Let . Pairing consecutive terms shows and
The even partial sums increase and the odd partial sums decrease. They satisfy , while
Both are monotone bounded sequences and hence have limits, and the displayed difference makes their limits equal. Every partial sum belongs to one subsequence, so the alternating series converges. This is a proof of the alternating series test; the monotonicity and zero-limit hypotheses play separate roles.
The alternating series test says that if , , and , then converges.
To prove it, let be the partial sums. The even sums satisfy
so is increasing. The odd sums satisfy
so is decreasing. Also , so both are bounded and converge. Their difference is , hence their limits agree and the whole sequence converges.
Taking proves convergence of the alternating harmonic series. Its even partial sums lie below its limit , while its odd partial sums lie above it. Since
and
we obtain
The alternating series test says that if decreases to , then
converges. Taking proves convergence of the given series.
It is not absolute convergence, because the series of absolute values is the p-series
which diverges since . 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 , then take the first unused negative term. Next take positive terms until the sum exceeds , 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 tends to zero, while the preceding partial sum exceeds ; consequently these rearranged partial sums tend to . This is a divergent series with exactly the prescribed terms.