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The Alternating Series Test

Calculus 2 · Chapter 7: Convergence Tests for SeriesSection 7.3 · lesson 34 of 5410 min · free

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In this lesson

When changing signs rescue a series.

  • Check both Leibniz conditions before concluding
  • Use the derivative when 'decreasing' is not obvious
  • Bound the truncation error by the first omitted term
  • Distinguish absolute from conditional convergence

Transcript

Everything said in the video, section by section (1,433 words). Read along, or search it.

MAT-265 - Calculus 2 · Section 7.3

The Alternating Series Test

Every test so far has required positive terms. The Integral Test, the comparison tests — all of them assume the series is adding up positive quantities. But many important series alternate in sign. Plus, minus, plus, minus. And something remarkable happens when they do. The alternating harmonic series — one minus one half plus one third minus one quarter — converges. Even though the ordinary harmonic series, with all the same magnitudes, diverges. The signs alone rescue it. This section explains when that rescue works, and gives a bonus no other test provides: a built-in error bound.

The Test

Two Leibniz Conditions

Here's the test, due to Leibniz. The alternating series, sum of minus one to the n minus one, times b n — with every b n positive — converges if two conditions hold. Condition one. Decreasing. The magnitudes are eventually non-increasing. b n plus one is at most b n, for all sufficiently large n. Condition two. Limit zero. The magnitudes tend to zero. Both conditions are required. Neither alone suffices. And the same statement holds for the form with minus one to the n, since the two differ only by an overall sign. The intuition behind the test is geometric, and the picture makes it clear. Because the terms alternate and shrink, each partial sum overshoots the true total. Then the next one undershoots it. By an ever-smaller amount. So the partial sums oscillate inward, like a tightening spiral. They trap the limit between consecutive partial sums, and converge to it. And the limit-zero condition is exactly what ensures the oscillations actually close in on a single value — rather than bouncing forever.

Worked Examples 1 and 2

Both Conditions Must Be Checked

Two examples, showing why both conditions matter. First. Show that the alternating harmonic series converges. The magnitude is b n equals one over n. Which is positive. Check condition one, decreasing. One over n plus one is less than one over n. Yes. Check condition two, limit zero. The limit of one over n is zero. Yes. Both conditions hold. So the series converges by the Alternating Series Test. And notice — this is the series from the opening. It converges even though its non-alternating cousin, the harmonic series, diverges. Second. Determine whether the sum of minus one to the n minus one, times n over n plus one, converges. The magnitude is n over n plus one. Check condition two first, because it's often quickest to rule out. The limit of n over n plus one is one. Not zero. So condition two fails, and the Alternating Series Test does not apply. And in fact — because the terms do not approach zero — the series diverges. By the Test for Divergence, from section six point two. Alternation cannot save a series whose terms fail to shrink to zero.

Worked Example 3

When 'Decreasing' Is Not Obvious

Sometimes the decreasing condition isn't obvious. Here's the reliable technique. Show that the sum of minus one to the n minus one, times n over n squared plus one, converges. Condition two is easy. The limit of n over n squared plus one is zero, since the denominator grows faster. But condition one — is n over n squared plus one decreasing? It's not obvious by inspection. The numerator grows, which pushes up. The denominator grows faster, which pushes down. So treat b n as a continuous function. f of x equals x over x squared plus one. And check its derivative. By the quotient rule, f prime is x squared plus one, times one, minus x times two x — all over x squared plus one, squared. The numerator simplifies to one minus x squared. Now read off the sign. For x greater than one, one minus x squared is negative. So f prime is negative, and f is decreasing. Which means b n plus one is less than b n, for n at least one. Both conditions hold, so the series converges. Using the derivative is the reliable way to confirm the decreasing condition whenever it can't be seen directly.

The Practical Payoff

A Built-In Error Bound

Now the most valuable practical feature of an alternating series. Its partial sums come with a built-in error guarantee. When the Leibniz conditions hold, stopping the sum after N terms produces an error no larger than the very next term. The first one omitted. Formally: the remainder, S minus S sub N, is bounded in size by b sub N plus one. And the reason is exactly that inward-spiraling behavior. Since the true sum lies between consecutive partial sums S N and S N plus one — and those differ by exactly b N plus one — the distance from S N to S cannot exceed b N plus one. No other elementary convergence test offers such a simple, sharp bound. Which is why alternating series are prized in numerical computation. A single glance at the next unused term certifies how accurate your current approximation is. Let's use it. Approximate the sum of minus one to the n minus one over n cubed, using the first three terms, and bound the error. The third partial sum is one, minus one eighth, plus one twenty-seventh. About zero point nine one two. The error is bounded by the first omitted term, b four. Which is one over four cubed. One sixty-fourth. About zero point zero one five six. So the true sum lies within about zero point zero one six of zero point nine one two. Three terms already pin it to two decimal places. Now the reverse question. How many terms of the alternating harmonic series are needed to approximate its sum, log two, with error less than zero point zero one? The estimation theorem says the error after N terms is at most one over N plus one. To force that below zero point zero one, you need N plus one greater than one hundred. So N greater than ninety-nine. One hundred terms. Note how slowly that series converges. A hundred terms for just two-decimal accuracy. That's a direct consequence of its conditional, cancellation-driven convergence. Faster-shrinking alternating series need far fewer terms.

