Question

[khanh nguyen dang]

$\frac{\mathrm{so\; sánh}}{}$

$\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}$+\(\dfrac{1}{3^{100}}\)  với $\frac{1}{2}$

Answer 1

[Nguyễn Thị Thương Hoài]

    A  =       \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+...+ \(\dfrac{1}{3^{99}}\) + \(\dfrac{1}{3^{100}}\)

   3A  = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+...+ \(\dfrac{1}{3^{99}}\)

3A - A = 1 - \(\dfrac{1}{3^{100}}\)

  2A     =  1 - \(\dfrac{1}{3^{100}}\)

     A     = \(\dfrac{1}{2}\) - \(\dfrac{1}{2.3^{100}}\) < \(\dfrac{1}{2}\)