Cho S = 1/3.6 + 1/6.9 + . . . + 1/( 3 x − 3 ) .3 x (với x ∈ N ∗ , x ≥ 2 ). Chứng minh rằng S < 1 9 .

Ta có \(\frac{3}{{\left( {3x - 3} \right) \cdot 3x}} = \frac{{3x - \left( {3x - 3} \right)}}{{\left( {3x - 3} \right) \cdot 3x}} = \frac{{3x}}{{\left( {3x - 3} \right) \cdot 3x}} - \frac{{3x - 3}}{{\left( {3x - 3} \right) \cdot 3x}} = \frac{1}{{3x - 3}} - \frac{1}{{3x}}\)

Có \[3S = \frac{3}{{3 \cdot 6}} + \frac{3}{{6 \cdot 9}} + ... + \frac{3}{{\left( {3x - 3} \right) \cdot 3x}}\]

\[ = \frac{1}{3} - \frac{1}{6} + \frac{1}{6} - \frac{1}{9} + ... + \frac{1}{{3x - 3}} - \frac{1}{{3x}}\]

\[ = \frac{1}{3} - \frac{1}{{3x}}\]

Vì \[x \in \mathbb{N}*,x \ge 2\] nên \[\frac{1}{{3x}} > 0\]

Suy ra \[\frac{1}{3} - \frac{1}{{3x}} < \frac{1}{3}\]

Hay \[3S < \frac{1}{3}\]

Vậy \(S < \frac{1}{9}\).