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So sánh 1/2+1/2^2+1/2^3+....+1/2^2014 và 1

$\frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + \frac{1}{{{2^4}}} + ... + \frac{1}{{{2^{2014}}}}$ $\frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + \frac{1}{{{2^4}}} + ... + \frac{1}{{{2^{2014}}}}$ và 1

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$\begin{matrix} {S_{2014}} = \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \dfrac{1}{{{2^4}}} + ... + \dfrac{1}{{{2^{2014}}}} \hfill \\ 2{S_{2014}} = 1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \dfrac{1}{{{2^4}}} + .... + \dfrac{1}{{{2^{2013}}}} \hfill \\ \Rightarrow 2{S_{2014}} - {S_{2014}} = \left( {1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \dfrac{1}{{{2^4}}} + .... + \dfrac{1}{{{2^{2013}}}}} \right) - \left( {\dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \dfrac{1}{{{2^4}}} + ... + \dfrac{1}{{{2^{2014}}}}} \right) \hfill \\ \Rightarrow {S_{2014}} = \dfrac{1}{2} - \dfrac{1}{{{2^{2014}}}} < \dfrac{1}{2} < 1 \hfill \\ \Rightarrow \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \dfrac{1}{{{2^4}}} + ... + \dfrac{1}{{{2^{2014}}}} < 1 \hfill \\ \end{matrix}$ $\begin{matrix} {S_{2014}} = \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \dfrac{1}{{{2^4}}} + ... + \dfrac{1}{{{2^{2014}}}} \hfill \\ 2{S_{2014}} = 1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \dfrac{1}{{{2^4}}} + .... + \dfrac{1}{{{2^{2013}}}} \hfill \\ \Rightarrow 2{S_{2014}} - {S_{2014}} = \left( {1 + \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \dfrac{1}{{{2^4}}} + .... + \dfrac{1}{{{2^{2013}}}}} \right) - \left( {\dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \dfrac{1}{{{2^4}}} + ... + \dfrac{1}{{{2^{2014}}}}} \right) \hfill \\ \Rightarrow {S_{2014}} = \dfrac{1}{2} - \dfrac{1}{{{2^{2014}}}} < \dfrac{1}{2} < 1 \hfill \\ \Rightarrow \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \dfrac{1}{{{2^4}}} + ... + \dfrac{1}{{{2^{2014}}}} < 1 \hfill \\ \end{matrix}$
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3 Trả lời 02/07/21