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

\noindent {\bf Question}

\noindent Determine the radius of convergence and the interval of
convergence of the power series
\[ \sum_{n=1}^\infty \left( 1 + \frac{1}{n}\right)^n (x-1)^n. \]

\medskip

\noindent {\bf Answer}

\noindent Apply the ratio test:
\[ \lim_{n\rightarrow\infty} \left| \frac{ \left( 1 + \frac{1}{n+1}
\right)^{n+1} (x-1)^{n+1}}{ \left( 1 + \frac{1}{n} \right)^n
(x-1)^n} \right| = |x-1| \lim_{n\rightarrow\infty} \frac{ \left( 1
+\frac{1}{n+1} \right)^{n+1}}{ \left( 1+ \frac{1}{n} \right)^n} =
|x-1| \frac{e}{e} = |x-1|. \] So, the radius of convergence is
$1$, and this series converges absolutely for $| x-1| <1$.  We
need to check the endpoints of this interval.

\medskip
\noindent At $x =0$, the series becomes $\sum_{n=1}^\infty \left(
1 + \frac{1}{n}\right)^n (-1)^n$, which diverges by the $n^{th}$
term test for divergence, since $\lim_{n\rightarrow\infty} \left(
1 + \frac{1}{n}\right)^n (1)^n$ does not exist, since
$\lim_{n\rightarrow\infty} \left( 1 + \frac{1}{n} \right)^n =e$.

\medskip
\noindent At $x =2$, the series becomes $\sum_{n=1}^\infty \left(
1 + \frac{1}{n}\right)^n$, which diverges since
$\lim_{n\rightarrow\infty} \left( 1 + \frac{1}{n}\right)^n =e$.

\medskip
\noindent So, the interval of convergence is $(0,2)$.


\end{document}
