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