Cho tổng gồm 2014 số hạng: \[S = \frac{1}{4} + \frac{2}{{{4^2}}} + \frac{3}{{{4^3}}} + \frac{4}{{{4^4}}} + ... + \frac{{2014}}{{{4^{2014}}}}\]. Chứng minh rằng S < 1.

\[ \Rightarrow 4.S = 1 + \frac{2}{4} + \frac{3}{{{4^2}}} + \frac{4}{{{4^3}}} + ... + \frac{{2014}}{{{4^{2013}}}}\]

\[ \Rightarrow 4.S - S = \left( {1 + \frac{2}{4} + \frac{3}{{{4^2}}} + \frac{4}{{{4^3}}} + ... + \frac{{2014}}{{{4^{2013}}}}} \right) - \left( {\frac{1}{4} + \frac{2}{{{4^2}}} + \frac{3}{{{4^2}}} + ... + \frac{{2014}}{{{4^{2014}}}}} \right)\]

\[ \Rightarrow 3.S = 1 + \left( {\frac{2}{4} - \frac{1}{4}} \right) + \left( {\frac{3}{{{4^2}}} - \frac{2}{{{4^2}}}} \right) + \left( {\frac{4}{{{4^3}}} - \frac{3}{{{4^3}}}} \right) + ... + \left( {\frac{{2014}}{{{4^{2013}}}} - \frac{{2013}}{{{4^{2013}}}}} \right) - \frac{{2014}}{{{4^{2014}}}}\]

\[ \Rightarrow 3.S = 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{4^{2013}}}} - \frac{{2014}}{{{4^{2014}}}}\]

Tính \[A = 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{4^{2013}}}}\]

\[ \Rightarrow 4.A = 4 + 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{4^{2012}}}}\]

\[ \Rightarrow 4.A - A = 4 - \frac{1}{{{4^{2013}}}} \Rightarrow A = \frac{4}{3} - \frac{1}{{{{3.4}^{2013}}}}\]

\[ \Rightarrow 3.S = \frac{4}{3} - \frac{1}{{{{3.4}^{2013}}}} - \frac{{2014}}{{{4^{2014}}}} \Rightarrow S = \frac{4}{9} - \frac{1}{{{{9.4}^{2013}}}} - \frac{{2014}}{{{4^{2014}}}}\].

Vậy S < 1.