Solution
= Solution
The inner <geometric series> has common ratio $2/3$, so
$$
\sum_{m=n}^{\infty}\frac{2^m}{3^m}=\frac{(2/3)^n}{1-2/3}=3(2/3)^n.
$$
The outer <geometric series> therefore converges as well, with value
$$
\boxed{\sum_{n=0}^{\infty}\sum_{m=n}^{\infty}(2/3)^m=3\sum_{n=0}^{\infty}(2/3)^n=9.}
$$
All summands are nonnegative, so there is no cancellation or conditional-convergence issue in evaluating the nested sums.