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\item \textbf{Section 3.3}
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\begin{enumerate}
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\item[2)] Let $x_1 > 1$ and $x_{n+1} := 2-1/x_n$ for $n \in \N$. Show that $(x_n)$ is bounded and monotone. Find the limit.
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\\\\ The first five terms of this sequence are $x_1 >1,x_2 > \frac{3}{2}, x_3 >\frac{4}{5}, x_4 > \frac{5}{4}, x_5 > \frac{6}{5}$. This sequence appears to be decreasing.
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\\\\ The first five terms of this sequence are $x_1 \geq 2,x_2 \geq \frac{3}{2}, x_3 \geq \frac{4}{5}, x_4 \geq \frac{5}{4}, x_5 \geq \frac{6}{5}, \dots$. This sequence appears to be decreasing.
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\\\\Recall the Monotone Sequence Property:
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\begin{theorem*}{Monotone Sequence Property}
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A monotone sequence of real numbers is convergent if and only if it is bounded. Further,
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@@ -69,15 +69,147 @@
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x_1 &\geq x_{1+1} \\
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x_1 &\geq x_2
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\end{align*}
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Since $x_1>1 \Rightarrow \frac{1}{x_1} < 1$, we have
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\[x_2 = 2-\frac{1}{x_1} > 1\]
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\[\Rightarrow 1 < x_2 < 2.\]
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Since $x_1>1$ and because $1 < x_2 < 2$, we have that $x_1 \geq x_2$.
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\\\\\textbf{Inductive Step:} Assume $1 < x_n < 2,\ \forall\ n \in \N$.
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\\\\\textbf{Show:} Now we want to show that $x_n \leq x_{n+1}$.
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\\So,
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\[1 < x_n <2\]
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\[1 > \frac{1}{x_n} > \frac{1}{2}\]
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\[-1 < -\frac{1}{x_n} < -\frac{1}{2}\]
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\[1 < 2-\frac{1}{x_n} < 2-\frac{1}{2} < 2\]
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\[1 < x_{n+1} < 2\]
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Thus we have that $(x_n)$ is bounded between 1 and 2.
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\end{proof}
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Now we need to show that $(x_n)$ is monotone decreasing; that is, we must show that $x_1 \geq x_2 \geq \dots \geq x_n$.
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\begin{proof}
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We want to show that $x_1 \geq x_2 \geq \dots \geq x_n,\ \forall\ n \in \N$. We prove it by method of mathematical induction.
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\\\\\textbf{Basis Step:} Let $n=1$. Then since $x_1 >1$ is given, we have that $\frac{1}{x_1} < 1$. This yields $x_2=2-\frac{1}{x_1}>1$, as was determined for the boundedness proof, and thus we have that $1 < x_2 < 2$. This means that $1 > \frac{1}{x_2} > \frac{1}{2}$, and since $\frac{1}{2} \leq \frac{1}{x_n}$, we have $x_2 \geq x_1$.
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\\\\\textbf{Inductive Step:} Assume $x_n \geq x_{n+1}\ \forall\ n \in \N$.
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\\\\\textbf{Show:} We now want to show that $x_{n+2} \leq x_{n+1}$.
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\\So,
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\[x_{n+2}=2-\frac{1}{x_{x+1}}\]
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Recall the inductive hypothesis, in that $x_n \geq x_{n+1} \Rightarrow \frac{1}{x_n} \leq \frac{1}{x_{n+1}}$. Thus,
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\[-\frac{1}{x_n} \geq -\frac{1}{x_{n+1}}\]
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\[\Rightarrow 2-\frac{1}{x_n} \leq 2 -\frac{1}{x_{n+1}}\]
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\[x_{n+1} \leq x_{n+2}\]
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$\therefore$ we have that $x_1 \geq x_2 \geq \dots \geq x_n,\ \forall\ n \in \N$.
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\end{proof}
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Thus $(x_n)$ is monotone decreasing.
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\\\\By the \textit{Monotone Sequence Property}, since we have shown that $(x_n)$ is both bounded (and thus converges), and that $(x_n)$ is monotone decreasing, we have that
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\begin{align*}
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\lim (x_n) &= \inf \{x_n: n \in \N\} \\
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&=\inf (1,2) \\
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&= 1
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\end{align*}
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Hence the sequence converges to the previously found possible limit of 1. \\
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\item[3)] Let $x_1 > 1$ and $x_{n+1} := 1 + \sqrt{x_n - 1}$ for $n \in \N$. Show that $(x_n)$ is decreasing and bounded below by $2$. Find the limit.
