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\def\absolute{\centerline{{\color{blue} Absolute}\qquad Conditional\qquad Divergent}}
\def\conditional{\centerline{Absolute\qquad{\color{blue} Conditional}\qquad Divergent}}
\def\divergent{\centerline{Absolute\qquad Conditional\qquad{\color{blue} Divergent}}}

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depth 0pt{\parfillskip0pt\medskip}}

\let\bu\bullet

{\bigbf\centerline{Holden Rohrer}\bigskip\centerline{MATH 4317\qquad
HOMEWORK \#5}}

\question{Are the series absolutely convergent, conditionally
convergent, or divergent? No proof required. (Cauchy Criteria can help
on some of these.)}

\tf
$$\sum_n {\sin n\over n^2}$$
\endtf

\absolute

\tf
$$\sum_n {(-1)^n\over n\log n}$$
\endtf

\conditional

\tf
$${3\over 4}-{4\over 6}+{5\over 8}-{6\over 10}+{7\over 12}-\cdots$$
\endtf

\divergent

\tf
$$\sum_n {(-1)^n\over n(\log n)^2}$$
\endtf

\conditional

\question{Show that the series $$\sum_n {(-1)^{n+1}\over n} = 1-{1\over
2}+{1\over 3}-{1\over 4}+\cdots$$ is convergent. (This is the simplest
instance of the Alternating Series Test (Thm 2.7.7). But you can't just
cite that, since the Theorem is not proved in the text. You can follow
any of the proof strategies outlined in Exercise 2.7.1. Or come up with
your own, tailored to this particular example.}

\proof
We start by stating $$s_n = 1-{1\over 2}+{1\over 3}-{1\over 4}+\cdots\pm
{1\over n}.$$
Next, we calculate, for $n\geq 2k,$ $$s_n-s_{2k} = {1\over 2k+1} -
{1\over 2k+2} +\cdots \pm {1\over n} = {1\over (2k+1)(2k+2)} + \cdots
\pm {1\over n} > -{1\over n} > -{1\over 2k}.$$
By similar reasoning, $s_n-s_{2k+1}<{1\over 2k}.$

Therefore, $s_{2k}-{1\over 2k} < s_n < s_{2k+1}+{1\over 2k},$
and since $|s_{2k+1}+{1\over 2k}-s_{2k}+{1\over 2k}| \leq {3\over 2k},$
we can show for both that
$$|s_n-s_{2k}| \leq {3\over 2k}\qquad |s_n-s_{2k+1}| \leq
{3\over 2k}.$$
This means the sequence is Cauchy and thus convergent.

\endproof

\question{(2.7.9, Ratio Test) Given a series $\sum_n a_n,$ with $a_n
\neq 0,$ the Ratio Test states that if
$$\limsup_n \left|{a_n\over a_{n+1}}\right| = r < 1,$$
then the series $$\sum_n a_n$$ converges absolutely. Take these steps to
verify this statement.}

\item{a.}  Show that for $r<s<1,$ there is $N$ so that for all $n\geq
N,$ we have
$$\left| {a_n\over a_{n+1}} \right| < s.$$

\proof
The definition of $\limsup$ tells us that for some $N,$ that for all
$n\geq N,$
$$\left| {a_{n+1}\over a_n} \right| \leq r < s.$$
\endproof

\item{b.} Conclude that for $n > N,$ that $|a_n| < |a_N|s^{n-N}.$

\proof
We will show this by induction.

Note that $$\left|{a_{n+1}\over a_n}\right| = {|a_{n+1}|\over |a_n|}.$$
(This can be shown by a sort of casework: the signs of $a_n$ and
$a_{n+1}$ can be independently flipped without affecting the expression,
and the positive case is trivial)

Using the same value of $N$ as in the previous question, we show the
base case for $n = N+1,$
$${|a_{N+1}|\over |a_N|} < s \Longrightarrow |a_n| < s|a_N|
\Longrightarrow |a_n| < |a_N|s^{n-N}.$$
For the inductive step, we assume for $n > N,$ that $|a_n| <
s^{n-N}|a_N|,$ and then use ${|a_{n+1}|\over |a_n|} < s$ to show $|a_{n+1}|
< s|a_n| < s^{n-N+1}|a_N|,$ completing the inductive step.
\endproof

\item{c.} Conclude that $\sum_n a_n$ converges absolutely.

\proof
Taking $N$ again as in previous questions,
$$\sum_{n>N} |a_n| < \sum_{n>N} |a_N|s^{n-N} = {|a_N|\over 1-s},$$ and
since the sequence is positive, the partial series are monotone
increasing, giving convergence by the Monotone Convergence Theorem.

The geometric sum is from the book.
\endproof

\bye