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10.10 Alternating Series Error Bound

♾️AP Calculus AB/BC
Unit 10 Review

10.10 Alternating Series Error Bound

Written by the Fiveable Content Team • Last updated September 2025
Verified for the 2026 exam
Verified for the 2026 examWritten by the Fiveable Content Team • Last updated September 2025
♾️AP Calculus AB/BC
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10.10 Alternating Series Error Bound

Welcome to AP Calc 10.10! In this lesson, you’ll learn how to estimate the accuracy of a partial sum for an alternating series.

🚧 This is an AP Calculus BC topic only! If you are taking Calculus AB, you can skip this material. If you’re taking AP Calculus BC, here you go! ⬇️


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➕ Alternating Series Error Bound Theorem

The error bound theorem for an alternating series states that for a convergent alternating series, n=1(1)nan\sum^\infty_{n=1}(-1)^n\cdot a_n, we can estimate its true value by using an error bound. The error bound is defined as aia_i. We can estimate the true value of the sum (ss) with the following equation: ssi1ai|s-s_{i-1}|\leq a_i where ii is the first omitted term of our estimation.

That’s a lot of information—now let’s break it down with an example!

🧱 Breaking Down the Theorem

The first thing we need is a convergent, alternating series. Let’s use this one:

n=1(1)nn24n\sum_{n=1}^\infty \frac{(-1)^n\cdot n^2}{4^n}

We’re going to find the error bound for an estimation made with the first 3 terms. That means our bound is defined as a4a_4. Notice that we need to find up to the fourth term to find the error bound. This is because four is our first omitted term. For now, let’s calculate this fourth term.

a4=(1)44244=4244=142=116a_4=\frac{(-1)^4\cdot 4^2}{4^4}=\frac{4^2}{4^4}=\frac{1}{4^2}=\boxed{\frac{1}{16}}

This is our error bound. That means that when we estimate the sum of the infinite series using the third term, our error won’t be greater than 116\frac{1}{16} in either direction. Let’s finish this example by using the sum of the first three terms and the error bound to estimate what our infinite series is equal to.

n=13(1)nn24n=964\sum_{n=1}^3 \frac{(-1)^n\cdot n^2}{4^n}=-\frac{9}{64}

We can now say that the error bound is:

s(964)116\left|s-\left(-\frac{9}{64}\right)\right|\leq \frac{1}{16}

We can simplify this like so:

116s+964116116964s1169641364s564-\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}}

This means that the true value of the infinite sum lies somewhere between about –0.203125 and –0.078125. You can also tell that this estimation isn’t very accurate, since the two bounds don’t share any digits. You can make more accurate estimations by calculating more terms.


📝 Alternating Series Error Bound Practice

Now it’s your turn to apply what you’ve learned!

❓Alternating Series Error Bound Problems

Find the error bound of n=15\sum_{n=1}^5 for the following infinite series and state which is a more accurate estimation.

  1. (1)nn\sum \frac{(-1)^n}{n}
  2. (1)n(1+n2)n6+6\sum \frac{(-1)^n(1+n^2)}{n^6+6}

💡 Alternating Series Error Bound Solutions

For each problem, first find a6a_6, the value of the first omitted term:

1. a6=(1)66=161. \ a_6=\frac{(-1)^6}{6}=\frac{1}{6} 2. a6=(1)6(1+62)66+6=1(1+36)46656+6=37466622. \ 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 (review Riemann sums for help calculating summation!).

1. n=15(1)nn=47600.7833331. \ \sum_{n=1}^5 \frac{(-1)^n}{n}=\frac{-47}{60}\approx 0.783333 2. n=15(1)n(1+n2)n6+60.2254102. \ \sum_{n=1}^5 \frac{(-1)^n(1+n^2)}{n^6+6}\approx −0.225410

Next, set up your inequalities.

1. ss5161. \ |s-s_5|\leq \frac{1}{6} 2. ss637466622. \ |s-s_6|\leq \frac{37}{46662}

Finally, simplify!

1. ss51616ss51616s0.783333160.616667s0.951. \ |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{0.616667\leq s\leq 0.95} 2. ss537466623746662s(0.225410)37466623746662s+0.22541037466620.226203s0.2246172. \ |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}

Now that we have both our answers, we can compare their accuracy. For the first problem, we see that the answers don’t share any decimal points—meaning it’s not very accurate. However, the second problem is accurate to two decimal places. Therefore, the second estimation is more accurate.


