Answer
Yes, the given infinite geometry series has a limit, and the value of the limit is ${{S}_{\infty }}=-4$.
Work Step by Step
The infinite geometry series is:
$-6+3-\frac{3}{2}+\frac{3}{4}-\cdots $
Here ${{a}_{1}}=-6,{{a}_{2}}=3$
The value of $\left| r \right|$ is,
$\begin{align}
& \left| r \right|=\left| \frac{{{a}_{2}}}{{{a}_{1}}} \right| \\
& =\left| \frac{3}{-6} \right| \\
& =\left| -\frac{1}{2} \right| \\
& =\frac{1}{2}
\end{align}$
Now, find the limit of the infinite geometry series ${{S}_{\infty }}=\frac{{{a}_{1}}}{1-r}$.
$\begin{align}
& {{S}_{\infty }}=\frac{{{a}_{1}}}{1-r} \\
& =\frac{-6}{1-\left( -\frac{1}{2} \right)} \\
& =\frac{-6}{1+\frac{1}{2}} \\
& =\frac{-6}{\frac{3}{2}}
\end{align}$
This implies,
$\begin{align}
& {{S}_{\infty }}=-6\cdot \frac{2}{3} \\
& =-2\cdot 2 \\
& =-4
\end{align}$
Thus, the limit of the infinite geometric series is ${{S}_{\infty }}=-4$.
