
\def\newp#1.{\par\noindent{\bf #1.}}
\def\endp{\par\noindent\vrule width\hsize depth 0pt height .1pt}
\def\frac#1#2{{{\textstyle#1}\over{\textstyle#2}}}
\def\leaderfill{\leaders\hbox to .5em{\hss.\hss}\hfil}

\def\ans{\par\noindent\hbox to \hsize{\leaderfill}
\par\medskip}

\magnification1200
\def\frac#1#2{{{#1}\over{#2}}}

\newp 13.
Determine whether the series
$$\sum_{n=1}^{\infty}\ (-1)^n\ \frac{n^2+1}{n^3-9}$$
is absolutely convergent, conditionally convergent or divergent.
\ans
We first check for absolute convergence.
In other words, does 
$\displaystyle\sum_{n=1}^{\infty}\ \frac{n^2+1}{n^3-9}$
converge or not?
Use the Limit Comparison Test to compare our series to the harmonic
series $\displaystyle\sum_{n=1}^{\infty}\ \frac{1}{n}$
which we know diverges.
Compute $\displaystyle\lim_{n\to\infty}
\frac{\frac{n^2+1}{n^3-9}}{\frac{1}{n}}=
\lim_{n\to\infty}\ \frac{n(n^2+1)}{n^3-9}$.
This is a rational function with the degree of the numerator ($3$) equal
to the degree of the denominator. Hence the limit is the ration of the
degree $3$ coefficients or $\frac{1}{1}=1$.
Since $0<1<\infty$, the Limit Comparison Test shows
$\displaystyle\sum_{n=1}^{\infty}\ \frac{n^2+1}{n^3-9}$ divergses.

The series 
$\displaystyle\sum_{n=1}^{\infty}\ (-1)^n\ \frac{n^2+1}{n^3-9}$
is alternating and $\displaystyle\lim_{n\to\infty} \frac{n^2+1}{n^3-9}=0$
since it is a limit of a rational function for which the degree of the
numerator is less than the degree of the denominator.
Finally, since $\displaystyle\frac{d\>\frac{x^2+1}{x^3-9}}{dx}=
\frac{2x(x^3-9)-(x^2+1)(3x^2)}{(x^3-9)^2}=
\frac{-x^4-3x^2-18x}{(x^3-9)^2}<0
$ for $x>0$, the sequence $\displaystyle\frac{n^2+1}{n^3-9}$
is decreasing. Hence the Alternating Series Test shows
$\displaystyle\sum_{n=1}^{\infty}\ (-1)^n\ \frac{n^2+1}{n^3-9}$
is convergent and therefore conditionally convergent.
\endp

\newp 14.
Find the radius of convergence and 
interval of convergence of the power series
$$\sum_{n=1}^{\infty}\ \frac{(-1)^n}{\sqrt n}\ (x-3)^n$$
\ans
Begin with the radius using either the Ratio method or the Root method.
For the Ration method, compute

\noindent$\displaystyle\lim_{n\to\infty}
\frac{\frac{\vert x-3\vert^{n+1}}{\sqrt{n+1}}}
{\frac{\vert x-3\vert^{n}}{\sqrt{n}}}=
\lim_{n\to\infty}\frac{\sqrt{n}}{\sqrt{n+1}}\cdot\vert x-3\vert=
\vert x-3\vert
$.

\noindent
For the Root method compute

\noindent$\displaystyle\lim_{n\to\infty}\root n\of
{\frac{\vert x-3\vert^{n}}{\sqrt{n}}}=
\lim_{n\to\infty}\frac{1}{\root n\of{\sqrt{n}}}\cdot\vert x-3\vert=
\frac{1}{\sqrt{\displaystyle\lim_{n\to\infty}\root n\of{n}}}\cdot\vert x-3\vert=
\vert x-3\vert$.

\noindent
Either of these calculations shows that the radius of convergence is $1$.

To calculate the interval of convergence we need to evaluate the
two series
\par\noindent
$\displaystyle\sum_{n=1}^{\infty}\ \frac{(-1)^n}{\sqrt n}\ (1)^n=
\sum_{n=1}^{\infty}\ \frac{(-1)^n}{\sqrt n}$
and
$\displaystyle\sum_{n=1}^{\infty}\ \frac{(-1)^n}{\sqrt n}\ (-1)^n
=\sum_{n=1}^{\infty}\ \frac{1}{\sqrt n}$.

The second series is a $p${--}series with $p=1/2\leq 1$ and so diverges.
The first series is an alternating $p${--}series with $p=1/2>0$ and
so converges.
Hence the interval of convergence is $(2,4]$.
\endp
 
\newp 15.
Find the power series centered at the origin for $f(x)=\frac{1}{(2-x)^2}$.
\par\noindent{\bf Hint:}$f(x)$ is related to the derivative of a series you should know.
\ans
First note that 
$\displaystyle\frac{d\>\frac{1}{2-x}}{dx}=\frac{-1}{(2-x)^2}\cdot(-1)=
\frac{1}{(2-x)^2}$.
Next recall that you can figure out a power series for 
$\displaystyle\frac{1}{2-x}=\frac{1}{2}\cdot \frac{1}{1-(x/2)}=
\frac12\cdot\sum_{n=0}^\infty \biggl(\frac x2\biggr)^n=
\sum_{n=0}^\infty \frac{x^n}{2^{n+1}}$.
Taking derivatives we see
$\displaystyle\frac{1}{(2-x)^2}=
\sum_{n=1}^\infty \frac{n\>x^{n-1}}{2^{n+1}}$,
or if you prefer
$\displaystyle\sum_{n=0}^\infty \frac{(n+1)\>x^n}{2^{n+2}}$.
\endp

\end

