 Solution

Start with all of $N_0$ and $N_1$, namely $-1,-1/2,-1/2$, whose sum is $-2$. For $n=0,1,2,\dots$ in order, append $2^{n+1}$ copies of the tripleThis uses all $2^{n+1}$ occurrences of $P_n$ and all $2^{n+2}$ occurrences of $N_{n+2}$, exactly once. Together with the initial terms, every original occurrence is used, establishing a rearrangement of a series. Its beginning isEach triple has sum zero. At its intermediate terms the partial sum is at most $2^{-(n+1)}$ above $-2$. This bound tends to zero, so the whole series, rather than just the subsequence of sums at completed triples, has sum $\boxed{-2}$.