\documentclass[a4paper,12pt]{article}
\begin{document}

\noindent {\bf Question}

\noindent {\bf The power series scavenger hunt:} for each of the
power series given below, determine the radius and interval of
convergence.
\begin{enumerate}
\item $\sum_{n=0}^\infty (-1)^n x^n / n!$;
\item $\sum_{n=1}^\infty 5^n x^n / n^2$;
\item $\sum_{n=1}^\infty x^n / (n(n+1))$;
\item $\sum_{n=1}^\infty (-1)^n x^n / \sqrt{n}$;
\item $\sum_{n=0}^\infty (-1)^n x^{2n+1} / (2n + 1)!$;
\item $\sum_{n=0}^\infty 3^n x^n / n!$;
\item $\sum_{n=0}^\infty x^n / (1+n^2)$;
\item $\sum_{n=1}^\infty (-1)^{n+1} (x+1)^n / n$;
\item $\sum_{n=0}^\infty 3^n (x+5)^n / 4^n$;
\item $\sum_{n=1}^\infty (-1)^n (x+1)^{2n+1} / (n^2+4)$;
\item $\sum_{n=0}^\infty \pi^n (x-1)^{2n} / (2n+1)!$;
\item $\sum_{n=2}^\infty x^n / (\ln(n))^n$;
\item $\sum_{n=0}^\infty 3^n x^n$;
\item $\sum_{n=0}^\infty n! x^n / 2^n$;
\item $\sum_{n=1}^\infty (-2)^n x^{n+1} / (n+1)$;
\item $\sum_{n=1}^\infty (-1)^n x^{2n} / (2n)!$;
\item $\sum_{n=1}^\infty (-1)^n x^{3n} / n^{3/2}$;
\item $\sum_{n=2}^\infty (-1)^{n+1} x^n / (n\ln^2(n))$;
\item $\sum_{n=0}^\infty (x-3)^n / 2^n$;
\item $\sum_{n=1}^\infty (-1)^n (x-4)^n / (n+1)^2$;
\item $\sum_{n=0}^\infty (2n+1)! (x-2)^n / n^3$;
\item $\sum_{n=1}^\infty \ln(n) (x-3)^n / n$;
\item $\sum_{n=0}^\infty (2x-3)^n / 4^{2n}$;
\item $\sum_{n=2}^\infty (x-a)^n / b^n$, where $b>0$ is arbitrary.
\item $\sum_{n=0}^\infty (n+p)! x^n / (n!(n+q)!)$, where $p$,
$q\in {\bf N}$;
\item $\sum_{n=1}^\infty x^{n-1} / (n 3^n)$;
\item $\sum_{n=1}^\infty (-1)^{n-1} x^{2n-1} / (2n-1)!$;
\item $\sum_{n=1}^\infty n! (x-a)^n$, where $a\in {\bf R}$ is arbitrary;
\item $\sum_{n=1}^\infty n (x-1)^n / (2^n (3n-1))$;
\end{enumerate}
\medskip

\noindent {\bf Answer}
\begin{enumerate}
\item {\bf radius of convergence is $\infty$, interval of convergence
is ${\bf R}$:} Apply the ratio test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{ (-1)^{n+1} x^{n+1} /
(n+1)!}{(-1)^n x^n / n!}\right|  = |x| \lim_{n\rightarrow\infty}
\frac{1}{n+1} =0. \] Hence, this series converges absolutely for
all values of $x$ (since this limit is $0$ for every value of
$x$).
\item {\bf radius of convergence is $\frac{1}{5}$, interval of
convergence is $\left[ -\frac{1}{5}, \frac{1}{5}\right]$:} Apply
the ratio test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{5^{n+1} x^{n+1} /
(n+1)^2}{5^n x^n / n^2}\right|  = |x|\lim_{n\rightarrow\infty}
\frac{5n^2}{(n+1)^2} =5 |x|. \] Hence, this series converges
absolutely for $5|x| <1$, that is for $|x| < \frac{1}{5}$, and so
the radius of convergence is $\frac{1}{5}$.  We now need to check
the endpoints of the interval $(-\frac{1}{5}, \frac{1}{5})$:

\medskip
\noindent At $x =-\frac{1}{5}$, the series becomes
$\sum_{n=1}^\infty 5^n (-1/5)^n / n^2 =\sum_{n=1}^\infty (-1)^n /
n^2$, which converges absolutely.