💫 Closing

As complicated as this might feel, this skill relies on techniques you’ve been learning for a long time! Make sure you keep practicing and you’re sure to ace any questions on this topic 💯

Vocabulary

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

TermDefinition
alternating seriesA series whose terms alternate in sign, typically written in the form Σ(-1)^n * a_n where a_n > 0.
alternating series error boundA 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 testA convergence test that determines whether an alternating series converges based on whether its terms decrease in absolute value and approach zero.
convergesA series converges when the sequence of partial sums approaches a finite limit as n approaches infinity.
partial sumThe sum of the first n terms of a series, denoted S_n.
seriesA sum of the terms of a sequence, often written as the sum of infinitely many terms.

Frequently Asked Questions

How do I find the error bound for an alternating series?

If an alternating series satisfies the Alternating Series Test (terms a_n ≥ 0 are monotone decreasing and a_n → 0), then the alternating series error bound (Leibniz criterion) says the remainder after n terms, R_n = S − S_n, satisfies |R_n| ≤ a_{n+1}. So to bound the error: 1) verify a_n is positive, decreasing, and →0; 2) compute a_{n+1}; that value is an upper bound on how far S_n can be from the true sum S. Also the sign of R_n equals the sign of the (n+1)-st term, so S_n is an over- or under-estimate accordingly. To get at most a given error ε, pick n so that a_{n+1} ≤ ε (solve inequality or check values). This is exactly the LIM-7.B objective on the AP BC exam. For a short study guide and examples see the Topic 10.10 study guide (https://library.fiveable.me/ap-calculus/unit-10/alternating-series-error-bound/study-guide/3jO1mS42e1MwgYafawP5). For more unit review and lots of practice problems go to (https://library.fiveable.me/ap-calculus/unit-10) and (https://library.fiveable.me/practice/ap-calculus).

What's the formula for alternating series error bound?

If an alternating series ∑(-1)^{k} a_k (with a_k ≥ 0) satisfies the Alternating Series Test (a_k decreases to 0), then the error (remainder) after n terms, R_n = S − S_n, is bounded in absolute value by the next term: |R_n| = |S − S_n| ≤ a_{n+1}. Equivalently, the true sum S lies between consecutive partial sums S_n and S_{n+1}, and the truncation error is at most the size of the first omitted term. This is the “next-term” or Leibniz error bound used on the AP BC exam for approximating alternating series (CED LIM-7.B). For a worked review, see the Topic 10.10 study guide (https://library.fiveable.me/ap-calculus/unit-10/alternating-series-error-bound/study-guide/3jO1mS42e1MwgYafawP5). For extra practice, try problems at Fiveable’s AP Calculus practice page (https://library.fiveable.me/practice/ap-calculus).

When can I use the alternating series error bound theorem?

You can use the Alternating Series Error Bound (Leibniz criterion) when all three AST conditions hold for your series ∑(-1)^{k} a_k (with a_k ≥ 0): (1) the signs alternate, (2) the sequence of term magnitudes a_k is monotone decreasing (eventually), and (3) a_k → 0. If those hold, the series converges by the Alternating Series Test and the remainder R_n = S − S_n after n terms satisfies |R_n| ≤ a_{n+1} (the next-term bound). That lets you say “using n terms, the truncation error is at most a_{n+1}” and choose n to get a desired accuracy. Note: this works even for conditionally convergent alternating series (not only absolutely convergent ones). This is exactly what LIM-7.B in the CED targets—use it to approximate a sum and bound the error. For a quick refresher, see the Topic 10.10 study guide (https://library.fiveable.me/ap-calculus/unit-10/alternating-series-error-bound/study-guide/3jO1mS42e1MwgYafawP5) and practice problems (https://library.fiveable.me/practice/ap-calculus).

I don't understand how to calculate how close my partial sum is to the actual sum - can someone explain?

