AP Calculus: Alternating Series Error Bound Theorem (AP)

1. What You Need to Know

Why this matters (AP-level)

When you approximate an alternating series (or an alternating Taylor/Maclaurin series) by a partial sum, the Alternating Series Error Bound Theorem gives you a fast, reliable way to guarantee how close your approximation is—often without complicated remainder formulas.

On AP Calculus BC, this shows up when you:

• estimate a value like \ln(2) or \arctan(1) using a few terms,
• choose how many terms you need to get within a specified accuracy,
• justify whether an approximation is an overestimate or underestimate.

The theorem (state it cleanly)

Consider an alternating series written in the standard form

\sum_{n=1}^{\infty} (-1)^{n-1} b_n \quad \text{with} \quad b_n > 0.

If:

1. b_{n+1} \le b_n for all n (the magnitudes decrease), and
2. \lim_{n\to\infty} b_n = 0,

then the series converges (Alternating Series Test), and the error after using the nth partial sum S_n satisfies:

|R_n| = |S - S_n| \le b_{n+1}.

So: the absolute error is at most the first omitted term’s magnitude.

Critical AP takeaway: If you approximate an alternating series with decreasing term magnitudes, your error is bounded by the very next term.

Extra fact you can use: “between two partial sums”

For such alternating series,

S_{n+1} \le S \le S_n \quad \text{or} \quad S_n \le S \le S_{n+1}.

In words: the true sum lies between two consecutive partial sums.

And the error has the same sign as the next term a_{n+1}:

R_n = S - S_n \text{ has the sign of } a_{n+1}.

(Where a_n = (-1)^{n-1}b_n.)

2. Step-by-Step Breakdown

A) Using the theorem to bound approximation error

1. Put the series in alternating form

   • Identify a_n and rewrite as a_n = (-1)^{n-1}b_n (or (-1)^n b_n), where b_n > 0.

2. Check the conditions (don’t skip this on FRQs)

   • Decreasing: verify b_{n+1} \le b_n (often by reasoning, derivative, or monotonicity).
   • Limit: verify \lim_{n\to\infty} b_n = 0.

3. Compute the partial sum you’re using

   • S_n = a_1 + a_2 + \cdots + a_n.

4. Bound the error using the next term

   • |S - S_n| \le b_{n+1}.

5. (Optional but common) Decide over/under estimate

   • If the next term a_{n+1} > 0, then S - S_n > 0 so S_n is an underestimate.
   • If the next term a_{n+1} < 0, then S - S_n < 0 so S_n is an overestimate.

B) Choosing the number of terms for a desired accuracy

Goal: guarantee

|S - S_n| < \varepsilon.

1. Confirm it’s an alternating series with decreasing b_n and b_n \to 0.
2. Use the theorem requirement:

b_{n+1} < \varepsilon.

3. Solve that inequality for n and choose the smallest integer that works.

Micro worked walkthrough (generic)

Suppose you have

\sum_{n=1}^{\infty} (-1)^{n-1}\frac{1}{n^2}.

• Here b_n = \frac{1}{n^2}, decreasing and \to 0.
• If you use S_4, then

|S - S_4| \le b_5 = \frac{1}{25} = 0.04.

If you want error < 0.001:

b_{n+1} = \frac{1}{(n+1)^2} 1000 \Rightarrow n+1 > \sqrt{1000}.
So n+1 \ge 32 (since \sqrt{1000} \approx 31.62), hence n \ge 31.

3. Key Formulas, Rules & Facts

Core rules (high-yield)

What counts as b_n?

If

a_n = (-1)^{n-1}b_n,
then

b_n = |a_n|.

You are bounding with the magnitude of the next term:

|S-S_n| \le |a_{n+1}| = b_{n+1}.

“Decreasing” nuance (edge case)

You often only need b_n to be decreasing eventually (after some index). On AP, you typically:

• show it decreases for all n \ge 1, or
• state it decreases for n \ge N and you’re taking partial sums beyond that.

If the series alternates but b_n is not decreasing, you cannot use this theorem until you justify monotonic decrease (at least eventually).

4. Examples & Applications

Example 1: Approximate \ln(2) with an error guarantee

A classic AP series:

\ln(2) = \sum_{n=1}^{\infty} (-1)^{n-1}\frac{1}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots

Use S_4:

S_4 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4}.

Here b_n = \frac{1}{n} is decreasing and \to 0.

Error bound:

|\ln(2) - S_4| \le b_5 = \frac{1}{5} = 0.2.

Over/under?

• Next term a_5 = +\frac{1}{5} is positive, so \ln(2) - S_4 > 0.
• Therefore S_4 is an underestimate.

