Alternating Series Test for Convergence

3. Visual Connection

5. Concept Development

BC-only topic. Alternating signs can create cancellation strong enough for convergence even when the corresponding positive series diverges. The Alternating Series Test makes that idea precise.

Recognize an alternating series

A standard alternating series has one of the forms

The factor \((-1)^n\) controls the sign, while \(b_n=|a_n|\) is the positive magnitude. Write the first several terms to verify that signs truly alternate every term.

Alternating Series Test

Both conditions must be addressed. “Eventually decreasing” means the inequality may begin after finitely many exceptional terms.

Why the test works

For a series beginning \(b_1-b_2+b_3-b_4+\cdots\), the even partial sums increase because each added pair \(b_{2k-1}-b_{2k}\) is nonnegative. The odd partial sums decrease because each update \(-b_{2k}+b_{2k+1}\) is nonpositive. The two sequences approach the same value because

Thus the full partial-sum sequence converges.

Geometric picture of the partial sums

When the first term is positive, the partial sums move back and forth across the sum with shrinking steps. Under the test's hypotheses,

This bracketing leads to the error estimate developed in Topic 10.10.

Check the zero limit first

If \(b_n\not\to0\), the signed terms \((-1)^n b_n\) do not approach zero, and the series diverges immediately by the nth term test. There is no need to investigate monotonicity.

Method 1: Compare consecutive formulas

Show \(b_{n+1}\le b_n\) by direct algebra. For reciprocal powers,

For positive quotients, cross-multiply only after confirming the denominators are positive.

Method 2: Differentiate a continuous extension

Choose \(f(x)\) with \(f(n)=b_n\). If \(f'(x)<0\) for all sufficiently large \(x\), then \(b_n\) is eventually decreasing. This is useful for logarithmic or rational magnitudes.

Method 3: Use a magnitude ratio

If

eventually, the magnitudes decrease. This can be efficient for expressions involving exponentials or factorials.

Decreasing need not be strict

The theorem permits \(b_{n+1}\le b_n\); occasional equal magnitudes do not cause failure. What matters is eventual nonincrease together with the limit zero.

The conditions are sufficient, not necessary

If monotonicity cannot be established, the Alternating Series Test is inconclusive rather than proof of divergence. Another convergence argument might still work. However, alternating signs plus \(b_n\to0\) alone also do not guarantee convergence.

Absolute versus conditional convergence

The Alternating Series Test proves convergence of the signed series. It does not determine whether

converges. Topic 10.9 distinguishes absolute from conditional convergence by testing the magnitude series separately.

Finite initial terms

Changing or removing finitely many terms does not affect convergence. If the magnitudes become decreasing after \(n=N\), apply the test to the tail and then restore the finite initial sum.

AST checklist

1. Write the first terms and verify one-by-one sign alternation.
2. Define the positive magnitude \(b_n=|a_n|\).
3. Compute \(\lim b_n\); if it is not zero, conclude divergence.
4. Prove \(b_{n+1}\le b_n\) eventually.
5. Conclude convergence by the Alternating Series Test.
6. If requested, separately investigate absolute or conditional convergence.

6. Detailed Worked Example and Error Check

Example 1: Alternating harmonic series

For

the magnitudes \(b_n=1/n\) decrease and approach zero. Therefore the series converges by the Alternating Series Test.

Example 2: Alternating reciprocal squares

For \(\sum(-1)^{n+1}/n^2\),

The series converges by the AST. In fact, its magnitude series also converges, so it is absolutely convergent.

Example 3: Alternation cannot rescue nonzero terms

For \(\sum(-1)^{n+1}n/(n+1)\), the magnitudes approach \(1\), not \(0\). The signed terms do not approach zero, so the series diverges by the nth term test.

Example 4: Prove decrease with a derivative

Let \(b_n=(n+2)/(n^2+1)\). The extension \(f(x)=(x+2)/(x^2+1)\) satisfies

and \(b_n\to0\). Therefore \(\sum(-1)^n(n+2)/(n^2+1)\) converges.

Example 5: Eventually decreasing logarithmic magnitudes

For \(\sum_{n=2}^{\infty}(-1)^n\ln n/n\), let \(f(x)=\ln x/x\). Then

and \(\ln n/n\to0\). The finite terms before decrease begins do not matter, so the series converges by the AST.

Example 6: A fractional-power magnitude

The magnitudes \(b_n=1/\sqrt n\) decrease to zero. Hence \(\sum(-1)^{n+1}/\sqrt n\) converges by the AST, even though \(\sum1/\sqrt n\) diverges.

Example 7: Alternating signs and zero limit are not enough

Define \(b_n=(2+(-1)^n)/n\). Then \(b_n>0\) and \(b_n\to0\), but the magnitudes repeatedly rise from an odd term to the next even term. For the signed series \(\sum(-1)^n b_n\), pair consecutive terms:

The paired positive lower bound forms a divergent harmonic multiple, so the series diverges. Monotone decrease was not decorative.

Example 8: Irregular beginning, regular tail

Suppose \(b_1=1\), \(b_2=4\), and \(b_n=1/n\) for \(n\ge3\). The full magnitude sequence is not decreasing at the start, but the tail decreases to zero. Therefore \(\sum(-1)^n b_n\) converges.

Example 9: Use a ratio to show decrease

For \(b_n=n/2^n\),

and \(n/2^n\to0\). Thus \(\sum(-1)^n n/2^n\) converges by the AST.

Example 10: A shifted reciprocal magnitude

For \(b_n=1/(n+\ln n)\), the denominator increases and tends to infinity. Hence \(b_n\) decreases to zero, so \(\sum_{n=1}^{\infty}(-1)^{n+1}/(n+\ln n)\) converges.

Common errors

• Concluding convergence from alternating signs alone.
• Checking \(b_n\to0\) but omitting monotonicity.
• Checking monotonicity while the term limit is nonzero.
• Using the signed term \(a_n\) instead of its magnitude \(b_n\) in the decrease condition.
• Demanding decrease from the first term rather than eventually.
• Assuming AST convergence automatically means absolute convergence.
• Declaring divergence when AST is merely inconclusive.
• Failing to verify the displayed signs actually alternate every term.
• Cross-multiplying inequalities without checking denominator signs.
• Using the alternating error bound before verifying the AST hypotheses.