Cho tổng gồm 2014 số hạng A = 1/4 + 2/4^2 + 3/4^3 + 2014/4^2014. Chứng minh
Cho tổng gồm 2014 số hạng \[S = \frac{1}{4} + \frac{2}{{{4^2}}} + \frac{3}{{{4^3}}} + ... + \frac{{2014}}{{{4^{2014}}}}\]. Chứng minh \[S < \frac{1}{2}\].
\[4S = 1 + \frac{2}{4} + \frac{3}{{{4^2}}} + ... + \frac{{2014}}{{{4^{2013}}}}\]
\[4S - S = \left( {1 + \frac{2}{4} + \frac{3}{{{4^2}}} + ... + \frac{{2014}}{{{4^{2013}}}}} \right) - \left( {\frac{1}{4} + \frac{2}{{{4^2}}} + \frac{3}{{{4^3}}} + ... + \frac{{2014}}{{{4^{2014}}}}} \right)\]
\[3S = 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{4^{2013}}}} - \frac{{2014}}{{{4^{2014}}}}\]
Đặt \[A = 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{4^{2013}}}}\]
\[ \Rightarrow 4A = 4 + 1 + \frac{1}{4} + ... + \frac{1}{{{4^{2012}}}}\]
\[4A - A = 4 - \frac{1}{{{4^{2013}}}} \Rightarrow A = \frac{4}{3} - \frac{1}{{{{3.4}^{2013}}}}\]
\[ \Rightarrow 3S = \frac{4}{3} - \frac{1}{{{{3.4}^{2013}}}} - \frac{{2014}}{{{4^{2014}}}}\]\[ \Rightarrow \]\[S = \frac{4}{9} - \frac{1}{{{{9.4}^{2013}}}} - \frac{{2014}}{{{{3.4}^{2014}}}}\]
Vì \[\frac{4}{9} < \frac{1}{2}\] nên \[\frac{4}{9} - \frac{1}{{9 \times {4^{2013}}}} - \frac{{2014}}{{3 \times {4^{2014}}}} < \frac{1}{2}\]\[ \Rightarrow \]S < \[\frac{1}{2}\].
Vậy \[S < \frac{1}{2}\].