Answer
$\{a_{n}\}$ is convergent.
Work Step by Step
$ a_{n}=1+\displaystyle \frac{1}{n}\geq 1\quad$ ... bounded below by 1.
$\displaystyle \frac{1}{n}$ is the greatest when n is the smallest (n=1)
$\{a_{n}\}$ is bounded above by $1+1=2$
$\{a_{n}\}$ is bounded. If we can show that it is monotonic, we can apply Th.6.
\begin{align*}
a_{n+1}-a_{n}&=1+\displaystyle \frac{1}{n+1}-(1+\frac{1}{n}) \\
& =\displaystyle \frac{1}{n+1}-\frac{1}{n}\\
& =\displaystyle \frac{n-(n+1)}{n(n+1)}\\
& =\displaystyle \frac{-1}{n(n+1)}\leq 0\\
a_{n+1}&\leq a_{n} \\ \end{align*}
$\{a_{n}\}$ is nonincreasing, that is, monotonic. Th.6 applies.