Compute the sum

P.S.

$$\left[\ 1+\frac{1}{3}+\frac{2}{3}+\frac{2}{9}+\frac{4}{9}+\frac{4}{27}+\frac{8}{27} +\frac{8}{81} +\frac{16}{81}+\dots \ \right] \\\\ \\
= 1+ \frac{1}{3} + \frac{2}{3}
+ \underbrace{\frac{1}{3}*\frac{2}{3} }_{=\frac{2}{9}}
+ \underbrace{\frac{2}{3}*\frac{2}{3} }_{=\frac{4}{9}}
+ \underbrace{\frac{1}{3}*\frac{4}{3^2} }_{=\frac{4}{27}}
+ \underbrace{\frac{2}{3}*\frac{4}{3^2} }_{=\frac{8}{27}}
+ \underbrace{ \frac{1}{3}* \frac{8}{3^3}}_{=\frac{8}{81}}
+ \underbrace{\frac{2}{3}*\frac{8}{3^3}}_{=\frac{16}{81}}
+\dots \ \\\\\\
= 1+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2}{3}}
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{4}{3^2}}
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{8}{3^3}}
+\dots \ \\\\\\
= 1+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2}{3}}
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2^2}{3^2}}
+ ( \underbrace{\frac{1}{3} + \frac{2}{3} }_{=1} )*\frac{2^3}{3^3}}
+\dots \ \\\\\\
= 1+ 1
+ \frac{2}{3}}
+ \frac{2^2}{3^2}}
+ \frac{2^3}{3^3}}
+\dots \ \\\\\\
s=
1+1*( \frac{2}{3} ) ^0
+1*(\frac{2}{3})^1
+1*(\frac{2}{3})^2
+1*(\frac{2}{3})^3
+1*(\frac{2}{3})^4+
\dots \$$

$$\small{\text{ sum $ s= 1+1*( \frac{2}{3} ) ^0+ +1*(\frac{2}{3})^1+ +1*(\frac{2}{3})^2+ +1*(\frac{2}{3})^3+ +1*(\frac{2}{3})^4+ \dots $ }}\\ a= 1 \\ r = \frac{2}{3} \\ s = 1 + \frac{a}{1-r} = 1 + \frac{1}{1-\frac{2}{3}} = 1 + \frac{1} {\frac{1}{3}} = 1 + 3 = 4 \\ \boxed{s= 4}$$

There are 2 GPs here

$$\\\[ 1+\frac{2}{3}+\frac{4}{9}+\frac{8}{81}+\dotsb \]\\\\ S_{\infty}=\frac{a}{1-r}=\frac{1}{1-\frac{2}{3}}=\frac{1}{\frac{1}{3}}=3\\\\$$

-----------------------------------------

$$\\\[ \frac{1}{3}+\frac{2}{9}+\frac{4}{27}+\frac{8}{81}+\dotsb \]\\\\ S_{\infty}=\frac{a}{1-r}=\frac{\frac{1}{3}}{1-\frac{2}{3}} =\frac{\frac{1}{3}}{\frac{1}{3}}=1\\\\\\ Total=3+1=4$$

Again, we are both correct but heureka's method is preferable.        Mine is pretty silly really   

Thanks Heureka      

Here's another possibility...

Notice that the 2nd and 3rd terms sum to 1

And the 4th and 5th terms sum to 2/3

And the 6th and 7th terms sum to 4/9

So we have....adding in the first term.....

1 + 1/[1-(2/3)] = 1 + 1/(1/3)  = 1 + 3  = 4

There you go "All roads lead to Rome"    

True, Melody........but not all zeroes do.....!!!

(The Troll can attest to this...)

Very nice Heureka