Calculus II · Unit 4A · lesson

Alternating-Series Error Estimates

Concept

Learning objectives

use the first omitted term to control approximation error for a convergent alternating series.

Alternating-Series Error Estimates

Explanation

The next term measures the remaining uncertainty

For a decreasing alternating series, successive partial sums lie on opposite sides of the limit. The true sum is trapped between two consecutive partial sums, so the distance from either one to the limit cannot exceed the next term magnitude. This unusually clean estimate makes alternating series computationally attractive.

The bound is a guarantee, not an exact error. To meet a tolerance, solve $b_{N+1}<\varepsilon$ . Then check indexing carefully: the first omitted term after $s_N$ is term $N+1$ . If the series begins at another index, translate the term count rather than mechanically substituting $N+1$ .

Bridge

The next term controls the entire remaining error

For an alternating series satisfying the test, the true sum lies between every pair of consecutive partial sums. The error after $N$ terms is therefore no larger than the distance to the next endpoint of that bracket, namely $b_{N+1}$ .

This unusually simple bound turns convergence into a practical approximation method. To guarantee a tolerance, solve $b_{N+1}<\varepsilon$ . The sign of the first omitted term also tells which side of the true sum the approximation lies on.

Proof idea

The true sum stays inside the next bracket

The alternating partial sums nest around $S$ . Since $S$ lies between $s_N$ and $s_{N+1}$ , its distance from $s_N$ is at most $|s_{N+1}-s_N|=b_{N+1}$ .

Concept

Alternating remainder estimate

Under the Alternating Series Test hypotheses,

|S-s_N|\le b_{N+1}.

Guided walkthrough

Approximate $\ln2$

Using

\ln2=1-\frac12+\frac13-\cdots,

we want error below $0.01$ . Since the next term after $N$ terms has magnitude $1/(N+1)$ , require $1/(N+1)<0.01$ , so $N\ge100$ guarantees the tolerance.

Worked example

Guarantee four-decimal accuracy

For

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

how many terms guarantee error below $10^{-4}$ ? The alternating remainder satisfies

|R_N|\le \frac{1}{(N+1)^3}.

Require $(N+1)^3>10^4$ . Since $22^3=10648$ , $N+1=22$ works, so $N=21$ terms guarantee the requested accuracy.

Common mistake

Use the first omitted term

After summing through $n=N$ , the bound is $b_{N+1}$ , not $b_N$ . This one-index error can invalidate an accuracy claim.

Interactive checku4a-alternating_series_error-01

What bounds the error after $N$ terms of the alternating harmonic series?

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Show hint

Use the magnitude of the first omitted term.

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Exercise

How many terms guarantee error below $10^{-3}$ ?

Exercise

Approximate $\sum(-1)^{n-1}/n^2$ with four terms and state an error bound.

Exercise

Explain why the true sum lies between consecutive partial sums.

Exercise

Compare this error estimate with the integral remainder estimate.

After the explanation

Use the section idea

Reading lens

Separate sign behavior from magnitude, then use ratios or roots when powers and factorials dominate.

Mental model

Absolute convergence controls magnitude strongly enough to imply convergence; conditional convergence relies on cancellation.

Decision

Test absolute values first when practical, use the alternating-series hypotheses explicitly, and reserve ratio or root tests for matching algebraic structure.

Common trap

Calling any alternating-looking series convergent or treating a ratio/root limit of one as a verdict.

Check yourself

Can you state whether convergence is absolute, conditional, divergent, or still undecided?

Source & rights

Original instruction with traceable references.

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