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\noindent {\bf Question}

\noindent Let $\{ a_n\}$ be a sequence converging to $a$.  Show
that the following hold:
\begin{enumerate}
\item {\bf square roots:} if $a >0$, then $\{\sqrt{a_n}\}$ converges
to $\sqrt{a}$;
\item $\{ |a_n| \}$ converges to $|a|$;
\item if $a =\infty$, then $\{\frac{1}{a_n}\}$ converges to $0$.
\item If $a\neq 0$, then $\{ (-1)^n a_n\}$ diverges;
\item If $a =0$, then $\{ (-1)^n a_n\}$ converges to $0$.
\end{enumerate}

\medskip

\noindent {\bf Answer}
\begin{enumerate}
\item since $a >0$, we can apply the definition of
$\lim_{n\rightarrow\infty} a_n =a$ with $\varepsilon =\frac{1}{2}
a$ to see that there exists $P$ so that $a_n >0$ for $n >P$ (since
the interval of radius $\frac{1}{2} a$ centered at $a$ contains
only positive numbers), and so for $n >P$, $\sqrt{a_n}$ makes
sense.

\medskip
\noindent We need to get our hands on $| \sqrt{a_n} -\sqrt{a}|$,
which we do with our usual trick for handling differences of
square roots:
\[ | \sqrt{a_n} -\sqrt{a}| =| \sqrt{a_n} -\sqrt{a}| \frac{| \sqrt{a_n}
+\sqrt{a}|}{| \sqrt{a_n} +\sqrt{a}|} =\frac{|a_n -a|}{\sqrt{a_n}
+\sqrt{a}}. \] (Here we're using that both $\sqrt{a_n} >0$ and
$\sqrt{a} >0$ to say that $| \sqrt{a_n} +\sqrt{a}| =\sqrt{a_n}
+\sqrt{a}$.)  Since $\sqrt{a_n} +\sqrt{a} > \sqrt{a}$ for $n >P$,
we have that
\[ | \sqrt{a_n} -\sqrt{a}| = \frac{|a_n -a|}{\sqrt{a_n} +\sqrt{a}} <
\frac{|a_n -a|}{\sqrt{a}} \] for $n >P$. Since $\{ a_n\}$
converges to $a$, for every $\varepsilon >0$, we can choose $M >P$
so that $|a_n -a| <\varepsilon \sqrt{a}$ for $n >M$. For this
choice of $M$, we have that
\[ | \sqrt{a_n} -\sqrt{a}| = \frac{|a_n -a|}{\sqrt{a_n} +\sqrt{a}} <
\frac{|a_n -a|}{\sqrt{a}} <\frac{\varepsilon \sqrt{a}}{\sqrt{a}}
=\varepsilon, \] and so $\{\sqrt{a_n}\}$ converges to $\sqrt{a}$.
\item this one, we break into three cases.  If $a >0$, then (applying
the definition of $\lim_{n\rightarrow\infty} a_n =a$ with
$\varepsilon =a$) there exists $M_0$ so that $a_n >0$ for $n
>M_0$.  In this case, we have $|a_n| =a_n$ for $n >M_0$ and $|a|
=a$, and so $|| a_n| -|a|| =|a_n -a|$.  Since there is $M_1$ so
that $|a_n -a| <\varepsilon$ for $n >M_1$, we have that $|| a_n|
-|a|| <\varepsilon$ for $n > M ={\rm max}(M_0, M_1)$, and so
$\lim_{n\rightarrow\infty} |a_n| =|a|$.

\medskip
\noindent If $a <0$, then (applying the definition of
$\lim_{n\rightarrow\infty} a_n =a$ with $\varepsilon = |a|$) there
exists $M_0$ so that $a_n <0$ for $n >M_0$.  In this case, we have
$|a_n| = -a_n$ for $n >M_0$ and $|a| = -a$, and so $|| a_n| -|a||
=|-a_n +a| =|a_n -a|$.  Since there is $M_1$ so that $|a_n -a|
<\varepsilon$ for $n >M_1$, we have that $|| a_n| -|a||
<\varepsilon$ for $n > M ={\rm max}(M_0, M_1)$, and so
$\lim_{n\rightarrow\infty} |a_n| =|a|$.

