Calculus 8th Edition

The series converges to $2 + \sqrt 2$
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 $r = \frac{1}{\sqrt 2}$. 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 $\frac{ 2 +\sqrt 2}{(\sqrt 2)^2 - 1}$ = $\frac{ 2 +\sqrt 2}{1}$ = $2 + \sqrt 2$