\medskip
\noindent At $x =\frac{1}{5}$, the series becomes
$\sum_{n=1}^\infty 5^n (1/5)^n / n^2 =\sum_{n=1}^\infty 1 / n^2$,
which converges absolutely.

\medskip
\noindent So the series converges absolutely for all $x$ in the
closed interval $\left[ -\frac{1}{5}, \frac{1}{5}\right]$, and
diverges elsewhere.
\item {\bf radius of convergence is $1$, interval of convergence
is $[-1,1]$:} Apply the ratio test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{ x^{n+1} / ((n+1)(n+2))}{x^n
/ (n(n+1))} \right| = |x| \lim_{n\rightarrow\infty} \frac{n}{n+2}
=|x|. \] Hence, this series converges absolutely for $|x| <1$, and
so the radius of convergence is $1$.  We now need to check the
endpoints of the interval $(-1,1)$:

\medskip
\noindent At $x =-1$, the series becomes $\sum_{n=1}^\infty (-1)^n
/ (n(n+1))$, which converges absolutely.

\medskip
\noindent At $x =1$, the series becomes $\sum_{n=1}^\infty 1 /
(n(n+1))$, which converges absolutely.

\medskip
\noindent So, the series converges absolutely for all $x$ in the
closed interval $[-1,1]$, and diverges elsewhere.
\item {\bf radius of convergence is $1$, interval of convergence is
$[-1,1)$:} Apply the ratio test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{(-1)^{n+1}  x^{n+1} /
\sqrt{n+1}}{(-1)^n x^n / \sqrt{n}} \right|
=|x|\lim_{n\rightarrow\infty} \sqrt{\frac{n}{n+1}} =|x|. \] Hence,
this series converges absolutely for $|x| <1$, and so the radius
of convergence is $1$.  We now need to check the endpoints of the
interval $(-1,1)$:

\medskip
\noindent At $x =-1$, the series becomes $\sum_{n=1}^\infty (-1)^n
/ \sqrt{n}$, which converges conditionally.  (The alternating
series test yields convergence, but this series does not converge
absolutely, by comparison to the harmonic series.)

\medskip
\noindent At $x =1$, the series becomes $\sum_{n=1}^\infty 1 /
\sqrt{n}$, which diverges.

\medskip
\noindent So, the series converges absolutely for all $x$ in the
open interval $(-1,1)$, converges conditionally at $x =-1$, and
diverges elsewhere.
\item {\bf radius of convergence is $\infty$, interval of convergence
is ${\bf R}$:} Apply the ratio test and calculate:
\begin{eqnarray*} \lim_{n\rightarrow\infty} \left| \frac{(-1)^{n+1}
x^{2(n+1)+1} / (2(n+1) + 1)!}{(-1)^n x^{2n+1} / (2n + 1)!} \right|
& = & |x|^2 \lim_{n\rightarrow\infty} \frac{(2n+1)!}{(2n+3)!} \\
 & = & |x|^2 \lim_{n\rightarrow\infty} \frac{1}{(2n+3)(2n+2)} = 0.
\end{eqnarray*}
Hence, this series converges absolutely for all values of $x$
(since this limit is $0$ for every value of $x$).
\item {\bf radius of convergence is $\infty$, interval of convergence
is ${\bf R}$:} Apply the ratio test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{3^{n+1} x^{n+1} / (n+1)!}{
3^n x^n / n!} \right| =|x|\lim_{n\rightarrow\infty} \frac{3}{n+1}
=0. \] Hence, this series converges absolutely for all values of
$x$ (since this limit is $0$ for every value of $x$).
\item {\bf radius of convergence is $1$, interval of convergence
is $[-1,1]$:} Apply the ratio test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{x^{n+1} / (1+(n+1)^2)}{x^n /
(1+n^2)} \right| =|x| \lim_{n\rightarrow\infty} \frac{1+n^2}{2 +2n
+n^2} =|x|. \] Hence, this series converges absolutely for $|x|
<1$, and so the radius of convergence is $1$.  We now need to
check the endpoints of the interval $(-1,1)$:

\medskip
\noindent At $x =-1$, the series becomes $\sum_{n=0}^\infty (-1)^n
/ (1+n^2)$, which converges absolutely.

