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\title{Math 220.201 Class 23 Questions}
\begin{document}
\maketitle
\begin{definition}
Let $\{a_n\}=a_1, a_2, a_3, \ldots$ be a sequence of real numbers, and let $L$ be another real number. $\{a_n\}$ is said to {\bf converge to $L$} if the following holds:
\begin{center}
For every real number $\epsilon>0$, there exists some $N\in \mathbb{N}$, such that for every natural number $n>N$, $|a_n-L|<\epsilon$.
\end{center}
\end{definition}
\begin{enumerate}
\item [1.] Prove that the sequence $\{\frac{n}{2n+1}\}$ converges.
\vskip 0.2in
{\bf Scrapwork:} Intuitively, as $n\to\infty$, $\frac{n}{2n+1}\to 1/2$. To prove this, we need to show that for any $\epsilon$, there exists some $N$ such that if $n>N$, the inequality $|\frac{n}{2n+1}-\frac{1}{2}|<\epsilon$ holds true. We have to figure out how big $n$ must be (in terms of $\epsilon$) to make this inequality happen. Some simple algebra shows that
$$|\frac{n}{2n+1}-\frac{1}{2}|=|\frac{2n-(2n+1)}{2(2n+1)}|=|\frac{-1}{4n+2}|=\frac{1}{4n+2}$$
so we must have $\frac{1}{4n+2}<\epsilon$. This happens iff $n>\frac{1}{4\epsilon}-\frac{1}{2}$. So if we take $N=\lceil 1/4\epsilon\rceil$, then this inequality will always hold true if $n>N$.
\begin{proof}
We'll show that this sequence converges to $1/2$. Let $\epsilon$ be any positive real number. Then we claim that for any $n>\lceil1/4\epsilon\rceil$,
$$|\frac{n}{2n+1}-\frac{1}{2}|<\epsilon$$
that is, we may take $N=\lceil1/4\epsilon\rceil$
This is because
$$|\frac{n}{2n+1}-\frac{1}{2}|=|\frac{-1}{4n+2}|=\frac{1}{4n+2}<\frac{1}{1/\epsilon}=\epsilon$$
This completes the proof.

\end{proof}
\vskip 0.2in
\item [2.] Prove that the sequence $\{n\}$ diverges.
\vskip 0.2in
\begin{proof}
Suppose, for a contradiction, that this sequence has some limit $L$. Then for any $\epsilon$, there is some $N$ such that for any $n>N$, $|n-L|<\epsilon$. Pick $\epsilon=1/3$, and let $N$ be the number satisfying the condition for this $\epsilon$. Then $|(N+1)-L|<1/3$ and $|(N+2)-L|<1/3$. By the Triangle Inequality,
$$|(N+2)-(N+1)|\le |(N+2)-L|+|L-(N+1)|$$
But the left side equals $1$, while the right side is less than $2/3$. This is a contradiction! Thus, no such $L$ exists.
\end{proof}
\vskip 0.2in
\item [3.] Does the sequence $\{\frac{2^n-1}{2^n}+\frac{(-1)^n}{n^2}\}$ converge?
\item [4.] Suppose that $\{a_n\}$ is a sequence of real numbers which converges to $L$, and suppose that $a_n>1$ for every $n$. Is it true that $\{1/a_n\}$ converges to $1/L$?
\vskip 0.2in
These are both answered by the proof of the following proposition.
\begin{proposition}
Let $a_n\to a$ and $b_n\to b$ be two convergent sequences. Then
\begin{itemize}
\item $a_n+b_n\to a+b$
\item $a_nb_n\to ab$
\item For any $c\in\mathbb{R}$, $ca_n\to ca$
\item $1/b_n\to 1/b$ as long as $b_n\neq 0, b\neq 0$
\end{itemize}
\end{proposition}
\begin{proof}
\begin{itemize}
\item For any $\epsilon$, $\epsilon/2$ is another positive real number, and so there exist natural numbers $M, N$ such that if $m>M$, $|a_m-a|<\epsilon/2$ and if $n>N$, $|b_n-b|<\epsilon/2$. Therefore, if $k> \mathrm{max}(M,N)$,
$$|a_k-a|+|b_k-b|<\epsilon$$
By the Triangle Inequality, $|(a_k+b_k)-(a+b)|<|a_k-a|+|b_k-b|$, and so $|(a_k+b_k)-(a+b)|<\epsilon$. Thus, $\mathrm{max}(M,N)$ satisfies the required condition for the sequence $a_k+b_k$.
