jmc

Let $S=1+2x+3x^2+\cdots .$ Then $$xS=x+2x^2+3x^3+\cdots .$$Subtracting these equations, we get $$(1-x)S=1+x+x^2+\cdots =\frac{1}{1-x},$$so $S=\frac{1}{(1-x{)}^2}.$ Thus, we want to solve $$\frac{1}{(1-x{)}^2}=9.$$then $(1-x{)}^2=\frac{1}{9},$ so $1-x=\pm \frac{1}{3}.$ Since $x$ must be less than 1, $1-x=\frac{1}{3},$ so $x=\boxed{\frac{2}{3}}.$