\medskip
\noindent At $x =1$, the series becomes $\sum_{n=0}^\infty 1 /
(1+n^2)$, which converges absolutely.

\medskip
\noindent So, the series converges absolutely for all $x$ in the
closed interval $[-1,1]$, and diverges elsewhere.
\item {\bf radius of convergence is $1$, interval of convergence
is $(-2,0]$:} Apply the ratio test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{(-1)^{(n+1)+1} (x+1)^{n+1} /
(n+1)}{(-1)^{n+1} (x+1)^n / n} \right| =|x+1|
\lim_{n\rightarrow\infty} \frac{n}{n+1} =|x+1|. \] Hence, this
series converges absolutely for $|x +1| <1$, and so the radius of
convergence is $1$.  We now need to check the endpoints of the
interval $(-2,0)$:

\medskip
\noindent At $x =-2$, the series becomes $\sum_{n=1}^\infty
(-1)^{n+1} (-1)^n / n = -\sum_{n=1}^\infty 1/n$, which diverges,
being a constant multiple of the harmonic series.

\medskip
\noindent At $x =0$, the series becomes $\sum_{n=1}^\infty
(-1)^{n+1} / n$, which converges conditionally, as it is the
alternating harmonic series.

\medskip
\noindent So, the series converges absolutely for all $x$ in the
open interval $(-2,0)$, converges conditionally at $x =0$, and
diverges elsewhere.
\item {\bf radius of convergence is $\frac{4}{3}$, interval of convergence
is $(-\frac{19}{3}, -\frac{11}{3})$:} Apply the ratio test and
calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{3^{n+1} (x+5)^{n+1} /
4^{n+1}}{3^n (x+5)^n / 4^n} \right| = \frac{3}{4} |x+5|. \] Hence,
this series converges absolutely for $\frac{3}{4} |x+5| <1$, that
is for $|x+5| < \frac{4}{3}$, and so the radius of convergence is
$\frac{4}{3}$.  We now need to check the endpoints of the interval
$(-\frac{19}{3}, -\frac{11}{3})$.

\medskip
\noindent At $x =-\frac{19}{3}$, the series becomes
\[ \sum_{n=0}^\infty \frac{3^n \left(-\frac{19}{3} +5\right)^n}{4^n}
=\sum_{n=0}^\infty (-1)^n, \] which diverges (being, for instance,
a divergent geometric series).

\medskip
\noindent At $x =-\frac{11}{3}$, the series becomes
\[ \sum_{n=0}^\infty \frac{3^n \left( -\frac{11}{3}+5 \right)^n}{4^n}
=\sum_{n=0}^\infty 1, \] which diverges (again being, for
instance, a divergent geometric series).

\medskip
\noindent So, the series converges absolutely for all $x$ in the
open interval $(-\frac{19}{3}, -\frac{11}{3})$, and diverges
elsewhere.
\item {\bf radius of convergence is $1$, interval of convergence
is $[-2,0]$:} Apply the ratio test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{(-1)^{n+1} (x+1)^{2(n+1)+1}
/ ((n+1)^2+4)}{(-1)^n (x+1)^{2n+1} / (n^2+4)} \right| =|x+1|^2
\lim_{n\rightarrow\infty} \frac{n^2 +4}{n^2 +2n +5} =|x+1|^2. \]
Hence, this series converges absolutely for $|x+1|^2 <1$, that is
for $|x+1| < 1$, and so the radius of convergence is $1$.  We now
need to check the endpoints of the interval $(-2,0)$.

