Answer
$\dfrac{3}{2}+\dfrac{9}{8}+\dfrac{27}{32}+ \space ...$
Work Step by Step
The sum of an infinite geometric series converges if $\mid r\mid<1$, and the sum of an infinite series can be found using the formula:
$$S_\infty=\dfrac{a_1}{1-r}$$
Here: $S_{\infty}=6$ and $r=\dfrac{3}{4}$
So, substituting these values into the formula above gives:
\begin{align*}
6&=\dfrac{a_1}{1-\frac{3}{4}}\\
\\6&=\dfrac{a_1}{\frac{1}{4}}\\
\\6&=4a_1\\
\\\frac{6}{4}&=a_1\\
\\\dfrac{3}{2}&=a_1
\end{align*}
Therefore, the infinite series is:
$$ \dfrac{3}{2}+\dfrac{3}{2}\left(\dfrac{3}{4}\right))+\dfrac{3}{2}\left(\dfrac{3}{4}\right)^2+\dfrac{3}{2}\left(\dfrac{3}{4}\right)^3+ ... =\dfrac{3}{2}+\dfrac{9}{8}+\dfrac{27}{32}+ ...$$
