The alternating series error bound says that when a convergent alternating series is truncated, the size of the error is no larger than the absolute value of the first term you leave out. In symbols, if you stop after $n$ terms, then $|s-s_n|\leq a_{n+1}$ , where $a_{n+1}$ is the next omitted term.

What Is the Alternating Series Error Bound?

The alternating series error bound says that if an alternating series converges by the alternating series test, then the error from using a partial sum is at most the absolute value of the first omitted term. If you use $s_n$ to approximate the true sum $s$ , then $|s-s_n|\leq a_{n+1}$ .

For AP Calculus BC, the routine is: verify the alternating series test conditions, identify the next omitted term, take its absolute value as the bound, and write an interval around the partial sum. This is a justification tool, not just a computation shortcut.

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Why This Matters for the AP Calculus Exam

This is a BC-only topic from Unit 10. It shows up when you need to approximate the sum of a series and then justify how close your approximation is. The skill you are practicing is approximating the value of a series and bounding the error, which connects directly to Taylor and Maclaurin work later in the unit.

On the exam you may be asked to compute a partial sum, find the bound, and write an inequality that pins down the true value. Clean notation and a clear inequality matter for showing your reasoning, especially on questions where you must state why the error is within a given amount.

Key Takeaways

The error bound only applies to alternating series that converge by the alternating series test, so confirm the terms are positive, decreasing, and approaching 0 first.
The error is bounded by the absolute value of the first omitted term: $|s - s_n| \leq a_{n+1}$ .
The first omitted term is $a_{n+1}$ , so if you use $n$ terms, you need the value of term $n+1$ for the bound.
Build the inequality $s_n - a_{n+1} \leq s \leq s_n + a_{n+1}$ to trap the true sum.
A smaller next term means a tighter bound, so adding more terms makes the estimate more accurate.
Use absolute value when finding the bound, since the bound itself is always a positive magnitude.

Alternating Series Error Bound Theorem

For a convergent alternating series $\sum_{n=1}^\infty (-1)^n \cdot a_n$ , you can estimate its true value and control how far off your estimate is. If $s$ is the true sum and $s_{i-1}$ is the partial sum using the first $i-1$ terms, then

$|s - s_{i-1}| \leq a_i$

where $a_i$ is the first omitted term. The error is never larger than the magnitude of that first term you dropped.

Breaking Down the Theorem

Start with a convergent, alternating series. Use this one:

\sum_{n=1}^\infty \frac{(-1)^n\cdot n^2}{4^n}

To find the error bound for an estimate using the first 3 terms, the bound is $a_4$ , since the fourth term is the first omitted term. Calculate it:

a_4=\frac{(-1)^4\cdot 4^2}{4^4}=\frac{4^2}{4^4}=\frac{1}{4^2}=\boxed{\frac{1}{16}}

This is the error bound. When you estimate the infinite sum using the first three terms, the error will not be greater than $\frac{1}{16}$ in either direction. Now find the partial sum of the first three terms and combine it with the bound:

\sum_{n=1}^3 \frac{(-1)^n\cdot n^2}{4^n}=-\frac{9}{64}

The error bound statement becomes:

\left|s-\left(-\frac{9}{64}\right)\right|\leq \frac{1}{16}

Simplify to trap the true value:

-\frac{1}{16}\leq s+\frac{9}{64}\leq \frac{1}{16}\rightarrow -\frac{1}{16}-\frac{9}{64}\leq s\leq \frac{1}{16}-\frac{9}{64}\rightarrow \boxed{-\frac{13}{64}\leq s \leq -\frac{5}{64}}

The true value of the infinite sum lies between about $-0.203125$ and $-0.078125$ . This estimate is not very precise, since the two bounds do not share digits. Adding more terms gives a tighter interval.

How to Use This on the AP Calculus Exam

Problem Solving

Work through it in a fixed order:

Confirm the series is alternating and converges by the alternating series test.
Decide how many terms you are keeping, then identify the first omitted term $a_{n+1}$ .
Compute that omitted term and take its absolute value. That number is your bound.
Compute the partial sum $s_n$ .
Write the inequality $|s - s_n| \leq a_{n+1}$ and solve to get an interval for $s$ .

Common Trap

A frequent slip is bounding with the last term you used instead of the first one you dropped. If you keep $n$ terms, the bound is $a_{n+1}$ , not $a_n$ . Index carefully so you pull the correct term.