\medskip
\noindent At $x =-2$, the series becomes $\sum_{n=1}^\infty (-1)^n
(-1)^{2n+1} / (n^2+4) =\sum_{n=1}^\infty (-1)^{n+1} / (n^2+4)$,
which converges absolutely.

\medskip
\noindent At $x =0$, the series becomes $\sum_{n=1}^\infty (-1)^n
/ (n^2+4)$, again which converges absolutely.

\medskip
\noindent So, the series converges absolutely for all $x$ in the
closed interval $[-2,0]$, and diverges elsewhere.
\item {\bf radius of convergence is $\infty$, interval of convergence
is ${\bf R}$:} Apply the ratio test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{\pi^{n+1} (x-1)^{2(n+1)} /
(2(n+1)+1)!}{\pi^n (x-1)^{2n} / (2n+1)!} \right| =|x-1|^2
\lim_{n\rightarrow\infty} \frac{\pi}{(2n+2)(2n+3)} =0. \] Hence,
this series converges absolutely for all values of $x$ (since this
limit is $0$ for every value of $x$).
\item {\bf radius of convergence is $\infty$, interval of convergence
is ${\bf R}$:} This time, since the coefficients are $n^{th}$
powers, we apply the root test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{x^n}{(\ln(n))^n}
\right|^{1/n} = |x| \lim_{n\rightarrow\infty} \frac{1}{\ln(n)} =0.
\] Hence, this series converges absolutely for all values of $x$
(since this limit is $0$ for every value of $x$).
\item {\bf radius of convergence is $\frac{1}{3}$, interval of
convergence is $( -\frac{1}{3}, \frac{1}{3} )$:} Apply the ratio
test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{3^{n+1} x^{n+1}}{3^n x^n}
\right| =3|x|. \] Hence, this series converges absolutely for
$3|x| <1$, that is $|x| <\frac{1}{3}$, and so the radius of
convergence is $\frac{1}{3}$.   We now need to check the endpoints
of the interval $(-\frac{1}{3}, \frac{1}{3})$.

\medskip
\noindent At $x =-\frac{1}{3}$, the series becomes
\[ \sum_{n=0}^\infty 3^n \left(-\frac{1}{3}\right)^n
=\sum_{n=0}^\infty (-1)^n, \] which diverges (being, for instance,
a divergent geometric series).

\medskip
\noindent At $x =\frac{1}{3}$, the series becomes
\[ \sum_{n=0}^\infty 3^n \left( \frac{1}{3} \right)^n
=\sum_{n=0}^\infty 1, \] which diverges (again being, for
instance, a divergent geometric series).

\medskip
\noindent So, the series converges absolutely for all $x$ in the
open interval $(-\frac{1}{3}, \frac{1}{3})$, and diverges
elsewhere.
\item {\bf radius of convergence is $0$, interval of convergence
is $\{ 0\}$:} Apply the ratio test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{(n+1)! x^{n+1} / 2^{n+1}}{n!
x^n / 2^n} \right| =|x| \lim_{n\rightarrow\infty} \frac{n +1}{2}
=\infty. \] Hence, this series converges only for $x =0$ and
diverges elsewhere.
\item {\bf radius of convergence is $\frac{1}{2}$, interval of convergence
is $(-\frac{1}{2}, \frac{1}{2}]$:} Apply the ratio test and
calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{ (-2)^{n+1} x^{(n +1)+1} /
((n+1)+1)}{(-2)^n x^{n+1} / (n+1)} \right| = |x|
\frac{2(n+1)}{n+2} =2|x|. \] Hence, this series converges
absolutely for $2|x| <1$, that is $|x| <\frac{1}{2}$, and so the
radius of convergence is $\frac{1}{2}$.   We now need to check the
endpoints of the interval $(-\frac{1}{2}, \frac{1}{2})$.

\medskip
\noindent At $x =-\frac{1}{2}$, the series becomes
\[ \sum_{n=1}^\infty \frac{(-2)^n \left( -\frac{1}{2}\right)^{n+1}}{n+1}
=-\frac{1}{2}\sum_{n=1}^\infty \frac{1}{n+1}, \] which diverges,
as it is a constant multiple of the harmonic series.