\medskip
\noindent If $a =0$, then the definition of
$\lim_{n\rightarrow\infty} a_n =a$ becomes: for every $\varepsilon
>0$, there exists $M$ so that $|a_n -0| =|a_n| <\varepsilon$ for
$n >M$.  Since $|\: |a_n|\: | =|a_n|$, we have that the definition
of $\lim_{n\rightarrow\infty} |a_n| =0$ is satisfied without any
further work.
\item since $\lim_{n\rightarrow\infty} a_n =\infty$, for each
$\varepsilon >0$, there exists $M$ so that $a_n >\varepsilon$ for
$n
>M$.  Inverting both sides, we see that $\frac{1}{a_n} <
\frac{1}{\varepsilon}$ for $n >M$.  So, given $\mu >0$, choose
$\varepsilon >0$ so that $\frac{1}{\varepsilon} <\mu$, which can
be done by taking $\varepsilon$ large enough.  Then, there exists
$M$ so that $\left| \frac{1}{a_n} -0\right| =\frac{1}{a_n}
<\frac{1}{\varepsilon} <\mu$ for $n >M$, as desired.
\item if $a\ne 0$, consider the definition of
$\lim_{n\rightarrow\infty} a_n =a$ with $\varepsilon =\frac{1}{2}
|a|$: there exists $M$ so that $|a_n -a| < \frac{1}{2} |a|$ for $n
>M$.  That is, $a_n$ lies in the interval centered at $a$ with radius
$\frac{1}{2} |a|$, and so $|a_n| > \frac{1}{2} |a|$.

\medskip
\noindent Now consider the sequence $\{ (-1)^n a_n \}$.  For $n
>M$ and $n$ even, $(-1)^n a_n =a_n$ lies in the interval centered
at $a$ with radius $\frac{1}{2} |a|$.  For $n >M$ and $n$ odd,
$(-1)^n a_n =-a_n$ lies in the interval centered at $-a$ with
radius $\frac{1}{2} |a|$. In particular, we have, regardless of
whether $n$ is odd or even, that $|(-1)^n a_n - (-1)^{n+1}
a_{n+1}| > |a|$ for $n >M$, since $(-1)^n a_n$ and $(-1)^{n+1}
a_{n+1}$ lie on opposite sides of $0$ and are both distance at
least $\frac{1}{2} |a|$ from the origin.  Hence, $\{ (-1)^n a_n\}$
violates the Cauchy criterion (see Theorem below), and so
diverges.
\item if $a =0$, the definition of $\lim_{n\rightarrow\infty} a_n =0$
becomes: for every $\varepsilon >0$, there exists $M$ so that
$|a_n -0| =|a_n| <\varepsilon$ for $n >M$.  However, note that
$|(-1)^n a_n -0| =|a_n|$ as well, and so the definition of
$\lim_{n\rightarrow\infty} (-1)^n a_n =0$ is satisfied without any
further work.

{\bf Tests for convergence and divergence of sequences.}  Let $\{
a_n\}$, $\{ b_n\}$, and $\{ c_n\}$ be sequences.
\begin{enumerate}
\item {\bf Comparison test:} If $a_n\le b_n$ for all $n$ and if
$a_n\rightarrow\infty$ as $n\rightarrow\infty$, then
$b_n\rightarrow\infty$ as $n\rightarrow\infty$;
\item {\bf Limit comparison test:} If $\lim_{n\rightarrow\infty}
\frac{a_n}{b_n} =L$ with $0 <L <\infty$, then $\{ a_n\}$ converges
if and only if $\{ b_n\}$ converges.
\item {\bf l'Hopital's rule:} {\bf l'Hopital's rule:} Suppose that $f$ and $g$ are
differentiable on the union $I =(a-\varepsilon, a)\cup (a,a
+\varepsilon)$ for some $\varepsilon >0$, and that $g'(x)$ is
non-zero on $I$.  Suppose also that
\[ \lim_{x\rightarrow a} f(x) = \lim_{x\rightarrow a} g(x) = 0. \]
Then,
\[ \lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \lim_{x\rightarrow a}
\frac{f'(x)}{g'(x)}, \] provided that the right hand limit either
exists or is $\pm \infty$.
\item {\bf Squeeze rule:} If $a_n\le b_n\le c_n$ for all $n$ and if
$\{ a_n\}$ and $\{ c_n\}$ both converge with
$\lim_{n\rightarrow\infty} a_n =\lim_{n\rightarrow\infty} c_n$,
then $\{ b_n\}$ converges with $\lim_{n\rightarrow\infty} b_n
=\lim_{n\rightarrow\infty} c_n$.
\item {\bf Cauchy criterion:} if $\{ a_n\}$ converges, then for every
$\varepsilon >0$, there exists $M$ so that $|a_p -a_q|
<\varepsilon$ for all $p$, $q >M$.
\end{enumerate}
\end{enumerate}


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