Chứng minh P = 1/3^1 + 2/3^2 + 3/3^3 + + 100/3^100 < 3/4

Chứng minh $P = \frac{1}{{{3^1}}} + \frac{2}{{{3^2}}} + \frac{3}{{{3^3}}} + ... + \frac{{100}}{{{3^{100}}}} < \frac{3}{4}$.

$P = \frac{1}{{{3^1}}} + \frac{2}{{{3^2}}} + \frac{3}{{{3^3}}} + ... + \frac{{100}}{{{3^{100}}}}$

⇒ $\frac{1}{3}P = \frac{1}{{{3^2}}} + \frac{2}{{{3^3}}} + \frac{3}{{{3^4}}} + ... + \frac{{100}}{{{3^{101}}}}$

$P - \frac{1}{3}P = \frac{1}{{{3^1}}} + \left( {\frac{2}{{{3^2}}} - \frac{1}{{{3^2}}}} \right) + \left( {\frac{3}{{{3^3}}} - \frac{2}{{{3^3}}}} \right) + + ... + \left( {\frac{{100}}{{{3^{100}}}} - \frac{{99}}{{{3^{100}}}}} \right) - \frac{{100}}{{{3^{101}}}}$

$\frac{2}{3}P = \frac{1}{{{3^1}}} + \frac{1}{{{3^2}}} + \frac{1}{{{3^3}}} + ... + \frac{1}{{{3^{100}}}} - \frac{{100}}{{{3^{101}}}}$ (*)

Đặt $S = \frac{1}{{{3^1}}} + \frac{1}{{{3^2}}} + \frac{1}{{{3^3}}} + ... + \frac{1}{{{3^{100}}}}$

(*) trở thành: $\frac{2}{3}P = S - \frac{{100}}{{{3^{101}}}}$(1)

Xét $S = \frac{1}{{{3^1}}} + \frac{1}{{{3^2}}} + \frac{1}{{{3^3}}} + ... + \frac{1}{{{3^{100}}}}$

⇒$\frac{1}{3}S = \frac{1}{{{3^2}}} + \frac{1}{{{3^3}}} + \frac{1}{{{3^4}}} + ... + \frac{1}{{{3^{101}}}}$

⇒ $S - \frac{1}{3}S = \frac{1}{{{3^1}}} - \frac{1}{{{3^{101}}}}$

⇔ $S - \frac{1}{3}S = \frac{1}{{{3^1}}} - \frac{1}{{{3^{101}}}}$

⇔ $\frac{2}{3}S = \frac{1}{{{3^1}}} - \frac{1}{{{3^{101}}}}$

$S = \frac{3}{2}.\left( {\frac{1}{{{3^1}}} - \frac{1}{{{3^{101}}}}} \right)$(2)

Thay (2) vào (1) ta có:

$\frac{2}{3}P = \frac{3}{2}.\left( {\frac{1}{{{3^1}}} - \frac{1}{{{3^{101}}}}} \right) - \frac{{100}}{{{3^{101}}}}$

$P = \frac{9}{4}.\left( {\frac{1}{{{3^1}}} - \frac{1}{{{3^{101}}}}} \right) - \frac{{100}}{{{3^{101}}}} = \frac{3}{4} - \frac{9}{{{{4.3}^{101}}}} - \frac{{100}}{{{3^{101}}}} < \frac{3}{4}$

Vậy P < $\frac{3}{4}$.