\medskip
\noindent At $x =\frac{1}{2}$, the series becomes
\[ \sum_{n=1}^\infty \frac{(-2)^n \left( \frac{1}{2}\right)^{n+1}}{n+1}
=\frac{1}{2} \sum_{n=1}^\infty \frac{(-1)^n}{n+1} , \] which
converges, as it is a constant multiple of the alternating
harmonic series.

\medskip
\noindent So, the series converges absolutely for all $x$ in the
open interval $(-\frac{1}{2}, \frac{1}{2})$, converges
conditionally at $x =\frac{1}{2}$, and diverges elsewhere.
\item {\bf radius of convergence is $\infty$, interval of convergence
is ${\bf R}$:} Apply the ratio test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{(-1)^{n+1} x^{2(n+1)} /
(2(n+1))!}{ (-1)^n x^{2n} / (2n)!} \right| =|x|^2
\lim_{n\rightarrow\infty} \frac{1}{(2n+2)(2n+1)} =0. \] Hence,
this series converges absolutely for all values of $x$ (since this
limit is $0$ for every value of $x$).
\item {\bf radius of convergence is $1$, interval of convergence
is $[-1,1]$:} Apply the ratio test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{ (-1)^{n+1} x^{3(n+1)} /
(n+1)^{3/2}}{ (-1)^n x^{3n} / n^{3/2}} \right| =|x|^3
\lim_{n\rightarrow\infty} \frac{n^{3/2}}{(n+1)^{3/2}} =|x|^3. \]
Hence, this series converges absolutely for $|x|^3 <1$, that is
$|x| <1$, and so the radius of convergence is $1$.  We now need to
check the endpoints of the interval $(-1,1)$.

\medskip
\noindent At $x = -1$, the series becomes $\sum_{n=1}^\infty
(-1)^n (-1)^n / n^{3/2} =\sum_{n=1}^\infty 1 /n^{3/2}$, which
converges, by Note 1.

\medskip
\noindent At $x = 1$, the series becomes $\sum_{n=1}^\infty (-1)^n
/ n^{3/2}$, which converges absolutely, by Note 1.

\medskip
\noindent So, the series converges absolutely for all $x$ in the
closed interval $[-1,1]$, and diverges elsewhere.
\item {\bf radius of convergence is $1$, interval of convergence
is $[-1,1]$:} Apply the ratio test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{ (-1)^{(n+1)+1} x^{n+1} /
((n+1)\ln^2(n+1))}{(-1)^{n+1} x^n / (n\ln^2(n))} \right| =|x|
\lim_{n\rightarrow\infty} \frac{n\ln^2(n)}{(n+1)\ln^2(n+1)} =|x|.
\] Hence, this series converges absolutely for $|x| <1$, and so
the radius of convergence is $1$.  We now need to check the
endpoints of the interval $(-1,1)$.

\medskip
\noindent At $x = -1$, the series becomes $\sum_{n=2}^\infty
(-1)^{n+1} (-1)^n / (n\ln^2(n)) = -\sum_{n=2}^\infty 1 /
(n\ln^2(n))$, which converges by the integral test: take $f(x) =
1/(x \ln^2(x))$.  Then,
\[ f'(x) = \frac{-(\ln^2(x) + 2\ln(x))}{x^2 \ln^4(x)} <0 \]
for $x \ge 2$, and so $f(x)$ is decreasing.  Then, we evaluate
\begin{eqnarray*}
\int_2^\infty f(x) {\rm d}x & = & \lim_{M\rightarrow\infty}
\int_2^M \frac{1}{x\ln^2(x)} {\rm d}x \\
 & = & \lim_{M\rightarrow\infty} \frac{-1}{\ln(x)} \left|_2^M \right.
\\
 & = & \lim_{M\rightarrow\infty} \left( \frac{-1}{\ln(M)}
+\frac{1}{\ln(2)} \right) =\frac{1}{\ln(2)},
\end{eqnarray*}
which converges.  Hence, by the integral test, the series
converges.

