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{\bf Question}

Find the limit of the sequences:
\begin{description}
\item[(a)]$\ds u_n = \frac{(n + 1)^2}{n^2 + 1}$
\item[(b)]$\ds u_n = \frac{x^n}{n!}$
\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}


\item[(a)]
$\ds u_n = \frac{(n + 1)^2}{n^2 + 1} = \frac{\left(1 +
\frac{1}{n}\right)^2}{1 + \frac{1}{n^2}}$

$\ds \lim_{n \to \infty} u_n = \frac{\displaystyle \lim_{n \to
\infty}\left(1 + \frac{1}{n}\right)^2}{\displaystyle \lim_{n \to
\infty}{1 + \frac{1}{n^2}}} = \frac{1}{1} = 1$



\item[(b)]
$\ds \frac{u_{n+1}}{u_n} =
\frac{\frac{x^{n+1}}{(n+1)!}}{\frac{x^n}{n!}} = \frac{x}{n+1}$

For $m > n$  $\ds \frac{u_{m+1}}{u_m} =  \frac{x}{m+1} <
\frac{x}{n+1}$

For any given value of $x$,  $\ds \frac{x}{n+1} = r
< 1$ provided $n+1>x$

Hence $\ds u_{n+p} = u_n \times \frac{u_{n+1}}{u_n} \times
\frac{u_{n+2}}{u_{n+1}} \times ... \times
\frac{u_{n+p}}{u_{n+p-1}} < u_nr^p$

If r $<$ 1 $\ds \lim_{p \to \infty} r^p = 0$

Hence $\displaystyle \lim_{p \to \infty} u_{n+p} = 0$

Hence $\displaystyle \lim_{m \to \infty} u_m = 0$
\end{description}



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