[tex]a)\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}<\frac{1}{3}[/tex]
[tex]b)\frac{1}{3}-\frac{2}{3^{2}}+\frac{3}{3^{3}}-\frac{4}{3^{4}}+...+ \frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}[/tex]
Đặt $A=\dfrac{1}{3}-\dfrac{2}{3^{2}}+\dfrac{3}{3^{3}}-\dfrac{4}{3^{4}}+...+ \dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}$
$A=\dfrac{1}{3}-\dfrac{2}{3^{2}}+\dfrac{3}{3^{3}}-\dfrac{4}{3^{4}}+...+ \dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}$
$\Rightarrow 3A=1-\dfrac{2}{3}+\dfrac{3}{3^{2}}-\dfrac{4}{3^{3}}+...+ \dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}$
$\Rightarrow 4A=A+3A=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}$
$\Rightarrow 12A=3.4A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}$
$\Rightarrow 16A=12A+4A=3-\dfrac{1}{3^{99}}-\dfrac{100}{3^{99}}-\dfrac{100}{3^{100}}$
$\Leftrightarrow 16A=3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}$
$\Leftrightarrow A=\dfrac{3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}}{16}=\dfrac{3}{16}-\dfrac{\dfrac{101}{3^{99}}+\dfrac{100}{3^{100}}}{16}< \dfrac{3}{16}$
$\Rightarrow A<\dfrac{3}{16}$
P/s: mk cũng không bik có đúng ko.Nếu sai chỗ nào mong bạn thông cảm nha