Answer
The series converges to $2 + \sqrt 2 $
Work Step by Step
We can rewrite the initial series as the geometric series $\sum\limits^{\infty}_{k=0} (\frac{1}{\sqrt 2})^k$
We can see that the first term is $ a = 1$ and the common ratio is $r = \frac{1}{\sqrt 2}$.
Since $|r| \lt 1 $, we know the series converges and can use the formula $$\frac{a}{1-r} $$ to evaluate the sum
$\frac{1}{1-\frac{1}{\sqrt 2}}$ = $\frac{\sqrt 2}{\sqrt 2 - 1} $
The sum can be simplified further by multiplying both the numerator and denominator by $1 + \sqrt 2 $ to get $\frac{ 2 +\sqrt 2}{(\sqrt 2)^2 - 1} $ = $\frac{ 2 +\sqrt 2}{1} $ = $2 + \sqrt 2 $