Alternating Series Error Bound Practice

Find the error bound of $\sum_{n=1}^5$ for the following infinite series and state which is a more accurate estimation.

$\sum \frac{(-1)^n}{n}$
$\sum \frac{(-1)^n(1+n^2)}{n^6+6}$

Solutions

For each problem, first find $a_6$ , the value of the first omitted term:

1. \ a_6=\frac{(-1)^6}{6}=\frac{1}{6}

2. \ a_6=\frac{(-1)^6(1+6^2)}{6^6+6}=\frac{1(1+36)}{46656+6}=\frac{37}{46662}

Then find the value of the series up to five terms:

1. \ \sum_{n=1}^5 \frac{(-1)^n}{n}=\frac{-47}{60}\approx -0.783333

2. \ \sum_{n=1}^5 \frac{(-1)^n(1+n^2)}{n^6+6}\approx −0.225410

Next, set up your inequalities:

1. \ |s-s_5|\leq \frac{1}{6}

2. \ |s-s_5|\leq \frac{37}{46662}

Finally, simplify:

1. \ |s-s_5|\leq \frac{1}{6}\rightarrow -\frac{1}{6}\leq s-s_5\leq \frac{1}{6}\rightarrow -\frac{1}{6}\leq s+0.783333\leq \frac{1}{6}\rightarrow \boxed{-0.95\leq s\leq -0.616667}

2. \ |s-s_5|\leq \frac{37}{46662}\rightarrow -\frac{37}{46662}\leq s-(−0.225410)\leq \frac{37}{46662}\rightarrow \\ -\frac{37}{46662}\leq s+0.225410\leq \frac{37}{46662}\rightarrow \boxed{-0.226203\leq s \leq -0.224617}

Compare the accuracy. In the first problem, the bounds do not share any decimal places, so it is not very precise. The second problem is accurate to two decimal places, so the second estimation is more accurate.

Common Misconceptions

The error bound is not the actual error. It is an upper limit on how far off your partial sum can be. The true error is usually smaller.
You cannot apply this bound to just any series. The terms must be alternating, positive in absolute value, decreasing, and approaching 0. If those fail, use a different approach.
The bound is the next omitted term, not the last term you summed. Keeping 5 terms means the bound is the 6th term.
A bigger partial sum does not mean a better estimate. Accuracy depends on how small the next term is, not on how many digits the partial sum has.
The bound is always a positive magnitude. Take the absolute value of the omitted term, even if that term is negative in the original series.

Vocabulary

The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.

Term	Definition
alternating series	A series whose terms alternate in sign, typically written in the form Σ(-1)^n * a_n where a_n > 0.
alternating series error bound	A method for estimating the maximum error between a partial sum and the actual sum of a convergent alternating series, equal to the absolute value of the first omitted term.
alternating series test	A convergence test that determines whether an alternating series converges based on whether its terms decrease in absolute value and approach zero.
converges	A series converges when the sequence of partial sums approaches a finite limit as n approaches infinity.
partial sum	The sum of the first n terms of a series, denoted S_n.
series	A sum of the terms of a sequence, often written as the sum of infinitely many terms.

Frequently Asked Questions

What is the alternating series error bound?

The alternating series error bound says that if an alternating series converges by the alternating series test, then the error from a partial sum is no greater than the absolute value of the first omitted term.

When can you use the alternating series error bound?

You can use it only after the alternating series test conditions are satisfied: the terms alternate, the positive term sizes decrease, and the terms approach 0.

What is the first omitted term?

The first omitted term is the next term after the partial sum you used. If you approximate with s_n, the first omitted term is term n plus 1, and its absolute value gives the error bound.

How do you write an interval estimate with the alternating series error bound?

If the partial sum is s_n and the first omitted term has magnitude a_{n+1}, then the true sum s is between s_n minus a_{n+1} and s_n plus a_{n+1}.

Is the alternating series error bound the exact error?

No. It is an upper bound on the possible error, so the actual error may be smaller. The bound tells you how far off the approximation can be at most.

How is the alternating series error bound tested on AP Calculus BC?

AP Calculus BC questions may ask you to approximate a series, justify the error is below a value, or create an interval for the true sum. Show the alternating series test conditions and use the next omitted term.