Aim for FiveCalculus BC

Unit 10 · Topic 10.7

BC only

10.7 Alternating Series Test for Convergence

BC only. An alternating series switches between positive and negative terms. It converges if the sizes of its terms decrease and approach 0. That's why the alternating harmonic series 1 − 1/2 + 1/3 − 1/4 + … converges even though the harmonic series doesn't.

Key terms

  • alternating series
  • alternating series test
  • decreasing terms
  • alternating harmonic series

Alternating series

This whole unit is BC only. An alternating series has the form Σ (−1)ⁿ bₙ or Σ (−1)ⁿ⁺¹ bₙ, where every bₙ > 0. The (−1)ⁿ factor flips the sign each time. Sometimes it's hidden: cos(nπ) = (−1)ⁿ, so Σ cos(nπ)/n is alternating.

The alternating series test

If an alternating series Σ (−1)ⁿ bₙ, with bₙ > 0, satisfies both conditions below, it converges:

  • The sizes decrease: bₙ₊₁ ≤ bₙ for all n (or for all n past some point).
  • The sizes approach 0: lim (n → ∞) bₙ = 0.

Why it works

Picture the partial sums on a number line. You step forward b₁, back b₂, forward b₃, and so on. Each step is shorter than the one before, so you never get back to where you were, and the partial sums bounce back and forth in a narrower and narrower zone. Since the steps shrink to 0, the zone shrinks to a single point: the sum.

This picture also explains the error bound in 10.10. The true sum always lies between any two consecutive partial sums.

Alternating series you'll see often

  • The alternating harmonic series 1 − 1/2 + 1/3 − 1/4 + … converges, even though the harmonic series diverges. It's the classic example of conditional convergence (10.9).
  • Σ (−1)ⁿ/ln n (from n = 2) converges: 1/ln n is positive, decreasing and approaches 0, even though it shrinks very slowly.
  • Taylor series for sin x and cos x alternate at every x ≠ 0 (10.14), which is why the alternating series error bound (10.10) is so useful for them.

Checking the conditions

To show bₙ decreases, show bₙ₊₁ ≤ bₙ directly, or show that the related function f(x) has f′(x) < 0 for large x. If lim bₙ ≠ 0, the series diverges by the nth term test, not by the alternating series test. If the terms go to 0 but don't decrease, the alternating series test doesn't apply, and you'd need another method.

The alternating series test can only prove convergence. Failing it doesn't prove divergence by itself.

When the alternating series test works, it says nothing about absolute convergence. To classify a series fully, also check Σ bₙ (10.9).

Worked examples

Try each one yourself first, then open the solution.

  1. Example 1

    A standard alternating series

    Does Σ (n = 1 to ∞) (−1)ⁿ⁺¹/√n converge?

    Show the solution
    1. Step 1: It's alternating with bₙ = 1/√n > 0.
    2. Step 2: Decreasing: √(n + 1) > √n, so 1/√(n + 1) < 1/√n.
    3. Step 3: Limit: lim (n → ∞) 1/√n = 0.
    4. Step 4: Both conditions hold, so the series converges by the alternating series test.

    Answer: Converges (alternating series test).

  2. Example 2

    Trap: terms that don't go to 0

    Does Σ (n = 1 to ∞) (−1)ⁿ n/(n + 1) converge?

    Show the solution
    1. Step 1: bₙ = n/(n + 1) → 1, not 0.
    2. Step 2: The alternating series test doesn't apply. Students sometimes say “it alternates, so it converges.”
    3. Step 3: The terms (−1)ⁿ n/(n + 1) don't approach 0 (they approach ±1), so the series diverges by the nth term test.

    Answer: Diverges (nth term test).

  3. Example 3

    Showing decrease with a derivative

    Does Σ (n = 1 to ∞) (−1)ⁿ n/(n² + 1) converge?

    Show the solution
    1. Step 1: bₙ = n/(n² + 1) > 0. Limit: n/(n² + 1) → 0.
    2. Step 2: Decreasing: let f(x) = x/(x² + 1). Then f′(x) = (1 − x²)/(x² + 1)², which is ≤ 0 for x ≥ 1. So bₙ is decreasing for n ≥ 1.
    3. Step 3: Both conditions hold, so the series converges by the alternating series test.

    Answer: Converges (alternating series test).

Common mistakes

  • Concluding convergence just because the series alternates.
  • Checking only that bₙ → 0 and skipping the decreasing condition, or the reverse.
  • Saying a series diverges “by the alternating series test.” That test never proves divergence.
  • Including the (−1)ⁿ in bₙ. Test the positive part only.

On the exam

  • On free response, list both conditions explicitly: “bₙ = 1/√n is positive and decreasing, and lim bₙ = 0, so the series converges by the alternating series test.”
  • Alternating series often show up as the endpoint of an interval of convergence (10.13).

Connected topics

Videos

Check yourself

4 questions on 10.7 Alternating Series Test for Convergence. Pick an answer to see if you got it, and why.

Question 1 of 4

Which of the following series converge? I. Σ (n = 1 to ∞) (−1)ⁿ n/(n + 1) II. Σ (n = 2 to ∞) (−1)ⁿ/ln n III. Σ (n = 1 to ∞) (−1)ⁿ/∛n

Question 2 of 4

Which of the following correctly describes the series Σ (n = 1 to ∞) (−1)ⁿ⁺¹·n/(n² + 1)?

Question 3 of 4

Which of the following series converge? I. Σ (n = 3 to ∞) (−1)ⁿ (ln n)/n II. Σ (n = 1 to ∞) (−1)ⁿ (n + 1)/n III. Σ (n = 1 to ∞) (−1)ⁿ/n^(1/4)

Question 4 of 4

For which values of p does the series Σ (n = 1 to ∞) (−1)ⁿ/nᵖ converge?

0 of 4 answered