Solution
= Solution
For each fixed $n$, the inner <geometric series> sums to $2^{-n}/(1-1/2)=2^{1-n}$. Summing this second <geometric series> gives
$$
\boxed{\sum_{n=0}^\infty\sum_{m=n}^\infty2^{-m}=\sum_{n=0}^\infty2^{1-n}=4.}
$$
All terms are nonnegative, so there is no conditional-rearrangement issue; the iterated sum is finite and the <double series> converges. Equivalently, collecting equal values of $m$ gives $\sum_{m=0}^\infty(m+1)2^{-m}=4$.