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\begin{document}

\hbox to \hsize{\textbf{Math 331\hfill Homework 9}}
\nointerlineskip
\vskip 2pt
\hrule height 1pt

\medskip

{\narrower\narrower \textit{
   This homework is due on Tuesday, November 17, by 11:59 PM.  No rewrites are possible
   on this homework.  Sample solutions for homeworks 8 and 9 will be published on the
   course web page on Wednesday, November 18.  There is a test on Friday, November 20.
   A study guide for the test will be published on the course web page by Monday,
   November 16.
}\par}

\bigskip

\def\ds{\displaystyle}

\def\seq#1{\{#1_n\}_{n=1}^\infty}
\def\slim#1{{\ds\lim_{n\to\infty} #1_n}}


\begin{problem}  % problem 1
\ \ {\bf (a)}\ \ Suppose that $\sum_{k=1}^\infty a_k$ is a convergent series, and $\sum_{k=1}^\infty b_k$ 
is a divergent series.  Show that $\sum_{k=1}^\infty(a_k+b_k)$ diverges.  [Hint:  Proof by contradiction 
will work.]

\smallskip
\ \ {\bf (b)}\ \ Suppose that $\sum_{k=1}^\infty a_k$ and $\sum_{k=1}^\infty b_k$
are both divergent.  Give a simple example to show that $\sum_{k=1}^\infty(a_k+b_k)$ does not 
necessarily diverge.
\end{problem}

\begin{answer}
%You can put your answer here for problem 1, 3 points
\end{answer}




\begin{problem}  % problem 2
A nonnegative sequence must either converge (absolutely) or diverge to $+\infty$.  Classify each of the
following nonnegative series as either convergent or divergent.  In all cases, explain your reasoning, being 
explicit about any convergence tests that you apply.

\bigskip
\dspparts{
   \sum_{k=1}^\infty \frac{3k^2}{7k^5 + 2k^2} &
   \sum_{k=1}^\infty \frac{k}{\sqrt{k^2+1}} &
   \sum_{n=0}^\infty \pi^{-n} &
   \sum_{n=1}^\infty \frac{2^n+3^n}{4^n} \cr\noalign{\bigskip\medskip}
   \sum_{m=1}^\infty \frac{1+\sin(m)}{5^m} &
   \sum_{n=1}^\infty \frac{n^6}{5^n} &
   \sum_{k=1}^\infty \frac{(k!)^2}{(2k)!} &
   \sum_{n=1}^\infty \frac{1}{n+\sqrt{n}}
}
\end{problem}

\begin{answer}
%You can put your answer here for problem 2, 16 points
\end{answer}




\begin{problem}  % problem 3
A series of positive and negative terms can either diverge, converge absolutely, or converge conditionally.
Classify each of the following series as one of divergent, absolutely convergent, or conditionally convergent.
In all cases, explain your reasoning, being explicit about any convergence tests that you apply.

\setcounter{partcounter}{0}
\bigskip
\dspparts{
    \sum_{n=0}^\infty\frac{(-1)^{n+1}}{3^n} &
    \sum_{k=1}^\infty\frac{(-1)^k}{\sqrt{k+1}} \cr\noalign{\bigskip\medskip}
    \sum_{n=2}^\infty\frac{(-1)^n\ln(n)}{n} &
    \sum_{k=1}^\infty\frac{(-1)^k}{k^k}   
}
\end{problem}

\begin{answer}
%You can put your answer here for problem 3, 8 points
\end{answer}




\begin{problem}  % problem 4
The series $\sum_{k=2}^\infty \frac{1}{k\ln(k)}$ diverges, but none of the tests\vadjust{\smallskip}
that we have covered can prove it.  Note that ${\ds\lim_{n\to\infty}}\big(\int_2^n\frac{1}{x\ln(x)}dx\big)
={\ds\lim_{n\to\infty}}\big(\ln(\ln(x))-\ln(\ln(2))\big) = +\infty$.  Also\vadjust{\smallskip} note that
$f(x) = \frac{1}{x\ln(x)}$ is decreasing.  [You do not have to prove these facts.]
Show that\vadjust{\smallskip} the partial sum,
$s_n=\sum_{k=2}^n\frac{1}{k\ln(k)}$, satisfies $$s_n\ge \int_2^{n+1}\frac{1}{x\ln(x)}\,dx$$
by considering the upper sum using the partition $\{2,3,4,\dots,n+1\}$ of the interval $[2,n+1]$,
and conclude that $\sum_{k=2}^\infty \frac{1}{k\ln(k)}$ diverges.  (Note that this example is a special
case of something called the ``integral test.'')
\end{problem}

\begin{answer}
%You can put your answer here for problem 4, 3 points

\end{answer}







\end{document}
