When can I use the alternating series error bound instead of the Lagrange error bound?

Use it whenever your numerical series genuinely alternates in sign with terms decreasing in magnitude toward zero. Many Taylor approximations, like 1 or (1.5), produce such series, and the alternating bound |a_N+1| is usually simpler than Lagrange. If the terms do not strictly decrease, use Lagrange instead.

Does the Alternating Series Test prove absolute convergence?

No. The AST only proves the series converges—possibly conditionally. To determine absolute convergence, test the series of absolute values a_n separately using tools like the p-series or comparison tests. A series can pass the AST yet have a_n diverge, meaning it converges conditionally.

What if the terms increase at first before decreasing?

The AST and the error bound only require the terms to eventually decrease. Convergence still holds as long as the decrease begins after some index and the limit is zero. However, apply the error bound only from the point where the terms are actually decreasing; otherwise the bracketing argument fails.

Is the error bound the exact error?

No—it is an upper bound. The true error |S - S_N| is at most |a_N+1|, but usually smaller. On the exam, write it as an inequality with ≤, never as an equality. Claiming the error equals the next term is a common mistake that loses credit.

AP-CALCBC-10.7-10.10

U10.7 Alternating Series and Error Bound

Master the Alternating Series Test and the alternating series error bound |S − S_N| ≤ |a_{N+1}| for AP Calculus BC Unit 10.

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A voice-first session with the Crimsora tutor on U10.7 Alternating Series and Error Bound, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start U10.7 Alternating Series and Error Bound free

What this lesson covers

Many important series in AP Calculus BC—like

\sum \frac{(-1)^n}{n}

or the Maclaurin series for cosine—switch signs term after term. These alternating series have a beautifully simple convergence test and, even better, an error bound you can compute in one step. This lesson shows you exactly when the Alternating Series Test applies, how to check its conditions cleanly, and how to bound the error of a partial sum using nothing more than the size of the first term you leave off. Nailing this topic pays off directly on both multiple-choice and free-response questions, where you are often asked how many terms guarantee a target accuracy.

The Alternating Series Test (AST)

An alternating series has the form

\sum (-1)^n a_n

\sum (-1)^{n+1} a_n

where

a_n > 0

. The signs flip each term. The Alternating Series Test says the series converges if two conditions both hold:

First, the terms decrease in magnitude:

a_{n+1} \le a_n

for all

n

beyond some starting index. Second, the terms shrink to zero:

\lim_{n\to\infty} a_n = 0

.

Both conditions are about the positive part

a_n

, never the signed term. A common mistake is to test

(-1)^n a_n \to 0

; instead you must confirm

a_n \to 0

where

a_n = |(-1)^n a_n|

.

If the limit of

a_n

is not zero, the series diverges by the nth-Term Test—the AST is not even needed. If the limit is zero but the terms are not decreasing, the AST is inconclusive and you must try another approach.

To verify that terms decrease, you can compare

a_{n+1}

a_n

algebraically, or treat

a_n

as a function

f(x)

and show

f'(x) < 0

. On the exam, clearly stating both conditions is what earns the justification point.

Condition	What to check	Why it matters
Decreasing	$a_{n+1}\le a_n$	Partial sums bracket the true sum
Limit zero	$\lim a_n = 0$	Necessary for any convergence

Why the test works: bracketing partial sums

Understanding the mechanism makes the error bound obvious. Consider the partial sums

S_1, S_2, S_3, \dots

of an alternating series with decreasing terms. Because each term is smaller than the last and alternates sign, the partial sums oscillate—overshooting, then undershooting the true sum

S

.

The odd partial sums form one sequence closing in from one side, and the even partial sums close in from the other. Each new term is smaller, so the gap between consecutive partial sums shrinks toward zero. Since the terms go to zero, these two sequences squeeze together and trap the true value

S

between every consecutive pair

S_N

and

S_{N+1}

.

This bracketing is the whole reason the series converges, and it is also why the error is so easy to bound. The true sum always lies between the last two partial sums you computed. So whatever partial sum

S_N

you stop at, the exact answer is no farther away than the very next term.

