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Calculus II · Unit 4A · lesson

Alternating Series

Concept

Learning objectives

apply the Alternating Series Test and distinguish sign alternation from decreasing magnitude.

Alternating Series

Explanation

Cancellation can rescue a divergent positive series

A positive series may diverge because every term pushes partial sums in the same direction. Alternating signs can create cancellation: one partial sum overshoots, the next undershoots, and the oscillation narrows as term magnitudes shrink. This mechanism can produce convergence even when the corresponding positive series diverges.

Alternation alone is not enough. The magnitudes must eventually decrease and approach zero. These hypotheses make odd and even partial sums monotone in opposite directions and force them toward a common limit. The test establishes convergence but usually not the exact sum.

Decision

Three separate alternating-series questions

First verify that the magnitudes eventually decrease. Second verify that those magnitudes approach zero. That proves convergence by the Alternating Series Test. Only after that should you test ∑∣an∣\sum |a_n| to decide whether the convergence is absolute or conditional.

Bridge

Alternation works only when the overshoots shrink

An alternating series can converge because its partial sums approach the target from opposite sides. Odd partial sums form one monotone subsequence, even partial sums form another, and decreasing term magnitudes squeeze the two subsequences together.

Both conditions matter: the magnitudes must eventually decrease and must approach zero. Alternating signs alone create no guarantee. The theorem explains convergence through a geometric bracketing mechanism rather than through vague claims that positive and negative terms “cancel out.”

Alternating partial sums form shrinking brackets. Odd and even partial sums nesting around a common limit.
Read this graph as text

Alternating partial sums form shrinking brackets. Odd and even partial sums lie on opposite sides of the limit and the distance between them equals the next term magnitude. Odd and even partial sums nesting around a common limit.

Written labels, distinct line styles, markers, and fill patterns communicate every relationship in alternating partial sums form shrinking brackets; color is never the only cue.

Why it matters: Odd and even partial sums nesting around a common limit.

Alternating partial sums form shrinking brackets

Odd and even partial sums lie on opposite sides of the limit and the distance between them equals the next term magnitude.

Alternating partial sums form shrinking brackets. Odd and even partial sums nesting around a common limit.

Proof idea

Two subsequences close in on the same number

With decreasing bn→0b_n\to0, even partial sums move in one direction and odd partial sums in the other. Their difference is the next term magnitude, which tends to zero, so both subsequences converge to the same limit.

Alternating partial sums bracket the limit. Odd and even partial sums nesting around a common limit.
Read this graph as text

Alternating partial sums bracket the limit. When alternating term magnitudes decrease to zero, odd and even partial sums approach the same value from opposite sides. Odd and even partial sums nesting around a common limit.

Written labels, distinct line styles, markers, and fill patterns communicate every relationship in alternating partial sums bracket the limit; color is never the only cue.

Why it matters: Odd and even partial sums nesting around a common limit.

Alternating partial sums bracket the limit

When alternating term magnitudes decrease to zero, odd and even partial sums approach the same value from opposite sides.

Alternating partial sums bracket the limit. Odd and even partial sums nesting around a common limit.

How to read the visual

Odd partial sums decrease while even partial sums increase. The next omitted term bounds the remaining vertical gap to the limit.

Concept

Alternating Series Test

If bn≥0b_n\ge0, bn+1≤bnb_{n+1}\le b_n eventually, and bn→0b_n\to0, then

∑n=1∞(−1)n−1bn\sum_{n=1}^{\infty}(-1)^{n-1}b_n

converges.

Guided walkthrough

The alternating harmonic series

For bn=1/nb_n=1/n, the magnitudes decrease to zero. Therefore

1−12+13−14+⋯1-\frac12+\frac13-\frac14+\cdots

converges, even though ∑1/n\sum1/n diverges.

Worked example

Decrease may begin after a few terms

Test

∑n=2∞(−1)nln⁡nn.\sum_{n=2}^{\infty}(-1)^n\frac{\ln n}{n}.

Let bn=ln⁡n/nb_n=\ln n/n. We have bn→0b_n\to0. For x>ex>e,

ddxln⁡xx=1−ln⁡xx2<0,\frac{d}{dx}\frac{\ln x}{x}=\frac{1-\ln x}{x^2}<0,

so bnb_n is decreasing for all sufficiently large nn. Therefore the series converges by the Alternating Series Test.

Common mistake

Check magnitudes, not signed terms

The sequence (−1)nbn(-1)^n b_n is usually not decreasing. The theorem requires the positive magnitudes bnb_n to decrease eventually.

Interactive checku4a-alternating_series_test-01

Does ∑n=1∞(−1)n−1/n\sum_{n=1}^{\infty}(-1)^{n-1}/n converge?

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Show hint

Check decreasing magnitudes and the limit of 1/n1/n.

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Exercise

Classify ∑(−1)n/n\sum(-1)^n/\sqrt n.

Exercise

Explain why ∑(−1)n\sum(-1)^n diverges.

Exercise

Find an alternating series for which magnitudes are not initially decreasing but eventually are.

Exercise

Describe the behavior of odd and even partial sums.

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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