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\noindent
{\bf Math 3220}
\hfill {\Large \bf $\{(-1)^n\}_{n=1}^\infty$  and $\{ \sin(n) \}_{n=1}^\infty$ Diverge} \hfill 
{\bf Intro. to Real Analysis}

\vspace{0.2in}

\DEF The sequence $\{ a_n \}_{n=1}^\infty$ {\it converges} to the real number $L$ iff for every $\epsilon>0$ there exists some $N>0$ such that for all $n > N$ we have that $|a_n-L| < \epsilon$. We write $\displaystyle\lim_{n \to \infty} a_n = L$ or $a_n \to L$ and call $L$ the {\it limit}\footnote{\fbox{\bf Theorem:}  If a sequence converges, its limit is unique.Thus we are justified saying ``the'' limit not just ``a'' limit.} of the sequence. If a sequence does not converge to any real number, we say it {\it diverges}.

\vspace{0.1in}

Considering the above definition, if we are to prove that a sequence diverges, we must show the $L$ is not a limit no matter what $L$ is. Recall that negating a logical statement flips quantifiers from $\forall$ to $\exists$ and vice-versa. Thus we must show that for every $L \in \mathbb{R}$ there is some $\epsilon>0$ such that for every $N>0$ there is some $n > N$ such that $|a_n-L| \geq \epsilon$. 

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\EX $\{ (-1)^n \}_{n=1}^\infty$ diverges.

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\bproof
We will use proof by contradiction. Let $L \in \mathbb{R}$ and suppose $(-1)^n \rightarrow L$. 

Before getting too much further, we will need the ``ceiling'' function: $\lceil N \rceil$. This is the closest integer $k = \lceil N \rceil$ such that $k \geq N$.  For example, $\lceil 3.2 \rceil = \lceil \pi \rceil =  4$ and $\lceil 4.999 \rceil = \lceil 5 \rceil = 5$. 

Consider $\epsilon = 1/2 > 0$. Since $(-1)^n \to L$ there exists some $N > 0$ such that $|(-1)^n-L|<\epsilon=1/2$ for every $n>N$. 
Notice that both $2\lceil N \rceil$ and $2 \lceil N \rceil +1$ are greater than $N$. Thus $|(-1)^{2\lceil N \rceil}-L| = |1-L|<1/2$. This implies that $1/2=1-1/2 < L < 1+1/2 = 3/2$. 
Likewise, $|(-1)^{2 \lceil N \rceil +1}-L| = |-1-L|<1/2$ so that $-3/2 = -1-1/2 < L < -1+1/2 = -1/2$. 
This implies that $L$ is both positive and negative (contradiction). Therefore, no such limit exists. We have that $\{ (-1)^n \}_{n=1}^\infty$ diverges.
\eproof

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This argument can be modified to show that $\{ \sin(n) \}_{n=1}^\infty$ diverges.
The idea is that we if we go down the sequence far enough, we can hit values above $1/2$ and below $-1/2$.
So the same argument (with a little finesse) will work.

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\EX $\{ \sin(n) \}_{n=1}^\infty$ diverges.

\vspace{0.05in}

\bproof
We will use proof by contradiction. Let $L \in \mathbb{R}$ and suppose $\sin(n) \rightarrow L$. 

Let $\epsilon = 0.25 > 0$. There exists some $N > 0$ such that 
$$|\sin(n) - L| < 0.25 \qquad \mbox{for all} \qquad n > N.$$ Now let's pick out values for $n$ such
that $n \geq N$ and $n$ is as close to $\pi/2+2\pi k$ and $3\pi/2 + 2\pi k$ ($k \in \mathbb{Z}$) as possible (this is
where $\sin$ takes on values $1$ and $-1$ respectively). Consider $$N_1 = \lceil \pi/2+2\pi \lceil N \rceil \rceil \geq \pi/2 + 2\pi N > N \qquad
\mbox{and} \qquad N_2 = \lceil 3\pi/2+2\pi \lceil N \rceil \rceil \geq 3\pi/2 + 2\pi N > N$$
again where $\lceil x \rceil$ is the ceiling function. Notice that for any $x \in \mathbb{R}$,  $\lceil x \rceil = x + \ell$
for some $0 \leq \ell < 1$. In particular, let $N_1 = \pi/2 + 2\pi \lceil N \rceil + \ell_1$ and $N_2 = 3\pi/2 + 2\pi \lceil N \rceil + \ell_2$ where $0 \leq \ell_1,\ell_2 < 1$.

Now $\sin(n)$ decreases on the interval $[\pi/2+2\pi \lceil N \rceil, 3\pi/2+2\pi \lceil N \rceil ]$ and increases on the interval $[3\pi/2+2\pi \lceil N \rceil, 5\pi/2+2\pi \lceil N \rceil ]$.
Thus $$\sin(N_1) = \sin(\pi/2+2\pi \lceil N \rceil + \ell_1) > \sin(\pi/2+ 2\pi \lceil N \rceil + 1) = \sin(\pi/2+1) \approx 0.54 > 0.5$$ 
and $$\sin(N_2) = \sin(3\pi/2+2\pi \lceil N \rceil + \ell_2) < \sin(3\pi/2+ 2\pi \lceil N \rceil + 1) = \sin(3\pi/2+1) \approx -0.54 < -0.5$$ 

Finally, recalling $N_1,N_2 > N$ and that $|\sin(n) - L| < 0.25$ for all $n > N$, we have that
$$ |\sin(N_1)-L|<0.25 \qquad \mbox{and} \qquad |\sin(N_2)-L|< 0.25$$
Suppose that $L \geq 0$. This means that $|\sin(N_2)-L| = -(\sin(N_2)-L) = L-\sin(N_2) > L + 0.5 \geq 0.5$ since $\sin(N_2)<-0.5$.
But this is impossible since $|\sin(N_2)-L|<0.25$.  Therefore, it cannot be the case that $L \geq 0$. Thus 
we must have that $L< 0$. This means that $-L>0$ and so $|\sin(N_1)-L|=\sin(N_1)-L>0.5-L>0.5$ since $\sin(N_1)>0.5$. But this
cannot be since $|\sin(N_1)-L|<0.25$. Therefore, $L<0$ is impossible as well. 
Thus $L$ must not exist. In other words, the sequence diverges.
\eproof

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