\magnification1200
\def\newp#1.{%
\par\noindent\vtop\bgroup{\noindent\bf #1.} }
\def\endp{\egroup\vfil}
\def\frac#1#2{{{#1}\over{#2}}}
\let\ds=\displaystyle
\def\SUM#1#2{\ds\sum_{#1=#2}^\infty}
\def\Sum#1{\SUM{#1}{0}}

\newp 11.
\item{a.}\itemitem{$\bullet$} The series is clearly alternating:
$a_n=(-1)^nu_n$ with $u_n=\frac{1}{n\ln n}$ and $u_n>0$ for $n\geq 2$.
\itemitem{$\bullet$} The function $\frac1{x\ln x}$ is decreasing since
$\ds f^\prime(x)=\frac{-(\ln x+x\cdot\frac{1}{x})}{(x\ln x)^2}=
\frac{-1-\ln x}{(x\ln x)^2}<0$.
\itemitem{$\bullet$} $\ds \lim_{x\to \infty}\frac{1}{x\ln x}=0$

We have checked the three hypothesis for the Alternating Series Test to
hold and so the series converges.

\item{b.}
To determine whether or not the series absolutely converges, we must
check whether $\ds\sum_{n=2}^{\infty}\frac{1}{n\ln n}$ converges or not.
We wish to apply the Integral Test to the function
$f(x)=\frac{1}{x\ln x}$.
We checked in part a. that $f(x)$ is decreasing so  the hypotheses of
this test are satisfied.
Hence the series and the integral $\ds\int_2^\infty \frac{dx}{x\ln x}$
converge or diverge together.

The indefinite integral $\ds\int \frac{dx}{x\ln x}$ can be done via the
substitution $u=\ln x$: $du=\frac{dx}{x}$, so
$\ds\int \frac{dx}{x\ln x}=\int\frac{du}{u}=\ln u+C=\ln(\ln x)+C$.
Hence 
$\ds\int_2^\infty \frac{dx}{x\ln x}=\lim_{t\to\infty}\ln(\ln t)-\ln(\ln 2)$.
As $t\to\infty$, $\ln t\to\infty$ and hence $\ln(\ln t)\to\infty$.
Hence the integral diverges and therefore so does the series.
\endp

\newp 12.
To evaluate the terms in the MacLaurin series, we need to evaluate the
higher derivatives of $y$ at $x=0$.
By decree, $y(0)=1$.
From the differential equation
$y^\prime(0)-2y(0)=0$ (since it holds for all $x$) and hence
$y^\prime(0)=2$.
Differentiating the differential equation implicitly, we see
$y^{(2)}-2y^\prime=0$ so at $x=0$,
$y^{(2)}(0)-2y^\prime(0)=0$, or $y^{(2)}(0)=4$.

Differentiating the equation
$y^{(2)}-2y^\prime=0$ implicitly we see
$y^{(3)}-2y^{(2)}=0$, so at $x=0$,
$y^{(3)}(0)-2y^{(2)}(0)=0$, or $y^{(3)}(0)=8$.

Hence the first four non{--}zero terms in the MacLaurin series are
$$1+\frac{2}{1!}x+\frac{4}{2!}x^2+\frac{8}{3!}x^3=
1+2x+2x^2+\frac{4}{3}x^3$$
\endp

\newp 13.
First we calculate the radius of convergence via the $n$th root
(just to be different {--} calculating 
$\Big\vert\frac{a_{n+1}}{a_n}\Big\vert$ also works).

$$\lim_{n\to\infty}\root n\of{\frac{\vert x-3\vert^{2n}}{2^n}}=
\lim_{n\to\infty} \frac{\vert x-3\vert^{2}}{2}=
\frac{\vert x-3\vert^{2}}{2}\ .
$$
Solve $\frac{\vert x-3\vert^{2}}{2}=1$, or $\vert x-3\vert^2=2$ so
$\vert x-3\vert=\sqrt 2$ and the radius of convergence is $\sqrt 2$.
The two endpoints of our interval are $3\pm\sqrt 2$
and we need to check for convergence at the end points:
i.e. we need to check if the two series
$\ds\sum_{n=1}^{\infty}\frac{(-1)^n(3\pm\sqrt2 -3)^{2n}}{2^n}=
\sum_{n=1}^{\infty}\frac{(-1)^n(\pm\sqrt2)^{2n}}{2^n}$
converge or not.
But both the series are the same:
$\ds \sum_{n=1}^{\infty}\frac{(-1)^n 2^n}{2^n}=
\sum_{n=1}^{\infty}(-1)^n$ and this series diverges.
Hence the {\sl interval\/} of convergence is $(3-\sqrt 2,3+\sqrt 2)$.
\endp

