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\noindent {\bf Question}

\noindent For each of the following series, determine the values
of $x$ for which the series converges.
\begin{enumerate}
\item $\sum_{n=1}^\infty ((x+2)/(x-1))^n / (2n-1)$;
\item $\sum_{n=1}^\infty 1/((x+n)(x+n-1))$;
\end{enumerate}
\medskip

\noindent {\bf Answer}

\noindent We can use the same techniques that we have developed
for power series for other series, that are not strictly speaking
power series.  For instance, we can apply the ratio test to the
series, for all the values of $x$ for which the terms are defined.
\begin{enumerate}
\item first, we note that this series is not defined at $x =1$, but is
defined for all other values of $x$.  Applying the ratio test, we
calculate:
\[ \lim_{n\rightarrow\infty} \left| \frac{ ((x+2)/(x-1))^{n+1} /
(2(n+1)-1)}{ ((x+2)/(x-1))^n / (2n-1)} \right|
=\frac{|x+2|}{|x-1|} \lim_{n\rightarrow\infty} \frac{2n-1}{2n+1} =
\frac{|x+2|}{|x-1|}. \] Hence, this series converges absolutely
for $\frac{|x+2|}{|x-1|} < 1$, that is for $|x+2| <|x-1|$, which
is the open ray $(-\infty, -\frac{1}{2})$, and diverges for
$\frac{|x+2|}{|x-1|} > 1$, which is the union $(-\frac{1}{2},
1)\cup (1, \infty)$.

\medskip
\noindent At $x =-\frac{1}{2}$, the only remaining point at which
to test for convergence, the series becomes
\[ \sum_{n=1}^\infty \frac{1}{2n-1} \left( \frac{-\frac{1}{2} +2}{
-\frac{1}{2}-1} \right)^n  = \sum_{n=1}^\infty \frac{1}{2n-1}
(-1)^n,
 \]
which converges conditionally, by the alternating series test.
Hence, the series converges on the closed ray $(-\infty,
-\frac{1}{2}]$.
\item for this series, first note that the series is not defined at $x
=0$, $x=1$, $x=2$, et cetera, and so the domain of consideration
is the complement in ${\bf R}$ of the non-negative integers ${\bf
W} =\{ 0, 1, 2, \ldots\}$ (the {\bf whole numbers}).  Applying the
ratio test, we calculate
\[ \lim_{n\rightarrow\infty} \left| \frac{1/((x+n +1)(x+n +1-1))}{
1/((x+n)(x+n-1))} \right| =\lim_{n\rightarrow\infty} \left|
\frac{x+n-1}{x+n+1} \right| =1 \] for every (allowable) value of
$x$, and so yields no information. However, we are saved by the
observation that the series
\[ \sum_{n=1}^\infty \frac{1}{(n+\alpha)(n+\beta)} \]
converges for all $\alpha$, $\beta$, by limit comparison to the
series $\sum_{n=1}^\infty \frac{1}{n^2}$.  Hence, taking $\alpha
=x$ and $\beta =x-1$, we have that $\sum_{n=1}^\infty
1/((x+n)(x+n-1))$ converges at every value of $x$ for which it is
defined, namely the union
\[ (-\infty, 0)\cup (0,1)\cup (1,2)\cup (2,3) \cup \cdots ={\bf R}
-{\bf W}. \]
\end{enumerate}

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