College Algebra, Chapter 9, 9.3, Section 9.3, Problem 58

Determine the sum of the infinite geometric series $\displaystyle \frac{1}{\sqrt{2}} + \frac{1}{2} + \frac{1}{2 \sqrt{2}} + \frac{1}{4} + .....$.

Using the formula $\displaystyle S = \frac{a}{1 - r}$, here $\displaystyle a = \frac{1}{\sqrt{2}}$ and $\displaystyle r = \frac{1}{\sqrt{2}}$.

Thus, the sum of this infinite series is


$
\begin{equation}
\begin{aligned}

S =& \frac{\displaystyle \frac{1}{\sqrt{2}}}{\displaystyle 1 - \frac{1}{\sqrt{2}}}
&&
\\
\\
S =& \frac{\displaystyle \frac{1}{\cancel{\sqrt{2}}}}{\displaystyle \frac{\sqrt{2} - 1}{\cancel{\sqrt{2}}}}
&& \text{Cancel out like terms}
\\
\\
S =& \frac{1}{\sqrt{2} - 1} \cdot \frac{\sqrt{2} + 1}{\sqrt{2} + 1}
&& \text{Multiply by the conjugate of the radical}
\\
\\
S =& \frac{\sqrt{2} + 1}{2 - 1}
&& \text{Simplify}
\\
\\
S =& \sqrt{2} + 1
&&

\end{aligned}
\end{equation}
$