\medskip
\noindent At $x = 1$, the series becomes $\sum_{n=2}^\infty
(-1)^{n+1}/ (n\ln^2(n))$, which converges absolutely by the
argument just given.

\medskip
\noindent So, the series converges absolutely for all $x$ in the
closed interval $[-1,1]$, and diverges elsewhere.
\item {\bf radius of convergence is $2$, interval of convergence is
$(1,5)$:} Apply the ratio test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{(x-3)^{n+1} /
2^{n+1}}{(x-3)^n / 2^n} \right| =\frac{1}{2} |x-3|. \] Hence, this
series converges absolutely for $\frac{1}{2}|x -3| <1$, that is
$|x-3| < 2$, and so the radius of convergence is $2$.  We now need
to check the endpoints of the interval $(1,5)$.

\medskip
\noindent At $x = 1$, the series becomes $\sum_{n=0}^\infty (-2)^n
/ 2^n =\sum_{n=0}^\infty (-1)^n$, which diverges, being for
instance a divergent geometric series.

\medskip
\noindent At $x = 5$, the series becomes $\sum_{n=0}^\infty 1$,
which diverges, again being for instance a divergent geometric
series.

\medskip
\noindent So, the series converges absolutely for all $x$ in the
open interval $(1,5)$, and diverges elsewhere.
\item {\bf radius of convergence is $1$, interval of convergence
is $[3,5]$:} Apply the ratio test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{(-1)^{n+1} (x-4)^{n+1} /
((n+1)+1)^2}{(-1)^n (x-4)^n / (n+1)^2} \right| =|x-4|
\lim_{n\rightarrow\infty} \frac{(n+1)^2}{(n+2)^2} =|x-4|. \]
Hence, this series converges absolutely for $|x -4| <1$, and so
the radius of convergence is $1$.  We now need to check the
endpoints of the interval $(3,5)$.

\medskip
\noindent At $x = 3$, the series becomes $\sum_{n=1}^\infty (-1)^n
(-1)^n / (n+1)^2 =\sum_{n=1}^\infty 1/(n+1)^2$, which converges by
Note 1.

\medskip
\noindent At $x = 5$, the series becomes $\sum_{n=1}^\infty (-1)^n
/ (n+1)^2$, which converges absolutely, again by Note 1.

\medskip
\noindent So, the series converges absolutely for all $x$ in the
closed interval $[3,5]$, and diverges elsewhere.
\item {\bf radius of convergence is $0$, interval of convergence is
$\{ 2\}$:} Apply the ratio test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{ (2(n+1)+1)!\: (x-2)^{n+1} /
(n+1)^3}{ (2n+1)!\: (x-2)^n / n^3}\right|
=|x-2|\lim_{n\rightarrow\infty} \frac{(2n+3)! \: n^3}{(2n+1)!\:
(n+1)^3} =\infty \] for all $x\ne 2$.  Hence, the series converges
only for $x =2$.
\item {\bf radius of convergence is $1$, interval of convergence
is $[2,4)$:} Apply the ratio test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{\ln(n+1) (x-3)^{n+1} /
(n+1)}{\ln(n) (x-3)^n / n} \right| =|x-3|
\frac{n\ln(n+1)}{(n+1)\ln(n)} = |x-3|. \] Hence, this series
converges absolutely for $|x -3| <1$, and so the radius of
convergence is $1$.  We now need to check the endpoints of the
interval $(2,4)$.

\medskip
\noindent At $x = 2$, the series becomes $\sum_{n=1}^\infty \ln(n)
(-1)^n / n$, which converges by the alternating series test (but
does not converge absolutely).

\medskip
\noindent At $x = 4$, the series becomes $\sum_{n=1}^\infty \ln(n)
/ n$, which diverges by the first comparison test, since $\ln(n)/n
> 1/n$ for $n\ge 3$ and the harmonic series $\sum_{n=1}^\infty
1/n$ diverges.