\newp 14.
\item{a.}
We know that the MacLaurin series for 
$\ds\frac{1}{1+x}=\sum_{n=0}^\infty (-1)^n x^n$ 
with radius of convergence $1$.
Hence
$\ds\frac{1}{1+(-2x)}=
\Sum n(-1)^n (-2x)^n=\Sum n 2^n x^n$ and this representation is valid for
$\vert x\vert < 1/2$.
In particular, this must be the MacLaurin series for $\frac{1}{1-2x}$.

\item{b.}
Since $\ds\frac{d}{dx}\frac{1}{1-2x}=\frac{2}{(1-2x)^2}$
we know that the MacLaurin series for 
$=\frac{2}{(1-2x)^2}$ is the derivative of the series 
$\Sum n 2^n x^n$ which we know is
$\SUM n1 n\cdot 2^n x^{n-1}=\Sum n(n+1)2^{n+1}x^n$.
Hence $\Sum n(n+1)2^{n+1}x^n$ 
is the MacLaurin series for $\frac{2}{(1-2x)^2}$.

\item{c.}
Since $\ds\frac{2x}{(1-2x)^2}=x\cdot \frac{2}{(1-2x)^2}$
the MacLaurin series for 
$$\frac{2x}{(1-2x)^2}=x\Sum n(n+1)2^{n+1}x^n=\Sum n(n+1)2^{n+1}x^{n+1}=
\SUM n1 n\cdot 2^n\> x^n$$
\endp

\newp 15.
For any $x$ in the interval $[0,1]$ $\SUM n1(-1)^n\frac{nx^{n-1}}{11^n}$
is an alternating series with $\ds u_n=\frac{nx^{n-1}}{11^n}\geq0$.
The MacLaurin series for 
$F(x)=\SUM n1(-1)^n\frac{nx^{n}}{n\cdot 11^n}=\SUM n1(-1)^n\frac{x^{n}}{11^n}$.
Hence $F(x)$ is a geometric series, but this is not particularly relevant to
our current question.

More to the point, for $0<x\leq 1$, the series is alternating with
$u_n=\ds\frac{x^n}{11^n}$.
The $u_n$ decrease since $\ds\frac{d u_n}{d n}=
\biggl(\ln\Bigl(\frac{x}{11}\Bigr)\biggr)\cdot \biggl(\frac{x}{11}\biggr)^n$.
Since $0<x<11$, $\ln\Bigl(\frac{x}{11}\Bigr)<0$ and so is
$\ds\frac{d u_n}{d n}$.

Hence, for any $0<x\leq 1$, the Alternating Series Test AND its accuracy
result applies.
For $x=0$ $F(0)=0$ and all the MacLaurin polynomial approximations are also
$0$ so we get infinite accuracy with any approximation.
Otherwise, we know that the accuracy of the approximation is at worst the
first term neglected, which in our case is
$\ds\biggl(\frac{x}{11}\biggr)^4$.
This quantity is increasing as a function of $x$ so for $x$ in $[0,1]$ the
error is at most $\frac{1}{11^4}$.

\endp

\end

For a fixed $x$, $u_n=\frac nx\bigl(\frac{x}{11}\bigr)^n$ is eventually decreasing:
compute $\frac{d u_n}{dn}=
\frac 1x\bigl(\frac{x}{11}\bigr)^n+
\frac nx \bigl(\ln(\frac n{11}\bigr) \bigl(\frac{x}{11}\bigr)^n=
\frac 1x\bigl(1+n\cdot\ln(\frac n{11}\bigr) \bigr)
\bigl(\frac{x}{11}\bigr)^n$.
Since $x<11$, $\ln\frac{x}{11}$ is negative and so eventually
$1+n\cdot \ln(\frac n{11}\bigr)<0$.
Hence, for any $0<x\leq 1$, the Alternating Series Test AND its accuracy
result applies.
