Answer
convergent;
$S_{\infty}=\sqrt2 + 1$
Work Step by Step
RECALL:
(1) The sum of an infinite geometric series is convergent if $|r| < 1$. The sum is given by the formula:
$S_{\infty}=\dfrac{a}{1-r}$
(2) An infinite geometric series is divergent if $|r|\ge1$.
(3) The common ratio $r$ of a geometric series can be found by dividing any term by the term before it:
$r = \dfrac{a_n}{a_{n-1}}$
Solve for $r$ to obtain:
$\require{cancel}
r = \dfrac{\frac{1}{2}}{\frac{1}{\sqrt2}}
\\r=\dfrac{1}{2} \cdot \left(\dfrac{\sqrt2}{1}\right)
\\r=\dfrac{\sqrt2}{2}\approx 0.7071$
Since $\left|\dfrac{\sqrt2}{2}\right|<1$, then the series is convergent.
Solve for the sum using the formula above, with $a=\dfrac{1}{\sqrt2}$ and $r=\dfrac{\sqrt2}{2}$, to obtain:
$\require{cancel}
S_{\infty}=\dfrac{\frac{1}{\sqrt2}}{1-\frac{\sqrt2}{2}}
\\S_{\infty}=\dfrac{\frac{1}{\sqrt2}}{\frac{2}{2}-\frac{\sqrt2}{2}}
\\S_{\infty}=\dfrac{\frac{1}{\sqrt2}}{\frac{2-\sqrt2}{2}}
\\S_{\infty} = \dfrac{1}{\sqrt2} \cdot \dfrac{2}{2-\sqrt2}
\\S_{\infty}=\dfrac{2}{\sqrt2(2-\sqrt2)}
\\S_{\infty}=\dfrac{2}{2\sqrt2-2}
\\S_{\infty}=\dfrac{2}{2(\sqrt2-1)}
\\S_{\infty}=\dfrac{\cancel{2}}{\cancel{2}(\sqrt2-1)}
\\S_{\infty}=\dfrac{1}{(\sqrt2-1)}$
Rationalize the denominator by multiplying $\sqrt2+1$ to both the numerator and the denominator to obtain:
$S_{\infty}=\dfrac{1}{(\sqrt2-1)}\cdot \dfrac{\sqrt2+1}{\sqrt2+1}$
Simplify the denominator using the formula $(a-b)(a+b) = a^2-b^2$ to obtain:
$S_{\infty}=\dfrac{\sqrt2 + 1}{(\sqrt2)^2-1^2}
\\S_{\infty}=\dfrac{\sqrt2 + 1}{2-1}
\\S_{\infty}=\dfrac{\sqrt2 + 1}{1}
\\S_{\infty}=\sqrt2 + 1$