If your series is alternating, has terms a_n ≥ 0 that decrease to 0, and it converges by the Alternating Series Test, then the Alternating Series Error Bound (Leibniz criterion) tells you exactly how close a partial sum S_n is to the true sum S: - The remainder R_n = S − S_n satisfies |R_n| ≤ a_{n+1}. - In words: the absolute error after adding n terms is at most the size of the next term. So to get within a desired tolerance ε, pick n so that a_{n+1} < ε. Example: for the alternating harmonic series Σ(−1)^{k+1}/k, to have error < 0.01 you need 1/(n+1) < 0.01 ⇒ n ≥ 99. This is exactly the LIM-7.B item in the CED (use the next-term bound to approximate alternating series). For more explanation and examples see the Topic 10.10 study guide (https://library.fiveable.me/ap-calculus/unit-10/alternating-series-error-bound/study-guide/3jO1mS42e1MwgYafawP5) and try practice problems at (https://library.fiveable.me/practice/ap-calculus).

What's the difference between the alternating series test and the alternating series error bound?

The alternating series test (Leibniz criterion) is a convergence test: if the terms of an alternating series a1 − a2 + a3 − ... satisfy a_n ≥ 0, a_n+1 ≤ a_n (eventually monotone decreasing), and a_n → 0, then the series converges. The alternating series error bound is what you use once you know the series converges by that test: it gives a simple remainder (truncation) estimate. For the nth partial sum S_n, the error R_n = S − S_n satisfies |R_n| ≤ a_{n+1} (the next term’s absolute value). So the test tells you the series converges; the error bound tells you how far your partial sum is from the true sum (useful for approximations on the AP BC exam, LIM-7.B). Remember this applies only when the test’s conditions hold; it gives conditional/alternating-series approximation info, not absolute-convergence info. For the AP study guide on this topic, see Fiveable (https://library.fiveable.me/ap-calculus/unit-10/alternating-series-error-bound/study-guide/3jO1mS42e1MwgYafawP5). For broader review/practice, check the Unit 10 page (https://library.fiveable.me/ap-calculus/unit-10) and practice problems (https://library.fiveable.me/practice/ap-calculus).

How do I know if my alternating series converges enough to use the error bound?

Use the Alternating Series Test (Leibniz criterion) first. You can use the alternating-series error bound only when all three hold: - the terms alternate in sign (a_n = (−1)^n b_n or similar), - the absolute terms b_n are eventually monotone decreasing (b_{n+1} ≤ b_n for large n), - b_n → 0 as n → ∞. If those hold, the series converges by the AST and the truncation (remainder) after N terms satisfies |R_N| = |S − S_N| ≤ b_{N+1} (the next-term bound). “Eventually monotone” is fine—they don’t have to decrease from n=1, just from some N onward. Note: absolute convergence isn’t required; conditional convergence is okay if AST conditions hold. For AP exam language, justify convergence by the alternating series test (LIM-7.B) and then state the next-term error bound. For a quick refresher, see the Topic 10.10 study guide (https://library.fiveable.me/ap-calculus/unit-10/alternating-series-error-bound/study-guide/3jO1mS42e1MwgYafawP5). For more practice, try problems at (https://library.fiveable.me/practice/ap-calculus).

Step by step - how do I find the maximum error when approximating an alternating series?

Step-by-step: 1. Verify the series is alternating and meets the Alternating Series Test (Leibniz): terms a_n > 0, a_{n+1} ≤ a_n (eventually monotone decreasing), and lim_{n→∞} a_n = 0. If these hold, the series converges and the error bound applies. 2. Compute the partial sum S_n you’re using (sum of first n terms). 3. Use the Alternating Series Error Bound (next-term bound): the remainder R_n = S − S_n satisfies |R_n| ≤ a_{n+1}. So the maximum possible error when approximating the infinite sum by S_n is a_{n+1}. 4. If you want error ≤ ε, pick n so that a_{n+1} ≤ ε. Solve that inequality for n (sometimes need approximation or monotone bounds). Example: For ∑_{k=1}∞ (−1)^{k+1}/k, to get error ≤ 0.01, find n with 1/(n+1) ≤ 0.01 → n+1 ≥ 100 → n ≥ 99. So S_99 gives error ≤ 0.01. This is exactly the AP BC tool for approximating sums (CED LIM-7.B). For a refresher, see the Topic 10.10 study guide (https://library.fiveable.me/ap-calculus/unit-10/alternating-series-error-bound/study-guide/3jO1mS42e1MwgYafawP5). For more practice, use Fiveable’s AP Calc practice problems (https://library.fiveable.me/practice/ap-calculus).

Why does the error bound equal the absolute value of the next term in the series?