\medskip
\noindent So, the series converges absolutely for all $x$ in the
open interval $(2,4)$, converges conditionally at $x =2$, and
diverges elsewhere.
\item {\bf radius of convergence is $8$, interval of convergence
is $(-\frac{13}{2}, \frac{19}{2})$:} Apply the ratio test and
calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{ (2x-3)^{n+1} /
4^{2(n+1)}}{(2x-3)^n / 4^{2n}} \right| =\frac{1}{16} \left|
2x-3\right| = \frac{1}{8} \left| x -\frac{3}{2}\right|. \] Hence,
this series converges absolutely for $\frac{1}{8} \left| x
-\frac{3}{2}\right| <1$, that is for $\left| x-\frac{3}{2}\right|
< 8$, and so the radius of convergence is $8$.  We now need to
check the endpoints of the interval $(-\frac{13}{2},
\frac{19}{2})$.

\medskip
\noindent At $x = -\frac{13}{2}$, the series becomes
$\sum_{n=0}^\infty (2(-13/2)-3)^n / 4^{2n} = \sum_{n=0}^\infty
(-1)^n$, which diverges.

\medskip
\noindent At $x = \frac{19}{2}$, the series becomes
$\sum_{n=0}^\infty (2(19/2)-3)^n / 4^{2n} =\sum_{n=0}^\infty 1$,
which diverges.

\medskip
\noindent So, the series converges absolutely for all $x$ in the
open interval $(-\frac{13}{2}, \frac{19}{2})$, and diverges
elsewhere.
\item {\bf radius of convergence is $b$, interval of convergence
is $(a-b, a+b)$:} Apply the ratio test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{ (x-a)^{n+1} / b^{n+1}}{
(x-a)^n / b^n}\right| =\frac{1}{b} |x-a|. \] Hence, this series
converges absolutely for $\frac{1}{b} | x-a| <1$, that is for $|
x- a| < b$, and so the radius of convergence is $b$.  We now need
to check the endpoints of the interval $(a -b, a+b)$.

\medskip
\noindent At $x = a-b$, the series becomes $\sum_{n=2}^\infty
(a-b-a)^n / b^n =\sum_{n=2}^\infty (-1)^n$, which diverges.

\medskip
\noindent At $x = a+b$, the series becomes $\sum_{n=2}^\infty
(a+b-a)^n / b^n =\sum_{n=2}^\infty 1$, which diverges.

\medskip
\noindent So, the series converges absolutely for all $x$ in the
open interval $(a-b, a+b)$, and diverges elsewhere.  (Note that
the previous series is a specific example of this general
phenomenon, with $a =\frac{3}{2}$ and $b =8$.)
\item {\bf radius of convergence is $\infty$, interval of convergence
is ${\bf R}$:} Apply the ratio test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{ ((n+1)+p)! x^{n+1} /
((n+1)!((n+1)+q)!)}{(n+p)! x^n / (n!(n+q)!)} \right| =|x|
\lim_{n\rightarrow\infty} \frac{n+1+p}{(n+1)(n+1+q)} =0. \] Hence,
this series converges absolutely for all values of $x$ (since this
limit is $0$ for every value of $x$).
\item {\bf radius of convergence is $3$, interval of convergence
is $[-3,3)$:} Apply the ratio test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{ x^{(n+1)-1} / ((n+1)
3^{n+1})}{ x^{n-1} / (n 3^n)} \right|
=|x|\lim_{n\rightarrow\infty} \frac{n}{3(n+1)} =\frac{1}{3}|x|. \]
Hence, this series converges absolutely for $\frac{1}{3} | x| <1$,
that is for $| x| < 3$, and so the radius of convergence is $3$.
We now need to check the endpoints of the interval $(-3,3)$.

\medskip
\noindent At $x = -3$, the series becomes $\sum_{n=1}^\infty
(-3)^{n-1} / (n 3^n) =\frac{1}{3} \sum_{n=1}^\infty \frac{
(-1)^{n-1}}{n}$, which converges conditionally, as it is a
constant multiple of the alternating harmonic series.