Classification

Absolute Versus Conditional Convergence

The contrast between those two examples points to an important classification. An alternating series can converge for one of two fundamentally different reasons. If the series of magnitudes converges on its own, the original series is said to converge absolutely. Its convergence does not depend on the sign cancellation at all. The example: sum of minus one to the n minus one over n squared. The magnitudes give one over n squared, a convergent p-series. If instead the series of magnitudes diverges, but the alternating series still converges — thanks purely to the sign pattern — the convergence is conditional. The example: the alternating harmonic series. Its magnitudes give the divergent harmonic series. Now, why does this distinction matter in practice? Absolutely convergent series can be rearranged freely without changing the sum. Shuffle the terms however you like; the total is the same. Conditionally convergent series cannot. Rearranging their terms can change the sum to any value at all. Any real number you choose. That startling result is known as the Riemann rearrangement theorem. So conditional convergence is a genuinely more fragile kind of convergence.

Key Takeaways

Three Recurring Errors

Three errors recur with this test, and it's worth naming them. First. Students sometimes check only that the terms go to zero, and forget to confirm the magnitudes are decreasing. Both Leibniz conditions are required. The limit-zero condition alone does not guarantee convergence of an alternating series. Second. The test is a one-way criterion for convergence. If the conditions hold, the series converges. But if the decreasing condition fails, the test is simply inconclusive. It does not prove divergence. Only a failure of the limit-zero condition proves divergence — and that's the nth-term test doing the work, not the alternating series test. Third. The estimation bound uses the first omitted term. b sub N plus one. Not the last included term. Confusing the two overstates or understates the error. A reliable habit is to write out the conditions as a checklist, confirm both before claiming convergence, and index the error term carefully as one past where you stopped. Four takeaways. Check both conditions. Terms going to zero alone does not guarantee convergence of an alternating series. The test is one-way. A failed decreasing condition makes it inconclusive, not a proof of divergence. The error bound uses the first omitted term, indexed as one past where you stopped. And absolute convergence means the magnitudes converge on their own. Conditional convergence relies on cancellation, and forbids free rearrangement. Next up is section seven point four — Absolute and Conditional Convergence, developing that distinction in full.

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FAQ

Questions students ask us

About The Alternating Series Test

What are the two conditions of the Alternating Series Test?
The series \sum (-1)^{n-1} b_n with b_n > 0 converges if the magnitudes are eventually decreasing, b_{n+1} \le b_n, and if \lim_{n\to\infty} b_n = 0. Both conditions are required, because neither one alone is enough. For example, the alternating harmonic series \sum (-1)^{n-1}/n satisfies both and converges, even though the harmonic series \sum 1/n diverges.
How do you prove the terms of an alternating series are decreasing when it isn't obvious?
Treat b_n as a continuous function f(x) and show that f'(x) < 0. For b_n = n/(n^2+1), the quotient rule gives f'(x) = (1-x^2)/(x^2+1)^2. This is negative for x > 1, so f is decreasing and b_{n+1} < b_n for n \ge 1. Because b_n \to 0 as well, the series \sum (-1)^{n-1} n/(n^2+1) converges.
How do you estimate the error of an alternating series partial sum?
When both Leibniz conditions hold, the error is at most the first omitted term: |S - S_N| \le b_{N+1}. For \sum (-1)^{n-1}/n^3, the partial sum S_3 = 1 - 1/8 + 1/27 \approx 0.912. The error is at most b_4 = 1/64 \approx 0.0156. A common mistake is to use the last included term instead of the next one.
If an alternating series fails the Alternating Series Test, does it diverge?
Not necessarily, because the test only proves convergence. If the decreasing condition fails, the test is inconclusive. If b_n does not approach 0, the series diverges, but that conclusion comes from the Test for Divergence. For example, \sum (-1)^{n-1} n/(n+1) diverges because n/(n+1) \to 1. Alternating signs cannot rescue terms that fail to shrink to zero.

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