Answer
convergent
$S=6$
Work Step by Step
Given $$ \sum_{k=1}^{\infty}(-1)^{k-1}\frac{7}{6^{k-1}} = 7\sum_{k=1}^{\infty} \left(\frac{-1}{6}\right)^{k-1} $$
This is a geometric series with $|r|=\frac{1}{6}<1$ , so $\displaystyle \sum_{k=1}^{\infty}(-1)^{k-1}\frac{7}{6^{k-1}}$ is a convergent series.
To find the sum, since $a=(-1/6)^0=1$, then
\begin{align*}
S&=7\left(\frac{a}{1-r}\right)\\
&=\frac{7}{1+\frac{1}{6}}\\
&=6
\end{align*}
