ROUGH DRAFT authorea.com/81707
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  • Convergent sequences of real numbers

    Define the sequence of rational numbers p(n) recursively as follows:

    \[p(1)=1\]

    \[p(n)=1-\frac{p(n-1)}{2}\]

    • Compute and plot p(n) for \(n=1,2,\ldots , 20\). What inference do you draw about the terms p(n)?

    Replace this text with your caption

    Define the sequence of rational numbers a(n) recursively as follows:

    \[a(1)=1\]

    \[a(n)=\frac{1+1}{(1+a(n-1))}\]

    • Compute a(n) and \(a(n)^2-2\) for \(n=1,2,\ldots , 20\). What inference do you draw about the terms \(a(n)^2-2\)?

      The graph is different with points that are increasing and decreasing. \(a(n)=\frac{1+1}{(1+a(n-1))} \) is a horizontal line while \(a(n)^2-2\) is a more complex solution with a different inf(a(n)) and sup(a(n)).

    • From the recursive definition of p(n) find an equation for, and so evaluate, a possible limit p for p(n) as n increases without bound. \(p(1)=1\)

      \(p(n)=1-\frac{p(n-1)}{2}\)

      \[\lim_{n\to\infty} p(n) = p\]

      exists then p satisfies \[p = 1 - \frac{p}{2}\] \[p+\frac{p}{2} = 1\] \[\frac{3p}{2} = 1\] \[\frac{2}{3}*\frac{3p}{2} = 1*\frac{2}{3}\] \[p = \frac{2}{3}\]

    • Show, from the recursive definition of p(n) that \(\vert p(n) - p\vert = \frac{1}{2}\vert p(n-1) - p\vert\) for all \(n \geq 1\). \[p(n)=1-\frac{p(n-1)}{2}\]

      \[p(n)-\frac{2}{3} = 1 - \frac{p(n-1)}{2} - \frac{2}{3}\]

      \[\frac{1}{3} - \frac{p(n-1)}{2} = -\frac{1}{2}(p(n-1)-\frac{2}{3})\]

      Hence \[\vert p(n) - p \vert = \frac{1}{2} \vert p(n-1) - p \vert\]

    • So show that \(\vert p(n) - p\vert \leq (\frac{1}{2})^{n-1}\) for all \(n \geq\) 1.

      We have \[p(1) - p = \frac{1}{3}\] so its true for n = 1

      Assume that, for some \(n \geq 1\) \(\vert p(n) -p \vert \leq (\frac{1}{2})^{n-1}\) is true

      Then \[\vert p(n+1) - p \vert = \frac{1}{2}\vert p(n-1) - p \vert \leq \frac{1}{2}^{n}\]

      Then we have \[\vert p(n) - p \vert \leq \frac{1}{2}^{n-1}\] for all \(n \geq 1 \).

    • Prove that \(2^{n-1} \geq n\) for all \(n\geq 1\).

      This is true fr n = 1.
      Assume it’s true for some n.

      Then \(2^n = 2 * 2^{n-1} \geq 2n \geq n+1\) since \(n \geq 1.\)
      Therefore, by induction, \(2^{n-1} \geq n\) for all \(n \geq 1\)

    • Prove that the sequence p(n) converges to p.

      Notice that for all \(n \geq 1\), \[\vert p(n) - p \vert \leq \frac{1}{3}(\frac{1}{2})^{n-1} \leq \frac{1}{3n}\]

      so given \(\varepsilon > 0\) if we choose \(N > \frac{1}{3\varepsilon}\) then \(\vert p(n) -p\vert \leq\) all \(n \geq N\)

      which means that

      \[\lim_{n\to\infty} p(n) = p= \frac{2}{3}\]