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\\\\The first 5 terms of this sequence are $x_1 \geq 2, x_2 \geq 2, x_3 \geq 2, x_4 \geq 2, x_5 \geq 2, \dots$. Notice the following, however:
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\begin{align*}
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x_{n+1} \leq x_n &\iff 1+\sqrt{x_n - 1} \leq x_n \\
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&\iff \sqrt{x_n -1} \leq x_n -1
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\end{align*}
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which we know is always true since the square root function is a decreasing function.
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\\\\Now we must find the possible limit points (fixed points) of this sequence. So,
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\begin{align*}
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x &= 1 + \sqrt{x-1} \\
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x-1 &= \sqrt{x-1} \\
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x-1 &= (x-1)^2 \\
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x-1 &= x^2 -2x +1 \\
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(x-1)-(x^2-2x+1) &= 0 \\
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-x^2+3x-2 &=0 \\
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-(x^2-3x+2) &= 0 \\
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-(x-1)(x-2) &= 0 \\
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(x-1)(x-2) &= 0
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\end{align*}
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Thus $x=1$, or $x=2$. These are the possible limits of $(x_n)$. Since we hypothesized that $(x_n)$ is decreasing, then we say that $(x_n)$ is bounded below by 2, since we are given that $x_1 > 1$.
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\\\\Now we will prove that $(x_n)$ is bounded below by 2.\\
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\begin{proof}
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We want to show that $(x_n)$ is bounded below by 1; that is, we want to show that $1 \leq x_n,\ \forall\ n \in \N$. We prove it by method of mathematical induction.
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\\\\\textbf{Basis Step:} Let $n=1$. Then we are given that $x_1 \geq 2$.
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\\\\\textbf{Inductive Step:} Assume that $x_n \geq 2,\ \forall\ n \in \N$.
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\\\\\textbf{Show:} We now want to show that $x_{n+1} \geq 2,\ \forall\ n \in \N$.
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\\\\So,
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\begin{align*}
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x_{n+1} &= 1+\sqrt{x_n -1} \\
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&\geq 1+\sqrt{2 -1} \\
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&=1 + 1 \\
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&= 2
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\end{align*}
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Thus, $x_n \geq 2,\ \forall\ n \in \N$. By the definition of boundedness, we have that $(x_n)$ is bounded below by 2.
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\end{proof}
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Since we have also shown earlier that $(x_n)$ is monotone decreasing, we have that by the monotone sequence property, since $(x_n)$ is bounded, $(x_n)$ converges, and since $(x_n)$ is monotone decreasing, we have:
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\begin{align*}
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\lim (x_n) &= \inf \{x_n:n \in \N\} \\
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&=2
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\end{align*}
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\item[7)] Let $x_1 := a>0$ and $x_{n+1} := x_n+1/x_n$ for $n \in \N$. Determine whether $(x_n)$ converges or diverges.
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\\\\The first 5 terms of this sequence are $x_1 \geq 1, x_2 \geq 2, x_3 \geq \frac{5}{2}, x_4 \geq \frac{29}{10}, x_5 \geq \frac{941}{290}, \dots$. This sequence appears to be increasing. We show this to be true as follows:
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\begin{align*}
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x_{n+1} \geq x_n &\iff x_n + \frac{1}{x_n} \geq x_n \\
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&\iff x_n^2 + 1 \geq x_n^2 \\
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&\iff 1 \geq 0
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\end{align*}
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which is true.
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However, notice that one of the terms of the sequence is $x_n$. We know that $x_n$ is an unbounded sequence. Thus, we can infer that $(x_n)$ is unbounded above. We show this as follows:
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\begin{align*}
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x_{n+1}^2 &= \left(x_n + \frac{1}{x_n} \right)^2 \\
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&= x_n^2+2+\frac{1}{x_n^2} \\
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&> x_n^2 +2
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\end{align*}
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Since:
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\[x_{n+1}^2 > x_n^2+2 > x_{n-1}^2 +4 > \dots > x_1^2+2 \cdot n = a^2+2 \cdot n\]
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\[\Downarrow\]
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\[x_n > \sqrt{a^2 + 2 \cdot (n-1)}\]
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Since the right hand side of this inequality is unbounded, the left hand side is also unbounded.
