Alternating Series and Absolute Convergence

An alternating series changes sign at every term. For n ≥ 1 and b_n > 0, (-1)^(n+1)b_n starts positive; (-1)^n b_n starts negative. Either series converges if b_(n+1) ≤ b_n and b_n tends to 0. Nonincreasing permits equality. The conditions may start after finitely many terms. A missing condition makes this test inconclusive. If magnitudes do not tend to 0, neither do the terms, so the series diverges.

For an alternating series with nonincreasing magnitudes tending to 0, |S - S_N| ≤ b_(N+1), where S_N is the sum of the first N terms and S is the full sum. This bounds the size of the error by the first omitted magnitude. For magnitudes 1/n², eight terms give a bound of 1/9² = 1/81. If monotonicity starts later, stop within that decreasing tail before applying the bound.

Here Σ sums from n = 1 to infinity. Absolute convergence means Σ |a_n| converges. Conditional convergence means Σ a_n converges but Σ |a_n| diverges. Read p from 1/n^p; p > 1 gives convergence. For a_n = (-1)^n/n², |a_n| = 1/n² and p = 2, so convergence is absolute. The alternating harmonic series is conditional: 1/n decreases to 0, but Σ 1/n diverges. Removing finitely many terms preserves convergence or divergence, so harmonic tails diverge too.

Absolute convergence always implies convergence, but convergence need not imply absolute convergence. The alternating harmonic series is a counterexample to the reverse claim. Absolute convergence does not require alternating signs. If c_n is 1 or -1, then |c_n/(n²+4)| = 1/(n²+4) ≤ 1/n². Comparison proves absolute convergence for any choice of these signs.

The signed harmonic series converges. Its series of absolute values diverges.

Test Σ |a_n| with the positive-term method that fits its shape. For nonzero terms, a ratio limit below 1 proves absolute convergence, above 1 proves divergence, and 1 is inconclusive. For (-1)^n/n⁴, the absolute ratio tends to 1, yet p = 4 proves absolute convergence. If absolute convergence is not settled, use the alternating test only after checking its sign and magnitude conditions.

To certify a tolerance, make the remainder bound no larger than that tolerance. With magnitudes 1/n and allowed error 1/12, solve 1/(N+1) ≤ 1/12. Since N+1 is positive, this gives N+1 ≥ 12, or N ≥ 11. Eleven is the smallest N this bound certifies. To check minimality, substitute the preceding integer into the bound.

For alternating signs and nonincreasing magnitudes tending to 0, the remainder is zero or has the first omitted term's sign. If the series starts positive and N is even, S_N ≤ S ≤ S_N + b_(N+1). For example, S_6 = 0.72 and b_7 = 0.02 give 0.72 ≤ S ≤ 0.74. For odd N, S_N - b_(N+1) ≤ S ≤ S_N.

The endpoints bound S. The upper endpoint is not a claim that S equals 0.74.

Equal consecutive magnitudes do not prevent convergence. Check both kinds of partial sum: if the even and odd partial sums approach the same value, then every partial sum does. For terms +3/k, -3/k in that order for k ≥ 1, each complete pair cancels. Thus S_(2k) = 0 and S_(2k-1) = 3/k. Both tend to 0, so the series sums to 0.