Results 1 to 2 of 2

Thread: Convergence Proof

  1. #1
    Member
    Joined
    Sep 2008
    Posts
    86

    Convergence Proof

    I need to use the Archimedean Principle to prove that

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

    both converge to zero.

    Using this principle it tells me I can choose N $\displaystyle \epsilon$ N such that N > something $\displaystyle \varepsilon$ . This is where I get stuck because I dont know what to choose for something $\displaystyle \varepsilon$ in either of the cases. Maybe the fraction I could use $\displaystyle \frac{1}{\varepsilon}$ . Any suggestions??
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,782
    Thanks
    2824
    Awards
    1
    Here are strong hints for all three.

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

    $\displaystyle \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 \\
    \end{gathered} $

    $\displaystyle 2^n \geqslant n \Rightarrow \frac{1}{{2^n }} < \frac{1}{n} < \varepsilon $.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Convergence proof...
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Dec 20th 2011, 10:11 AM
  2. Convergence Proof
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Oct 12th 2009, 10:14 AM
  3. Proof of convergence
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Apr 13th 2008, 08:59 PM
  4. Convergence proof
    Posted in the Advanced Math Topics Forum
    Replies: 1
    Last Post: Oct 31st 2007, 01:17 PM
  5. Proof of Convergence
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Dec 4th 2006, 10:29 AM

Search Tags


/mathhelpforum @mathhelpforum