jmc

Let $$S=1+2\left(\frac{1}{1998}\right)+3{\left(\frac{1}{1998}\right)}^2+4{\left(\frac{1}{1998}\right)}^3+\cdots .$$Then $$1998S=1998+2+\frac{3}{1998}+\frac{4}{1998^2}+\cdots .$$Subtracting these equations, we get $$1997S=1998+1+\frac{1}{1998}+\frac{1}{1988^2}+\cdots =\frac{1998}{1-1/1998}=\frac{3992004}{1997},$$so $S=\boxed{\frac{3992004}{3988009}}.$