Answer
The series is convergent and the sum of the series is $\sqrt 2 +2$.
Work Step by Step
$\Sigma^{\infty}_{n=0} \frac{1}{(\sqrt 2)^{n}} =\Sigma^{\infty}_{n=0} (\frac{1}{\sqrt 2})^{n}$
$a=1$ and $r=\frac{1}{\sqrt 2} \lt 1$
Since $|r| \lt 1$
The series is convergent
$=\frac{a}{1-r}$
$=\frac{\sqrt 2}{\sqrt 2 -1}$
$=\sqrt 2 +2$