Example 2: How many terms for |\ln(2) - S_n| < 0.01?

Need

b_{n+1} = \frac{1}{n+1} 100.

So you can take n = 100 terms to guarantee error under 0.01.

Notice how simple: you didn’t compute the sum, just controlled the error.

Example 3: Alternating Taylor series (common FRQ style)

Maclaurin series:

\sin(x) = \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{(2n+1)!}.

Approximate \sin(1) by the 3-term polynomial:

S_2 = 1 - \frac{1}{3!} + \frac{1}{5!}.

Here

b_n = \frac{1^{2n+1}}{(2n+1)!} = \frac{1}{(2n+1)!},
which decreases and goes to 0.

Error bound uses the next omitted term (the 4th term overall):

|\sin(1) - S_2| \le \frac{1}{7!}.

Over/under?

• Next term would be negative: -\frac{1}{7!}.
• So \sin(1) - S_2 < 0 and S_2 is an overestimate.

Example 4: Trapping the exact sum between two partial sums

Consider

\sum_{n=1}^{\infty} (-1)^{n-1}\frac{1}{n^3} = 1 - \frac{1}{8} + \frac{1}{27} - \frac{1}{64} + \cdots

Compute consecutive partial sums:

S_3 = 1 - \frac{1}{8} + \frac{1}{27}, \quad S_4 = S_3 - \frac{1}{64}.

Because the conditions hold, the true sum S satisfies

S \in [S_4, S_3]
(since S_4 < S_3 here).

That’s a fast “interval answer” style bound that AP sometimes asks for.

5. Common Mistakes & Traps

1. Using the error bound without checking conditions

   • What goes wrong: You apply |S-S_n| \le b_{n+1} just because signs alternate.
   • Why wrong: The bound requires b_n decreasing and b_n \to 0.
   • Fix: Explicitly verify monotone decrease (at least eventually) and limit 0.

2. Bounding with the wrong term (using b_n instead of b_{n+1})

   • What goes wrong: You write |S-S_n| \le b_n.
   • Why wrong: The theorem uses the first omitted term.
   • Fix: After summing through term n, your bound is b_{n+1}.

3. Forgetting absolute value / mixing up sign

   • What goes wrong: You treat the remainder as exactly \pm b_{n+1}.
   • Why wrong: The theorem gives an inequality bound, not an exact remainder.
   • Fix: Write |S-S_n| \le b_{n+1}, then (optionally) use the next term’s sign for over/under.

4. Assuming “alternating” means “convergent”

   • What goes wrong: You conclude convergence from sign changes alone.
   • Why wrong: If \lim b_n \ne 0, the series diverges even if it alternates.
   • Fix: Always check \lim b_n = 0.

5. Not recognizing an alternating series because it’s not written with (-1)^n

   • What goes wrong: You miss alternation when it’s written like \sum \frac{(-1)^{n+1}}{2n-1} or with explicit + - + - pattern.
   • Fix: Rewrite to identify b_n = |a_n| and the alternating factor (-1)^n.

6. Claiming decreasing without justification

   • What goes wrong: You say “clearly decreasing” when it isn’t obvious.
   • Why wrong: FRQs often require a reason (comparison, derivative, monotonicity argument).
   • Fix: Use a quick argument like: for factorials/powers it decreases; for rational forms, compare b_{n+1} to b_n or check f'(x) < 0 for a continuous extension.

7. Index slip when the series starts at n=0

   • What goes wrong: You compute S_n but then bound with the wrong “next term” because you forgot the start index.
   • Fix: Be consistent: if your partial sum includes terms up through index n, the bound uses index n+1—regardless of whether you started at 0 or 1.

8. Using the theorem on a non-alternating Taylor series

   • What goes wrong: You use alternating error bound for something like e^x at positive x.
   • Why wrong: The alternating bound needs alternating signs.
   • Fix: Only use this theorem when the series truly alternates (many Taylor series do on certain inputs, e.g., e^{-x} at x>0 alternates).

6. Memory Aids & Quick Tricks

7. Quick Review Checklist

• [ ] Can you rewrite the series as \sum (-1)^{n-1}b_n with b_n>0?
• [ ] Did you verify b_{n+1} \le b_n (decreasing) and \lim b_n = 0?
• [ ] Do you know the main bound: |S-S_n| \le b_{n+1}?
• [ ] For a target error \varepsilon, did you solve b_{n+1} < \varepsilon for n?
• [ ] Can you state that S lies between S_n and S_{n+1}?
• [ ] Can you decide over/under using the sign of the next term a_{n+1}?
• [ ] Did you avoid the common index mistake (especially if starting at n=0)?

You’ve got this—if the series alternates nicely, the next term basically hands you the error bound for free.