The series to n terms can be written:
(10/3-1/30)+(10/3-1/300)+(10/3-1/3000)+...=
10n/3-(1/30)(1+0.1+0.01+...)=
10n/3-(1/30)(10/9)(1-10⁻ⁿ)=
10n/3-(1/27)(1-10⁻ⁿ).
This formula gives the sum to n terms, for example, n=2:
20/3-0.11/3=19.89/3=6.63=3.3+3.33.
As n→∞, this sum becomes 10n/3-1/27, which clearly approaches infinity because of the convergence of 1-10⁻ⁿ to 1.