\vskip 0.2in
\item Pick any $\epsilon$. First let's notice that
$$|a_nb_n-ab|=|a_nb_n-ab_n+ab_n-ab|\le |(a_n-a)b_n|+|a(b_n-b)|$$
Thus, if we want to have $|a_nb_n-ab|<\epsilon$, it suffices to have $|(a_n-a)b_n|<\epsilon/2$ and $|a(b_n-b)|<\epsilon/2$. We know that for $n$ larger than some $N_1$, $|b_n-b|<b/2$ (i.e. $b_n$ is between $b/2$ and $3b/2$) and for $n$ larger than some $N_2$, $|a_n-a|<\epsilon/4b$. Therefore, for $n$ larger than $\mathrm{max}(N_1, N_2)$, these both become true, which then implies that
$$|(a_n-a)b_n|<(\epsilon/4b)(3b/2)<\epsilon/2$$
Similarly, for $n$ larger than some $N_3$, $|b_n-b|<\epsilon/2a$ and thus
$$|a(b_n-b)|<\epsilon/2$$
So for $n>\mathrm{max}(N_1,N_2,N_3)$, both of the displayed inequalities are true, and thus $|a_nb_n-ab|<\epsilon$.
\vskip 0.2in
\item For any $\epsilon>0$, $\epsilon/c$ is another positive real number, and so there is some $N(\epsilon/c)$ such that if $n>N(\epsilon/c)$, then $|a_n-a|<\epsilon/c$. This means that $|ca_n-ca|<\epsilon$. Thus, $N(\epsilon/c)$ is the value of $N$ required for the sequence $ca_n$.
\vskip 0.2in
\item Pick any $\epsilon$. For $n$ larger than some $N_1$, $|b_n-b|<\frac{b^2}{2}\cdot\epsilon$. This means that
$$|bb_n\left(\frac{1}{b}-\frac{1}{b_n}\right)|<\frac{b^2}{2}\cdot\epsilon$$
For $n$ larger than some $N_2$, $|b_n-b|<b/2$, in which case
$$|\frac{b^2}{2}\left(\frac{1}{b}-\frac{1}{b_n}\right)|<|bb_n\left(\frac{1}{b}-\frac{1}{b_n}\right)|$$
Thus, for $n>\mathrm{max}(N_1,N_2)$, $|\frac{1}{b}-\frac{1}{b_n}|<\epsilon$, as desired.
\end{itemize}
\end{proof}
\vskip 0.2in
\item [5.] Prove that the sequence $\{a_n\}$ defined by $a_1=1$ and $a_{n+1}=a_n+\frac{1}{a_n}$ diverges. (Hint: Suppose that it converges, and use proof by contradiction.)
\vskip 0.2in
\begin{proof}
I'm going to save this for next class. Think about it!
\end{proof}
\vskip 0.2in
\item [6.] Prove that the sequence $\{a_n\}$ defined by $a_1=1$ and $a_{n+1}=\frac{a_n}{2}+\frac{1}{a_n}$ converges to $\sqrt{2}$. (Hint: Define a new sequence $\{b_n\}$ by $b_n=a_n-\sqrt{2}$. Now find a recursive formula for this new sequence and use that to show that it converges to 0.)
\vskip 0.2in
\begin{proof}
Define the sequence
$$b_n=a_n-\sqrt{2}$$
Then we can find a recursive formula
$$\begin{aligned}
b_{n+1}=a_{n+1}-\sqrt{2}&=\frac{a_n}{2}+\frac{1}{a_n}-\sqrt{2}\\
&=\frac{b_n+\sqrt{2}}{2}+\frac{1}{b_n+\sqrt{2}}-\sqrt{2}\\
&=\frac{(b_n+\sqrt{2})^2+2-2(b_n+\sqrt{2})\sqrt{2}}{2(b_2+\sqrt{2})}\\
&=\frac{b_n^2}{2(b_n+\sqrt{2})}
\end{aligned}$$
We'll first show by induction that $1-\sqrt{2}\le b_n\le \sqrt{2}-1$ for every $n\in\mathbb{N}$. Notice that $b_1=1-\sqrt{2}$ - this is the base case. And for any given $n$, if $1-\sqrt{2}\le b_n\le \sqrt{2}-1$, then
$$0\le \frac{b_n^2}{2(b_n+\sqrt{2})}<\frac{b_n^2}{2}<\frac{|b_n|}{2}$$
and therefore, $|b_{n+1}|<|b_n|/2$ ($\star$), which then implies that $1-\sqrt{2}\le b_{n+1}\le \sqrt{2}-1$. So $1-\sqrt{2}\le b_n\le \sqrt{2}-1$ for every $n\in\mathbb{N}$.

Note that we also showed $|b_{n+1}|<|b_n|/2$ for every $n\in\mathbb{N}$ (marked with a ($\star$)). From this, it is clear by induction that $|b_n|<1/2^n$, and so $b_n\to 0$. Since $b_n=a_n-\sqrt{2}$, it follows that $a_n\to\sqrt{2}$.
\end{proof}
\end{enumerate}
\end{document}