A frequent misconception: students think the AST tells you the series converges absolutely. It does not. The AST only guarantees convergence—possibly conditional. Whether it converges absolutely is a separate question answered by testing

\sum a_n

directly (covered in U10.8).

The Alternating Series Error Bound

If an alternating series satisfies the AST conditions and converges to

S

, then the error from stopping at the

N

th partial sum

S_N

satisfies

|S - S_N| \le |a_{N+1}|.

In words: the error is no bigger than the magnitude of the first term you did not include. This is remarkably clean—no derivatives, no factorials to estimate, unlike the Lagrange error bound in U10.11.

Be careful with indexing. If

S_N

includes terms up through index

N

, then

a_{N+1}

is the next term. Always identify which term comes after your last included term and take its absolute value.

A typical exam task: find how many terms guarantee accuracy within a tolerance

\varepsilon

. You solve the inequality

|a_{N+1}| \le \varepsilon

for

N

. For example, to approximate

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

within

0.01

, you need

\frac{1}{N+1} \le 0.01

, so

N+1 \ge 100

, meaning

N \ge 99

terms.

The bound only applies once the terms are actually decreasing. If the first few terms increase before settling, apply the bound starting from where monotonic decrease begins. Also, the bound requires

a_n \to 0

; without it, the series diverges and no partial sum approximates anything.

How the AP exam tests this

Expect three flavors of questions. First, justification: given a series, state that terms are positive and decreasing and that they tend to zero, then conclude convergence by the AST. Skipping either condition loses points even if the conclusion is right.

Second, error estimation: given a partial sum approximation, state a bound on the error using

|a_{N+1}|

, or determine the minimum number of terms needed for a target accuracy. These appear both as multiple-choice and as parts of FRQs involving Taylor and Maclaurin series.

Third, sign reasoning: because partial sums bracket the true value, you may be asked whether

S_N

overestimates or underestimates

S

. If the next term

a_{N+1}

is added with a positive sign, then

S_N

is an underestimate; if negative, an overestimate.

Task	Key move
Prove convergence	Verify decreasing and limit zero, cite AST
Bound error	Compute $\|a_{N+1}\|$
Find N for accuracy	Solve $\|a_{N+1}\|\le \varepsilon$
Over/under estimate	Check sign of first omitted term

Watch out: the error bound is a bound, not the exact error. Saying the error equals

|a_{N+1}|

is wrong—it is at most that.

Connecting to Taylor series

Many alternating series come from evaluating a Maclaurin series at a specific point, and this is where the error bound becomes a powerful shortcut. The series for

\cos x

\sin x

, and

\ln(1+x)

(for suitable

x

) all alternate.

For instance,

\cos 1 = 1 - \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!} + \cdots

is alternating with decreasing terms tending to zero. If you approximate

\cos 1

with the first three terms, the error is at most

\frac{1}{6!} = \frac{1}{720}

.

This often gives a tighter, easier bound than the Lagrange error bound from U10.11. When a Taylor approximation produces an alternating series with decreasing terms, you may use the alternating series bound instead of Lagrange—and AP scoring accepts either when applicable.

The key precondition is that the numerical series really does alternate with decreasing magnitudes. If plugging in a value produces terms that do not strictly decrease at first, the alternating bound may not apply immediately, and you would fall back on Lagrange. Recognizing which tool fits is part of what makes this topic worth mastering.

Key terms

Alternating series.: A series whose terms alternate in sign, written $\sum (-1)^n a_n$ or $\sum (-1)^{n+1} a_n$ with $a_n > 0$ .
Alternating Series Test (AST).: A test stating that $\sum (-1)^n a_n$ converges if $a_n$ is eventually decreasing and $\lim_{n\to\infty} a_n = 0$ .
Partial sum $S_N$ .: The sum of the first terms of a series up through index $N$ ; used to approximate the full sum $S$ .
Alternating series error bound.: For a convergent alternating series satisfying the AST, $|S - S_N| \le |a_{N+1}|$ , the size of the first omitted term.
Conditional convergence.: When a series converges but the series of absolute values diverges; alternating series often converge only conditionally.
nth-Term Test.: If $\lim_{n\to\infty} a_n \ne 0$ , the series diverges; used to rule out convergence before applying the AST.
Monotonic decrease.: The property $a_{n+1} \le a_n$ , required so partial sums bracket the true sum.

