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The alternating series test says that if $a_n\ge 0$ decreases to $0$, thenconverges. Taking $a_n=n^{-1/2}$ proves convergence of the given series.

It is not absolute convergence, because the series of absolute values is the p-serieswhich diverges since $p=1/2\le 1$. Thus the original series has conditional convergence.

For an explicit divergent rearrangement of a series, take unused positive terms, which are the even-indexed terms, until the partial sum exceeds $1$, then take the first unused negative term. Next take positive terms until the sum exceeds $2$, then the next unused negative term, and continue. Both the positive and negative subseries have infinite total magnitude, so this procedure uses every term. The negative term inserted at stage $m$ tends to zero, while the preceding partial sum exceeds $m$; consequently these rearranged partial sums tend to $+\infty$. This is a divergent series with exactly the prescribed terms.

Assume first that the original nonnegative series converges to $S$. Its partial sums $s_m$ increase to $S$. Write $t_N={\sum }_{n=1}^N{a}_{g(n)}$ for the partial sums of the rearrangement of a series. Since $g$ is an injection, the indices $g(1),\dots ,g(N)$ are distinct, soThus $(t_N)$ is an increasing bounded sequence, and the bounded monotone sequence theorem gives $t_N\to T\le S$.

For any fixed $m$, surjectivity of $g$ givesFor $N\ge N_m$, nonnegativity implies $t_N\ge s_m$. Taking limits of sequences first in $N$ and then in $m$ gives $T\ge S$. Hence $T=S$.

Conversely, $g^{-1}$ is a bijection, and the original series is the rearrangement of a series of $\sum b_n$ defined by $g^{-1}$. The same argument proves its convergence whenever $\sum b_n$ converges. Both series converge together, and their sums are equal.

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 $\boxed{-2}$.

Place the single term in $N_0$ first. For $n=0,1,2,\dots$ in order, append $2^{n+1}$ copies of the pairThis uses all of $P_n$ and all of $N_{n+1}$ exactly once, so it is an explicit rearrangement of a series. Its beginning isEvery 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 $\boxed{-1}$.