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\\\\Thus we have that this sequence $(x_n)$ is unbounded above.
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\\\\Since this sequence is increasing and unbounded above, we have that the sequence is divergent.\\
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\item[8)] Let $(a_n)$ be an increasing sequence, $(b_n)$ be a decreasing sequence, and assume that $a_n \leq b_n$ for all $n \in \N$. Show that $\lim (a_n) \leq \lim (b_n)$, and thereby deduce the Nested Intervals Property 2.5.2 from the Monotone Convergence Theorem 3.3.2.
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\\\\Since $(a_n)$ is an increasing sequence, we know that $(a_1 \leq a_2 \leq \dots \leq a_n)$, and since $(b_n)$ is a decreasing sequence, we know that $(b_1 \geq b_2 \geq \dots \geq b_n)$. Also, since we have that $a_n \leq b_n,\ \forall\ n \in \N$, we know that $(a_n)$ is bounded above by $(b_1)$. Thus, by the \textit{Monotone Convergence Theorem}, we know that
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\[\lim (a_n) = \sup \{a_n: n \in \N\}\]
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Also, since $(b_n)$ is a decreasing sequence such that it is bounded below by $(a_1)$, by the \textit{Monotone Convergence Theorem}, we have
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\[\lim (b_n) = \inf \{b_n: n \in \N\}\]
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Recall Theorem 3.2.5:
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\begin{theorem*}
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If $X=(x_n)$ and $Y=(y_n)$ are convergent sequences of real numbers and if $x_n \leq y_n\ \forall\ n \in \N$, then $\lim (x_n) \leq \lim (y_n)$.
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\end{theorem*}
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Also, recall the \textit{Nested Intervals Property}:
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\begin{theorem*}
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If $I_n=[a_n,b_n],\ n \in \N$, is a nested sequence of closed bounded intervals, then there exists a number $\xi \in \R \st \xi \in I_n\ \forall\ n \in \N$.
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\end{theorem*}
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Note that we have a nested sequence of closed, bounded intervals: $[a_n, b_n],\ n \in \N$. Since we showed that $\lim (a_n) \leq \lim (b_n)$, (and we are given that $(a_n)$ is increasing and $(b_n)$ is decreasing), we know that there exists $\xi$ such that
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\[\lim (a_n) \leq \xi \leq \lim (b_n)\]
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which means that $\xi \in [a_n, b_n],\ \forall\ n \in \N$.
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\end{enumerate}
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\item $a_1 = 1,\ a_{n+1}=\frac{a_n^2+5}{2a_n}$
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\\\\The first 5 terms of this sequence are $1, 3, \frac{7}{3}, \frac{47}{21}, \frac{2207}{987}, \dots$. This is an increasing sequence.
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\\\\First, we must find the possible limits (fixed points) of the sequence. So,
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\begin{align*}
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a&=\frac{a^2+5}{2a} \\
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2a^2 &= a^2+5 \\
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a^2 &= 5 \\
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a &= \pm \sqrt{5}
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\end{align*}
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Since we're given that $a_1=1$, we know that the most likely lower bound will be $\sqrt{5}$.
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\\\\Now we want to show that $(a_n)$ is bounded below by $-\sqrt{5}$.
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\begin{proof}
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We want to show that $a_n \leq \sqrt{5},\ \forall\ n \in \N$. We prove it by method of mathematical induction.
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\\\\\textbf{Basis Step:} Since $1 \geq -\sqrt{5}$, we have that $a_1 \geq -\sqrt{5}$
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\\\\\textbf{Inductive Step:} Assume that $a_n \geq -\sqrt{5}\ \forall\ n \in \N$.
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\\\\\textbf{Show:} We want to show that $a_{n+1} \geq -\sqrt{5}\ \forall\ n \in \N$. So,
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\begin{align*}
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a_{n+1} &= \frac{a_n^2 + 5}{2a_n}
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\end{align*}
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\end{proof}
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\item $a_1 = 5,\ a_{n+1}=\sqrt{4+a_n}$
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\end{enumerate}
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