Answer
Converges to $k$
Work Step by Step
The sum of a geometric series can be found as:
$S=\dfrac{a}{1-r}$
We have $\Sigma_{n=0}^{\infty} k(\dfrac{1}{2})^{n+1}$, which is a convergent geometric series with first term, $a=\dfrac{k}{2}$ and common ratio $r =\dfrac{1}{2}$
$S=\dfrac{\dfrac{k}{2}}{1-\dfrac{1}{2}}=k$
Hence, the given series converges to $k$.