\medskip
\noindent At $x = 3$, the series becomes $\sum_{n=1}^\infty
3^{n-1} / (n 3^n) =\frac{1}{3} \sum_{n=1}^\infty \frac{1}{n}$,
which diverges, as it is a constant multiple of the harmonic
series.

\medskip
\noindent So, the series converges absolutely for all $x$ in the
open interval $(-3,3)$, converges conditionally at $x =-3$, and
diverges elsewhere.
\item {\bf radius of convergence is $\infty$, interval of convergence
is ${\bf R}$:} Apply the ratio test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{ (-1)^{(n+1)-1} x^{2(n+1)-1}
/ (2(n+1)-1)!}{ (-1)^{n-1} x^{2n-1} / (2n-1)!} \right|
=|x|^2\lim_{n\rightarrow\infty} \frac{1}{2n(2n+1)} =0. \] Hence,
this series converges absolutely for all values of $x$ (since this
limit is $0$ for every value of $x$).
\item {\bf radius of convergence is $0$, interval of convergence is
$\{ a\}$:} Apply the ratio test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{ (n+1)! (x-a)^{n+1}}{ n!
(x-a)^n} \right| = |x-a|\lim_{n\rightarrow\infty} (n+1) =\infty \]
for all $x\ne a$.  Hence, the series converges only for $x =a$.
\item {\bf radius of convergence is $2$, interval of convergence
is $(-1,3)$:} Apply the ratio test and calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{ (n+1) (x-1)^{n+1} /
(2^{n+1} (3(n+1)-1))}{ n (x-1)^n / (2^n (3n-1))} \right| =|x-1|
\lim_{n\rightarrow\infty} \frac{(n+1)(3n-1)}{2n(3n+2)}
=\frac{1}{2}|x-1|. \] Hence, this series converges absolutely for
$\frac{1}{2} | x-1| <1$, that is for $| x -1| < 2$, and so the
radius of convergence is $2$.  We now need to check the endpoints
of the interval $(-1,3)$.

\medskip
\noindent At $x = -1$, the series becomes $\sum_{n=1}^\infty n
(-1-1)^n / (2^n (3n-1)) =\sum_{n=1}^\infty (-1)^n n/(3n-1)$, which
diverges by the $n^{th}$ term test for divergence, as
$\lim_{n\rightarrow\infty} \frac{n}{3n-1} =\frac{1}{3}$, and so
$\lim_{n\rightarrow\infty} \frac{(-1)^n n}{3n-1}$ does not exist.

\medskip
\noindent At $x = 3$, the series becomes $\sum_{n=1}^\infty n
(3-1)^n / (2^n (3n-1)) =\sum_{n=1}^\infty n/(3n-1)$, which again
diverges by the $n^{th}$ term test for divergence.

\medskip
\noindent So, the series converges absolutely for all $x$ in the
open interval $(-1,3)$, and diverges elsewhere.
\end{enumerate}

\noindent Note 1.

\noindent The series $\sum_{n=1}^\infty \frac{1}{n^s}$ converges
if and only if $s >1$.

\medskip
\noindent For $s =1$, this series is called the {\bf harmonic
series}, and we can prove directly that it diverges.  Note that
$\frac{1}{3} + \frac{1}{4} > \frac{1}{2}$, that $\frac{1}{5} +
\cdots + \frac{1}{8} > 4\frac{1}{8} = \frac{1}{2}$, and in general
that
\[ \frac{1}{2^{k-1} +1} + \frac{1}{2^{k-1} +2} +\cdots +\frac{1}{2^k}
> 2^{k-1}\frac{1}{2^k} =\frac{1}{2}. \]
Hence, the $(2^k)^{th}$ partial sum $S_{2^k}$ satisfies $S_{2^k}
>1 +k\frac{1}{2}$.  Since the terms in the harmonic series are all
positive, the sequence of partial sums is monotonically
increasing, and by the calculation done the sequence of partial
sums is unbounded, and so the sequence of partial sums diverges.
Hence, the harmonic series diverges.
\end{document}