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Math Help - Convergence Proof

  1. #1
    Junior Member
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    Convergence Proof

    I need to use the Archimedean Principle to prove that

    2^{-n} and \frac{1}{\sqrt{n}}

    both converge to zero.

    Using this principle it tells me I can choose N \epsilon N such that N > something \varepsilon . This is where I get stuck because I dont know what to choose for something \varepsilon in either of the cases. Maybe the fraction I could use \frac{1}{\varepsilon} . Any suggestions??
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  2. #2
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    Here are strong hints for all three.

    The Archimedean Principle says: \varepsilon  > 0 \Rightarrow \left( {\exists N \in \mathbb{N}} \right)\left[ {\frac{1}{N} < \varepsilon } \right].
    So that means n > N \Rightarrow \,\frac{1}{n} \leqslant \frac{1}{N} < \varepsilon .

    \begin{gathered}  \varepsilon  > 0 \Rightarrow \varepsilon ^2  > 0 \Rightarrow \left( {\exists N \in \mathbb{N}} \right)\left[ {\frac{1}{N} < \varepsilon ^2  \Rightarrow \frac{1}{{\sqrt N }} < \varepsilon } \right] \hfill \\  n > N \Rightarrow \,\frac{1}{{\sqrt n }} \leqslant \frac{1}{{\sqrt N }} < \varepsilon  \hfill \\ <br />
\end{gathered}

    2^n  \geqslant n \Rightarrow \frac{1}{{2^n }} < \frac{1}{n} < \varepsilon .
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