Worked example

Consider the series

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

. (a) Show it converges. (b) Approximate the sum using the first three terms and bound the error.

Part (a): Write the positive part as

a_n = \frac{1}{n^2}

. Check the two AST conditions.

Decreasing: since

n^2

increases,

\frac{1}{n^2}

decreases, so

a_{n+1} = \frac{1}{(n+1)^2} < \frac{1}{n^2} = a_n

for all

n \ge 1

. Limit:

\lim_{n\to\infty} \frac{1}{n^2} = 0

.

Both conditions hold, so by the Alternating Series Test the series converges.

Part (b): The first three terms give

S_3 = \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} = 1 - 0.25 + 0.1111\ldots = 0.8611\ldots

The error bound is the magnitude of the next term,

a_4 = \frac{1}{4^2} = \frac{1}{16} = 0.0625

.

So

|S - S_3| \le 0.0625

. The true sum lies within

0.0625

0.8611

. Since the fourth term

-\frac{1}{16}

is negative,

S_3

is an overestimate: the true sum is slightly less than

0.8611

Practice questions

How many terms of the series

\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^3}

are needed to approximate its sum with error less than

0.001

7 terms
8 terms
9 terms
10 terms

Answer: 9 terms

The error bound requires

|a_{N+1}| = \frac{1}{(N+1)^3} \le 0.001 = \frac{1}{1000}

. So

(N+1)^3 \ge 1000

, giving

N+1 \ge 10

, hence

N \ge 9

. Using

N=9

terms guarantees the error is at most

\frac{1}{10^3}=0.001

. The first omitted term is the 10th term, whose magnitude meets the tolerance.

The alternating series

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

converges to

S

. If you use the first four terms as an approximation

S_4

, state a bound on

|S - S_4|

and determine whether

S_4

is an overestimate or underestimate.

Answer: The error is at most $\frac{1}{9}$ , and $S_4$ is an underestimate.

Here

a_n = \frac{1}{2n-1}

, which decreases to zero, so the AST applies. The fifth term is

\frac{(-1)^{6}}{2(5)-1} = +\frac{1}{9}

. The error bound is

|a_5| = \frac{1}{9}

. Because the first omitted term is positive (it adds to the sum),

S_4

falls short of the true value, making it an underestimate.

Explain why the series

\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n+1}

diverges even though its terms alternate.

Answer: It diverges by the nth-Term Test because the terms do not approach zero.

For the AST to guarantee convergence, we need

a_n = \frac{n}{n+1} \to 0

. But

\lim_{n\to\infty} \frac{n}{n+1} = 1 \ne 0

. Since the terms fail to shrink to zero, the alternating series diverges by the nth-Term Test. Alternating signs alone never force convergence—the magnitude must tend to zero.

FAQ

When can I use the alternating series error bound instead of the Lagrange error bound?: Use it whenever your numerical series genuinely alternates in sign with terms decreasing in magnitude toward zero. Many Taylor approximations, like $\cos 1$ or $\ln(1.5)$ , produce such series, and the alternating bound $|a_{N+1}|$ is usually simpler than Lagrange. If the terms do not strictly decrease, use Lagrange instead.
Does the Alternating Series Test prove absolute convergence?: No. The AST only proves the series converges—possibly conditionally. To determine absolute convergence, test the series of absolute values $\sum a_n$ separately using tools like the p-series or comparison tests. A series can pass the AST yet have $\sum a_n$ diverge, meaning it converges conditionally.
What if the terms increase at first before decreasing?: The AST and the error bound only require the terms to eventually decrease. Convergence still holds as long as the decrease begins after some index and the limit is zero. However, apply the error bound only from the point where the terms are actually decreasing; otherwise the bracketing argument fails.
Is the error bound the exact error?: No—it is an upper bound. The true error $|S - S_N|$ is at most $|a_{N+1}|$ , but usually smaller. On the exam, write it as an inequality with $\le$ , never as an equality. Claiming the error equals the next term is a common mistake that loses credit.

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