chứng tỏ s<1/36

Lời giải:

Ta có: \(S = \frac{1}{{{7^2}}} + \frac{2}{{{7^3}}} + \frac{3}{{{7^4}}} + ... + \frac{{69}}{{{7^{70}}}}\)

\(\begin{array}{l}7S = \frac{1}{7} + \frac{2}{{{7^2}}} + ... + \frac{{69}}{{{7^{69}}}}\\7S - S = \left( {\frac{1}{7} + \frac{2}{{{7^2}}} + ... + \frac{{69}}{{{7^{69}}}}} \right) - \left( {\frac{1}{{{7^2}}} + \frac{2}{{{7^3}}} + ... + \frac{{69}}{{{7^{70}}}}} \right)\\6S = \frac{1}{7} + \frac{1}{{{7^2}}} + \frac{1}{{{7^3}}} + ... + \frac{1}{{{7^{69}}}} - \frac{{69}}{{{7^{70}}}}\end{array}\)

Đặt \(N = \frac{1}{7} + \frac{1}{{{7^2}}} + \frac{1}{{{7^3}}} + ... + \frac{1}{{{7^{69}}}}\,\,\,\left( {6S < N\,\,} \right)\,\,(1)\)

\(\begin{array}{l}7N = 1 + \frac{1}{7} + \frac{1}{{{7^2}}} + ... + \frac{1}{{{7^{68}}}}\\7N - N = \left( {1 + \frac{1}{7} + \frac{1}{{{7^2}}} + ... + \frac{1}{{{7^{68}}}}} \right) - \left( {\frac{1}{7} + \frac{1}{{{7^2}}} + \frac{1}{{{7^3}}} + ... + \frac{1}{{{7^{69}}}}} \right)\\6N = 1 - \frac{1}{{{7^{69}}}} < 1\\ \Rightarrow N < \frac{1}{6}\,\,\,\,\,\,\,\,\,\,(2)\end{array}\)

Từ (1) và (2) ta được: \(6S < N < \frac{1}{6} \Rightarrow 6S < \frac{1}{6} \Rightarrow S < \frac{1}{{36}}\,\,(dpcm)\)