Question 250559
Hint: Recall that the series {{{sum(a*r^n,n=0,infinity)}}} is convergent if {{{abs(r)<1}}} and it diverges otherwise. If it converges, then {{{sum(a*r^n,n=0,infinity)=a/(1-r)}}}


So for the first problem, let {{{a*r^n=(2/3)^n}}} which makes {{{a=1}}} and {{{r=2/3}}}. Because {{{abs(r)=abs(2/3)<1}}}, this means that {{{sum((2/3)^n,n=0,infinity)=1/(1-2/3)=1/(1/3)=3}}}. Use the same idea for the rest of the problems.