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Reading Time: 5 min
Last Updated: March 27, 2026
Main Ideas: 4
Reading Time: 5 min
Last Updated: March 27, 2026
Main Ideas: 4

Topic 10.10 Notes – Alternating Series Error Bound

Verified for 2027 AP® Calculus BC Exam
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If a series converges by the Alternating Series Test, there’s a simple and powerful way to control the error. The key idea is that the first omitted term tells you how wrong your approximation could be.

1. The Alternating Series Error Bound

We’re working with alternating series of the form

∑n=1∞(−1)nanor∑n=1∞(−1)n−1an \sum_{n=1}^\infty (-1)^n a_n \quad \text{or} \quad \sum_{n=1}^\infty (-1)^{n-1} a_n

where the signs switch back and forth.

Before you can use the error bound, the series must converge by the Alternating Series Test (AST). That means:

  • an>0a_n > 0
  • ana_n is decreasing
  • lim⁡n→∞an=0\lim_{n\to\infty} a_n = 0

If those are true, then not only does the series converge, but we also get this:

∣S−SN∣≤aN+1 |S - S_N| \le a_{N+1}

Where:

  • SS = the true infinite sum
  • SNS_N = the sum of the first NN terms
  • aN+1a_{N+1} = the first omitted term (without the alternating sign)

That’s the whole theorem. The error is at most the size of the next term.

And this only works for alternating series that pass AST.

2. How to Use the Error Bound

There are three ways this shows up on quizzes and FRQs.

A. “Find the error bound after N terms”

Suppose

∑n=1∞(−1)n−11n2 \sum_{n=1}^\infty (-1)^{n-1} \frac{1}{n^2}

Using 4 terms means the first omitted term is:

a5=152=125 a_5 = \frac{1}{5^2} = \frac{1}{25}

So:

∣S−S4∣≤125 |S - S_4| \le \frac{1}{25}

That fraction is your error bound. You don’t even need the partial sum unless they ask for it.

B. “Approximate the sum and give an interval”

Say you compute S3=1−14+19S_3 = 1 - \frac14 + \frac19.

The next term is a4=116a_4 = \frac{1}{16}.

So:

∣S−S3∣≤116 |S - S_3| \le \frac{1}{16}

Turn that into an interval:

S3−116≤S≤S3+116 S_3 - \frac{1}{16} \le S \le S_3 + \frac{1}{16}

That interval guarantees the true sum is inside.

On FRQs, writing the inequality clearly usually earns the point.

C. “How many terms for error < 0.001?”

This is the one that trips people up.

You use:

aN+1<0.001 a_{N+1} < 0.001

If an=1n2a_n = \frac{1}{n^2}, then:

1(N+1)2<0.001 \frac{1}{(N+1)^2} < 0.001

Solve it:

(N+1)2>1000 (N+1)^2 > 1000

N+1>1000≈31.6 N+1 > \sqrt{1000} \approx 31.6

N+1=32⇒N=31 N+1 = 32 \Rightarrow N = 31

Always round up. You’re guaranteeing the error is small enough.

Notice you didn’t compute any partial sums. You only needed the next term.

3. What’s Actually Happening

Alternating partial sums zig-zag toward the true value.

Here’s the picture to have in your head as you think about the alternating harmonic series:

Study guide illustration

Partial sums of the alternating harmonic series approaching ln 2

Each partial sum lands on opposite sides of the true sum, shown by the horizontal line.

Because:

  • Signs alternate
  • Terms get smaller

The next term represents the biggest possible overshoot. Once terms shrink, the error automatically shrinks.

That’s why the first omitted term controls everything.

Also, the true sum always lies between two consecutive partial sums. That’s a common AP multiple choice idea.

4. Common Exam Traps

  • Forgetting to verify AST first. If it’s not decreasing or the limit isn’t zero, the theorem doesn’t apply.
  • Using aNa_N instead of aN+1a_{N+1}. If you use S6S_6, the bound uses the 7th term.
  • Keeping the sign. The error bound is positive. Use the absolute value of the next term.
  • Rounding down when solving for N. That ruins the guarantee.
  • Mixing this up with Taylor polynomial error bounds. Totally different theorem.

On the AP exam, they often combine this with a “justify convergence” step first. If you don’t state decreasing and limit zero, you can lose points even if your bound is correct.

Key Takeaways

The alternating series error bound only works if the series converges by AST.
The maximum error after NN terms is aN+1a_{N+1}.
To guarantee error less than ε\varepsilon, solve aN+1<εa_{N+1} < \varepsilon and round up.
The true sum always lies between consecutive partial sums of an alternating series that passes AST.
If you’re using SNS_N, the bound comes from the next term, never the last one you added.

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