$\begin{array} { l }a_1={\left( \frac{ 2 }{ 3 } \right)}^{1-1},\\r=\frac{ 2 }{ 3 }\end{array}$
Simplify the expression$\begin{array} { l }a_1=1,\\r=\frac{ 2 }{ 3 }\end{array}$
Write the rule for the sum of a finite geometric series, $S_n=a_1 \times \frac{ 1-{r}^{n} }{ 1-r }$, for $n=6$$S_6=a_1 \times \frac{ 1-{r}^{6} }{ 1-r }$
Substitute $1$ for $a_1$ and $\frac{ 2 }{ 3 }$ for $r$$S_6=1 \times \frac{ 1-{\left( \frac{ 2 }{ 3 } \right)}^{6} }{ 1-\frac{ 2 }{ 3 } }$
Simplify the expression$\begin{align*}&S_6=\frac{ 665 }{ 243 } \\&\begin{array} { l }S_6=2 \frac{ 179 }{ 243 },& S_6\approx2.73663\end{array}\end{align*}$