The error is smaller than the first term you left out BC only

10.7 established that for a convergent alternating series the odd partial sums sit above the limit and the even ones below, so $S$ always lies between $S_n$ and $S_{n+1}$. The distance from $S_n$ to $S$ is therefore no more than the distance from $S_n$ to $S_{n+1}$, and that distance is exactly one term:

$$\left|S - S_n\right| \le b_{n+1}.$$

That is the entire theorem. It requires the same two conditions the convergence test did, since the bracketing depends on them, and it holds for every $n$ at once.

The figure shows the gap at $n = 4$. There $S_4 \approx 0.5833$, the limit is $\ln 2 \approx 0.6931$, and the actual error is $0.1098$. The bound is $b_5 = \frac{1}{5} = 0.2$. The bound is genuinely an over-estimate, which is what a bound is for, and it is available without ever knowing that the sum is $\ln 2$.

If you have used $b_1$ through $b_n$, the bound is $b_{n+1}$: the first term you did not use. Not $b_n$, which you did use.

Two reasons the off-by-one survives so long undetected:

1. It is wrong in the safe direction. $b_n \ge b_{n+1}$, so the last term used is also a valid bound, just a weaker one. Every answer bounded this way is technically true and none of them is what was asked for.
2. The arithmetic looks identical. Both are single terms of the same sequence, so nothing about the shape of the answer signals the error.

Where it stops being harmless is the how many terms question, which is the standard exam form. To make the error less than $0.01$ for $\sum \frac{(-1)^{n+1}}{n}$ you need $b_{n+1} = \frac{1}{n+1} < 0.01$, so $n + 1 > 100$, so $n \ge 100$: a hundred terms. Solving $\frac{1}{n} < 0.01$ instead gives $n > 100$, so $101$. One term out, every time, and in the direction that costs a mark.

A related trap: the bound is on the absolute error. The actual error has a sign, and that sign is known too, since $S$ lies on the far side of $S_n$ in the direction of the next term. Bounding is a separate question from locating.

Two error bounds appear in this unit and they answer the same question by different routes. The frozen taxonomy treats confusing them as a single error because they are confused in both directions.

1. Alternating series bound. Applies when the series alternates and satisfies 10.7's conditions. The bound is $b_{n+1}$, the next term. No derivatives, no factorials.
2. Lagrange bound (10.12). Applies to a Taylor polynomial approximation of a function, alternating or not. It needs a bound on the $(n+1)$st derivative, a factorial, and a power of $(x - a)$.

The giveaway is the amount of machinery. If the bound you are reaching for is just a term of the series, it is the alternating one; if it needs a derivative you have to bound over an interval, it is Lagrange.

They overlap: a Taylor series that happens to alternate at the point in question can be bounded either way, and the alternating bound is usually far easier. Approximating $\cos 1$ with its Maclaurin series is the standard case, since that series alternates and the next term is immediate, while the Lagrange route requires bounding a derivative of $\cos$ on $[0, 1]$.

Two standard question forms, and the procedure for each.

Bound the error of a given approximation. Identify $n$, write down $b_{n+1}$, and evaluate it. State it as $|S - S_n| \le b_{n+1}$ rather than as an equality.

Find how many terms are needed. Set $b_{n+1} < \varepsilon$, solve for $n$, and round up to the next integer that satisfies it. Then answer the question that was actually asked, which is sometimes the number of terms and sometimes the index of the last one, and those differ when the series does not start at $n = 1$.

Two conditions to confirm before using the bound at all, since they are the alternating series test's conditions and CA9's territory: the terms must alternate, and $b_n$ must be decreasing. Without the decrease the partial sums do not bracket the limit and the bound is simply false. The bound is a consequence of the bracketing, not a separate fact about alternating series, so anything that breaks the bracketing breaks it too.