= Solution
{c}
Start with all of $N_0$ and $N_1$, namely $-1,-1/2,-1/2$, whose sum is $-2$. For $n=0,1,2,\ldots$ in order, append $2^{n+1}$ copies of the triple
$$
\left(2^{-(n+1)},\ -2^{-(n+2)},\ -2^{-(n+2)}\right).
$$
This 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 is
$$
-1,-\frac12,-\frac12,\quad
\frac12,-\frac14,-\frac14,\frac12,-\frac14,-\frac14,\quad\ldots.
$$
Each 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 $\boxed{-2}$.
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