Alternating Series and Error Control

Cancellation can produce convergence

If $b_n\ge0$ decreases eventually to zero, the alternating series $\sum(-1)^{n-1}b_n$ converges. Pairing consecutive terms shows that even and odd partial sums approach a common limit from opposite sides. The eventual decrease and zero limit are both needed for this test.

Absolute convergence means $\sum|a_n|$ converges and guarantees convergence of $\sum a_n$ . Conditional convergence means the signed series converges but its absolute-value series diverges. This distinction affects operations such as rearranging terms: arbitrary rearrangements preserve absolutely convergent sums, but not generally conditionally convergent sums.

Visual guide

VISUAL GUIDEPartial sums bracket the limit

Limit ln 2

Alternating harmonic partial sums oscillate around ln 2. Even sums lie below and odd sums above. For decreasing term magnitudes tending to zero, the next omitted term bounds the error.

Worked example: alternating harmonic series

The magnitudes $1/n$ decrease to zero, so $1-1/2+1/3-\cdots$ converges. Its absolute-value series is the divergent harmonic series. Therefore convergence is conditional.

The alternating remainder bound is $|S-S_N|\le b_{N+1}$ when the alternating-test hypotheses apply from the relevant tail onward. To guarantee error at most $0.01$ , choosing $N=100$ suffices because the next magnitude is $1/101<0.01$ . This bound is conservative and does not require knowing the sum.

Worked example: absolute convergence first

For $\sum(-1)^{n-1}/n^2$ , taking absolute values produces a convergent $p$ -series. Thus the original series converges absolutely. The alternating bound also applies and is sharper here than the positive-series integral bound: after $N$ terms, its error is at most $1/(N+1)^2$ .

A sequence of small terms with changing signs is not automatically an alternating series. The theorem applies to a strict alternating tail with decreasing magnitudes. Check the pattern and hypotheses before quoting the next-term error bound.

Practice

Classify $\sum(-1)^n/\sqrt n$ .
Bound the error after $10$ terms of $\sum_{n=1}^\infty(-1)^{n-1}/n^2$ .
Does $\sum(-1)^n n/(n+1)$ converge?

Show worked solutions

Conditionally: the alternating test works, but the absolute series has $p=1/2$ and diverges.
At most $1/121$ . The next term is positive, so the sum is above the even partial sum.
No. Magnitudes approach $1$ , so the terms do not approach zero.