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%  Homework for the course "Calculus II", 
%  Spring semester, 2024, Wei-Cheng Wang.

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\noindent
{\scriptsize Calculus II, Spring 2024 
(http://www.math.nthu.edu.tw/\~\,wangwc/)} \hfill 

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\large
Homework 03   
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\vspace{.3in}

%           The questions!

\begin{enumerate}

\item % The Integral Test
Section 10.3: Problems 7, 27, 28, 31, 33, 37, 41, 55. %remove 38 next time. too complicated for integral test.
% add 51 next time (error estimate for integral test)

Hint for problem 41:
If both $\sum a_n$ and $\sum b_n$ diverge,
then $\sum ( a_n \pm b_n )$ could either converge or diverge in general.

To determine the convergence or divergence,
one can, for example, simplify $ a_n = 
\displaystyle {a \over n+2}
-
\displaystyle {1 \over n+4}
=
\displaystyle {\cdots \over (n+2)(n+4)}
$
and use one of the comparison methods in Section 10.4.

\item % Comparison Tests
Section 10.4: Problems 15, 16, 17, 27, 29, 31, 43, 45, 51, 61, 62.

Hints for problem 61: 
\begin{enumerate}
% \item
% It suffices to consider the case $q>0$.
% The case $q\leq 0$ is much easier. (Why?)
\item
% If you find it difficult to deal with general $p>1$ and $q>0$,
% try to start with, for example,
For simplicity, 
just take a fixed case $p=1.5$, $q=3$ and proceed.
% and proceed with $r=(1+p)/2$.
The same argument works for all $p>1$, $q>0$. 
The case $p>1$, $q \leq 0$ is easier (Why?).
\item
Instead of evaluating 
$\displaystyle{ \lim_{n\to \infty} { (\ln n)^3 \over n^{1.5-1.25} }}$,
it is and easier to evaluate 
$\left( \displaystyle{ \lim_{n\to \infty} { \ln n \over n^{1.5-1.25\over 3} }} \right)^3$.

\end{enumerate}

Hint for problem 62: 

For simplicity, 
just take a fixed case $p=0.5$, $q=-2$ and proceed.
The same argument works for all $0<p<1$, $q <0$. 
The case $0<p<1$, $q \geq 0$ is easier.

% Remark: Stirling's formula about (n!)^{1/n}, see Chap 8, 
%         additional and advanced exercises 32(a), p511
\item % The Ratio and Root Tests
Section 10.5: % (homework problems for next week, if you want to start early): 
Odd numbered problems in problem 17-43, 61, 65.

Hint: In general, the ratio test applies if you see the factorial
$(\cdots)!$ appearing in $a_n$.
% Similarly, the root test should work if you see the power
% $(\cdots)^n$
% appearing in $a_n$.
We did not have time for more examples in lecture 04.
Do your best as time permits. 
We will proceed with more examples (and the proof) for both tests in lecture 05.

Hint for problem 61: multiply by $2*4*\cdots*2n = 2^n n!$
both on the denominator and the numerator.

Hint for problem 65: Ratio test will be inconclusive, but
root test works.

% 
% \item % The Ratio and Root Tests
% Section 10.5: (optional, to answer a question raised in class)
% Take the following Stirling's formula
% for granted:
% \begin{equation} \label{Stirling}
 % \lim_{n\to\infty} {e^n n! \over n^n \sqrt{n} } = \sqrt{2\pi}
% \end{equation}
% You can easily find some readable proof via Google search if you are curious.
% Use \eqref{Stirling} to determine whether
% $\displaystyle{ \sum_{n=1}^\infty { (2n)! \over 4^n n! n! }}$
% converges or not.
% 
\end{enumerate}

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