Question 1107236
You know that {{{(9^x)(9^y)(9^z)=9^((x+y+z))}}} .
So,
{{{9^"1/3"*9^"1/9"*9^"1/27"*"..."=9^(("1/3 + 1/9 + 1/27 + ..."))=9^"1/2"=sqrt(9)=highlight(3)}}} ,

because
{{{1/3+1/9+1/27+"..."=1/2}}}
 
That sum is the sum of an infinite geometric sequence.
The "formula" for such a sum with a first term {{{b}}} and a common ratio {{{r<1}}} is
{{{sum=b/(1-r)}}} .
In the case of the sum above, with {{{b=1/3}}} and {{{r=1/3}}} ,
{{{sum="1/3"/(1-"1/3")="1/3"/"2/3"=1/2}}} .
 
There is no need to memorize formulas, if you can deduce them.
Consider the sum or {{{n}}} terms
{{{S=b+br+br^2+br^3+"..."+br^(n-1)}}} .
{{{rS=br+br^2+br^3+"..."+br^(n-1)+br^n}}} , and
{{{rS-S=br^n-b}}} --> {{{(r-1)S=b(r^n-1)}}} --> {{{S=b(r^n-1)/r-1}}} .
If {{{r<1}}} , we might like to write it as
{{{S=b(1-r^n)/(1-r)}}} ,
and as {{{n}}} approaches {{{infinity}}} , {{{r^n}}} approaches {{{0}}} ,
and {{{S}}} approaches {{{b/(1-r)}}} .