If you have an alternating series with terms a_k > 0 that decrease to 0 (the Leibniz/alternating-series-test hypotheses), then partial sums S_n alternate above and below the true sum S. Concretely, S_1 > S_3 > S_5 > ... > S > ... > S_6 > S_4 > S_2 (or the reverse sign case). That means S lies between S_n and S_{n+1} for every n. So the remainder R_n = S − S_n has the same sign as the next term and its magnitude is at most the size of that next term: |R_n| = |S − S_n| ≤ |S_{n+1} − S_n| = a_{n+1}. Intuitively: each new term “corrects” the previous partial sum but then overshoots in the opposite direction by less than the size of that term. This is the alternating-series (next-term) error bound used to approximate sums (CED LIM-7.B). For a quick review, see the Topic 10.10 study guide (https://library.fiveable.me/ap-calculus/unit-10/alternating-series-error-bound/study-guide/3jO1mS42e1MwgYafawP5). For more practice on error bounds and related series topics, check the AP practice questions (https://library.fiveable.me/practice/ap-calculus).

Can I use the alternating series error bound on any series or just alternating ones?

Short answer: you can only use the alternating series error bound for series that meet the Alternating Series Test—i.e., an alternating sign pattern, terms whose absolute values are eventually monotone decreasing, and terms that tend to 0. If those three conditions hold, the remainder after N terms satisfies |R_N| ≤ next term (the “next-term” or Leibniz bound). This is exactly LIM-7.B.1 in the CED. If a series is not alternating, the alternating-series error bound doesn’t apply. For absolutely convergent series you can use other error tools (like comparison with a tail integral or remainder estimates from a convergent positive-term test); for conditionally convergent but non-alternating series you need other techniques or to rearrange terms (but rearranging can change sums). For a quick refresher see the Topic 10.10 study guide (https://library.fiveable.me/ap-calculus/unit-10/alternating-series-error-bound/study-guide/3jO1mS42e1MwgYafawP5). For extra practice, try problems in the Unit 10 practice set (https://library.fiveable.me/practice/ap-calculus).

How many terms do I need to approximate the sum of an alternating series within 0.001?

Use the Alternating Series Error Bound (Leibniz criterion): if your series is ∑(-1)^{k} b_k with b_k positive, decreasing, and b_k → 0, then the error after N terms (|R_N| = |S − S_N|) is at most the next term b_{N+1}. So to get error ≤ 0.001, find the smallest N with b_{N+1} ≤ 0.001. Procedure: (1) Identify b_k (the absolute value of the k-th term). (2) Verify b_k is decreasing and → 0 (Alternating Series Test). (3) Solve b_{N+1} ≤ 0.001 for N and take the smallest integer N satisfying it. Example: for ∑(-1)^{k+1}/k, require 1/(N+1) ≤ 0.001 → N+1 ≥ 1000 → N ≥ 999, so use 999 terms. This is exactly the next-term bound in the CED (LIM-7.B). For a refresher and worked examples, see the Topic 10.10 study guide (https://library.fiveable.me/ap-calculus/unit-10/alternating-series-error-bound/study-guide/3jO1mS42e1MwgYafawP5). For extra practice problems, try Fiveable’s practice set (https://library.fiveable.me/practice/ap-calculus).

What does it mean when they ask for the error bound of the nth partial sum?

They’re asking: how far can the nth partial sum S_n differ from the true infinite sum S? For a convergent alternating series that meets the Alternating Series Test (terms alternate, a_n ≥ 0, a_n decreases to 0), the remainder R_n = S − S_n satisfies the Alternating Series Error Bound (Leibniz criterion): |R_n| ≤ a_{n+1}. So the “error bound of the nth partial sum” is the numerical upper bound |S − S_n| (usually given by the next term a_{n+1}). Practically, if you compute S_n and want guaranteed accuracy ε, pick n so a_{n+1} ≤ ε. This is exactly what the CED’s LIM-7.B learning objective covers (next-term/truncation error for alternating series). For a quick refresher and examples, see the Topic 10.10 study guide (https://library.fiveable.me/ap-calculus/unit-10/alternating-series-error-bound/study-guide/3jO1mS42e1MwgYafawP5). For more practice problems, use Fiveable’s practice bank (https://library.fiveable.me/practice/ap-calculus).

I'm confused about when the alternating series error bound applies - does the series have to pass the alternating series test first?

