{3^{100}}\)
\(4A=\frac{2}{3}+\frac{2}{3^3}+...+\frac{2}{3^{99}}-\frac{100}{3^{100}}\)
\(2A=\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{50}{3^{100}}\)
\(18A=3+\frac{1}{3}+...+\frac{1}{3^{97}}-\frac{450}{3^{100}}\)
\(\Rightarrow 18A-2A=3-\frac{1}{3^{99}}-\frac{450}{3^{100}}+\frac{50}{3^{100}}=3-\frac{1}{3^{99}}-\frac{400}{3^{100}}\)
\(\Leftrightarrow 16A=3-\frac{1}{3^{99}}-\frac{400}{3^{100}}<3\Rightarrow A< \frac{3}{16}\)
Đặt A=1/3-2/3^2+3/3^3-4/3^4+...+99/3^99-100/3^100
3A=1-2/3+3/3^2-4/3^3+...+99/3^98-100/3^99
3A+A=1-1/3+1/3^2-1/3^3+1/3^4-...+1/3^98-1/3^99-100/3^100
<1-1/3+1/3^2-1/3^3+1/3^4-...+1/3^98-1/3^99
Đặt S=1-1/3+1/3^2-1/3^3+1/3^4-...+1/3^98-1/3^99
3S=3-1+1/3-1/3^2+1/3^3-...-1/3^98
3S+S=3-1/3^99
S=(3-1/3^99) :4
S=3/4-1/4.3^99
\(\Rightarrow\)4A<3/4-1/4.3^99
\(\Rightarrow\)A<(3/4-1/4.3^99):4
\(\Rightarrow\)A<3/16-1/16.3^99<3/16
Vậy 1/3-2/3^2+3/3^3-4/3^4+...+99/3^99-100/3^100<3/16
Ta có:\(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}\)
\(=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+...+\left(1-\dfrac{1}{100}\right)\)
\(=\left(1-1\right)+\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+...+\left(1-\dfrac{1}{100}\right)\)\(=\left(1+1+...+1\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)\)
\(=100-\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)\)(đpcm)