= Solution
{c}
Place the single term in $N_0$ first. For $n=0,1,2,\ldots$ in order, append $2^{n+1}$ copies of the pair
$$
\left(2^{-(n+1)},\ -2^{-(n+1)}\right).
$$
This uses all of $P_n$ and all of $N_{n+1}$ exactly once, so it is an explicit <rearrangement of a series>. Its beginning is
$$
-1,\quad \frac12,-\frac12,\frac12,-\frac12,\quad
\frac14,-\frac14,\frac14,-\frac14,\frac14,-\frac14,\frac14,-\frac14,\quad\ldots.
$$
Every completed pair leaves the <partial sum> at $-1$. Within a level-$n$ pair the <partial sum> differs from $-1$ by at most $2^{-(n+1)}$. As $n\to\infty$ this tends to zero, proving that the rearranged <series> has sum $\boxed{-1}$.
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