Short answer: yes—you should first verify the Alternating Series Test (AST). The Alternating Series Error Bound (a.k.a. the next-term bound or Leibniz bound) only applies when the series meets the AST hypotheses: the terms alternate in sign, their absolute values are eventually monotone decreasing, and the terms go to 0. If those hold, the series converges and the error |R_n| = |S − S_n| is ≤ the magnitude of the next term, a_{n+1}. If any AST condition fails (e.g., terms don’t decrease or don’t tend to 0), you can’t use the alternating-series error bound—the series might diverge or converge by some other test, but the next-term bound won’t be justified. Also note: if a series is absolutely convergent you don’t need AST; other error estimates may apply. For a quick refresher and examples, check the AP Topic 10.10 study guide (https://library.fiveable.me/ap-calculus/unit-10/alternating-series-error-bound/study-guide/3jO1mS42e1MwgYafawP5) and try practice problems at (https://library.fiveable.me/practice/ap-calculus).

How do I use the error bound to determine how many terms I need for a given accuracy?

If an alternating series meets the alternating series test (terms decrease to 0), the Alternating Series Error Bound (Leibniz criterion) says the remainder after n terms, |R_n| = |S − S_n|, is ≤ the absolute value of the next term: |R_n| ≤ a_{n+1}. So to get a desired accuracy ε, pick n so that a_{n+1} ≤ ε. How to use it: 1. Identify a_k (positive decreasing term). 2. Solve a_{n+1} ≤ ε for n. 3. Use n terms (S_n)—the error is at most a_{n+1}. Quick example: for ∑ (−1)^{k+1}/k and ε = 0.001, require 1/(n+1) ≤ 0.001 ⇒ n+1 ≥ 1000 ⇒ use n = 999 terms (so S_999 is within 0.001). This is exactly LIM-7.B work on the AP BC exam (estimate sums/truncation error). For a short guide and practice, see the Topic 10.10 study guide (https://library.fiveable.me/ap-calculus/unit-10/alternating-series-error-bound/study-guide/3jO1mS42e1MwgYafawP5) and more Unit 10 resources (https://library.fiveable.me/ap-calculus/unit-10). For extra practice problems, use https://library.fiveable.me/practice/ap-calculus.

Why is the error bound always positive even though we're dealing with alternating series?

Good question—the error bound is always written positive because it’s an upper bound on the magnitude of the error, not its sign. For an alternating series that meets the Alternating Series Test (terms a_n ≥ 0, decreasing, a_n → 0), Leibniz’s/alternating-series error bound says the remainder R_n = S − S_n satisfies |R_n| ≤ a_{n+1}. Two things to notice: - R_n can be positive or negative (partial sums alternate above and below the true sum), but the bound uses absolute value so we get a nonnegative number that’s how far S_n could be from S. - a_{n+1} is positive (it’s the next term’s magnitude), so the bound is positive and easy to interpret: “the partial sum S_n is within a_{n+1} of the true sum.” This is exactly the next-term (truncation) bound in the CED (LIM-7.B). For extra practice and examples, see the Topic 10.10 study guide (https://library.fiveable.me/ap-calculus/unit-10/alternating-series-error-bound/study-guide/3jO1mS42e1MwgYafawP5) and the AP unit page (https://library.fiveable.me/ap-calculus/unit-10).

What's the connection between the alternating series test conditions and being able to use the error bound?

If you want to use the alternating series error bound (the “next-term” bound), the series must first meet the Alternating Series Test (Leibniz criterion): the terms alternate in sign, the absolute values of the terms are monotone decreasing eventually, and the terms tend to 0. Those conditions guarantee convergence (often conditional). Once they hold, the error after summing the first n terms (the remainder Rn = S − Sn) satisfies |Rn| ≤ |a_{n+1}|—i.e., the size of the next term bounds how far your partial sum is from the true sum. So the connection is direct: the AST conditions are the hypotheses that let you apply the error bound from LIM-7.B.1 on the CED. If the terms don’t decrease monotonically or don’t go to zero, you can’t use that simple next-term bound (and the series may diverge). For practice and a short guide, see the Topic 10.10 study guide (https://library.fiveable.me/ap-calculus/unit-10/alternating-series-error-bound/study-guide/3jO1mS42e1MwgYafawP5). For extra practice problems, check (https://library.fiveable.me/practice/ap-calculus).