Answer
convergent
$S=\dfrac{1}{2}$
Work Step by Step
Given $$ \sum_{k=1}^{\infty} \frac{1}{2^k}-\frac{1}{2^{k+1}}= \sum_{k=1}^{\infty} \frac{2-1}{2^{k+1}}=\sum_{k=1}^{\infty}\left( \frac{1}{2} \right)^{k+1}$$
This is a geometric series with $|r|=\frac{1}{2}<1$, so $\displaystyle \sum_{k=1}^{\infty}\left( \frac{1}{2} \right)^{k+1}$ is a convergent series.
To find the sum, since $a=(1/2)^2=\dfrac{1}{4}$, then
\begin{align*}
S&=\frac{a}{1-r}\\
&=\frac{\frac{1}{4}}{1-\frac{1}{2}}\\
&=\frac{1}{2}
\end{align*}
