Answer:
see the works below.
Step-by-step explanation:
We can convert a repeating decimal into a addition by splitting the repeated set of decimal:
For 0.666..., the repeating set of decimal is 6, therefore it will become:
[tex]0.666...=0.6+0.06+0.006+...[/tex]
[tex]=(0.6\times0.1^0)+(0.6\times0.1^1)+(0.6\times0.1^2)+(0.6\times0.1^3)+...[/tex]
We can also write this addition in the summation notation:
[tex]\boxed{\sum\limits_{n=1}^{\infty}0.6(0.1)^{n-1}}[/tex]
To find the sum of this summation, we notice that it forms a geometric series [tex]\left(\sum\limits_{n=1}^{\infty}ar^{n-1}\right)[/tex] with:
The formula of Convergent Infinite Geometric Series (a geometric series will converge providing -1 < r < 1):
[tex]\boxed{Sum\ (S)=\frac{a}{1-r} }[/tex]
[tex]\displaystyle S=\frac{0.6}{1-0.1}[/tex]
[tex]\displaystyle S=\frac{0.6}{0.9}[/tex]
[tex]\bf\displaystyle S=\frac{2}{3}[/tex]
