AP Calculus: Alternating Series Error Bound Theorem (AP)

What You Need to Know

Why this matters

On AP Calculus AB, you’re often asked to approximate a value using a partial sum of an alternating series (or a Taylor/Maclaurin series that alternates). The Alternating Series Error Bound Theorem tells you how far off your approximation can be without computing the whole remainder.

Core setup (alternating series form)

An alternating series is typically written as

\sum_{n=1}^{\infty} (-1)^{n-1} b_n \quad \text{or} \quad \sum_{n=0}^{\infty} (-1)^n b_n,

where b_n \ge 0.

Alternating Series Test (the needed conditions)

If

1. b_{n+1} \le b_n for all sufficiently large n (eventually nonincreasing), and
2. \lim_{n\to\infty} b_n = 0,

then the alternating series converges.

Alternating Series Error Bound Theorem (the main result)

Let

S = \sum_{n=1}^{\infty} (-1)^{n-1} b_n

be a convergent alternating series with b_n decreasing to 0. Let

S_n = \sum_{k=1}^{n} (-1)^{k-1} b_k

be the nth partial sum and define the remainder (error)

R_n = S - S_n.

Then:

1. Magnitude bound (what you use most):

|R_n| \le b_{n+1}.

2. Next-term sign / over-under fact (super useful):

R_n \text{ has the same sign as the } (n+1)\text{st term.}

So if the series starts positive:

• S_1 **overestimates**, S_2 **underestimates**, S_3 overestimates, etc.
• Odds and evens “trap” the true sum.

Big exam takeaway: If you want |S - S_n| < \varepsilon, it’s enough to make b_{n+1} < \varepsilon.

Step-by-Step Breakdown

A) Confirm you’re allowed to use the theorem

1. Rewrite the series (or Taylor series) into the form (-1)^n b_n with b_n \ge 0.
2. Check monotone decrease (eventually): verify b_{n+1} \le b_n for relevant n.
3. Check limit: compute \lim_{n\to\infty} b_n and confirm it equals 0.

If these hold, the series converges (AST) and the error bound applies.

If you don’t check decreasing-to-0, you can’t justify the error bound.

B) Find how many terms you need for a desired accuracy

Goal: choose n so that

|S - S_n| \le b_{n+1} < \varepsilon.

1. Identify b_n.
2. Solve the inequality b_{n+1} < \varepsilon for n.
3. Round up to the next integer (you need enough terms).

Mini worked example (term count):
Approximate \sum_{n=1}^{\infty} (-1)^{n-1}\frac{1}{n} within 0.001.

• Here b_n = \frac{1}{n}.
• Need b_{n+1} = \frac{1}{n+1} < 0.001.
• Solve: n+1 > 1000 \Rightarrow n \ge 1000.

So using S_{1000} guarantees error less than 0.001.

C) Decide whether your partial sum is an overestimate or underestimate

If the series starts with a positive term (common form (-1)^{n-1}b_n):

• S_n is an **overestimate** when n is odd.
• S_n is an **underestimate** when n is even.

Reason: R_n has the sign of term n+1.

Mini worked example (over/under):
For \sum_{n=1}^{\infty} (-1)^{n-1} b_n:

• If n=5, then term 6 is negative, so R_5 < 0 and S_5 = S - R_5 > S (overestimate).

Key Formulas, Rules & Facts

The core facts (keep these straight)

Practical inequality to pick n

To guarantee accuracy within \varepsilon:

b_{n+1} < \varepsilon \quad \Rightarrow \quad |S - S_n| < \varepsilon.

“Trapping” the true sum (interval estimate)

For a standard alternating series starting positive:

• If n is **odd**, then S_{n+1} < S < S_n.
• If n is **even**, then S_n < S < S_{n+1}.

So you can bound the true sum between two consecutive partial sums.

Examples & Applications

Example 1: Approximate \ln(2) using an alternating series

You should know the classic series:

\ln(1+x)=\sum_{n=1}^{\infty} (-1)^{n-1}\frac{x^n}{n}, \quad -1 < x \le 1.

Set x=1:

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

Question type: Find n so that the approximation error is less than 0.01.

• Here b_n=\frac{1}{n}, decreasing to 0.
• Need b_{n+1}=\frac{1}{n+1}100 \Rightarrow n\ge 100.

Extra exam twist (over/under):

• S_{100} is an **underestimate** because 100 is even.

Example 2: Maclaurin approximation with guaranteed error

Consider

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

Approximate e^{-1} using S_3 and bound the error.

• For x=1:

e^{-1} \approx S_3 = 1 - 1 + \frac{1}{2} - \frac{1}{6} = \frac{1}{3}.

• Here b_n=\frac{1}{n!} (decreasing), so

|R_3| \le b_4 = \frac{1}{4!} = \frac{1}{24}.

Over/under: Series starts positive with alternating signs. Since n=3 is odd, term 4 is positive, so R_3>0 and S_3 is an underestimate.

So:

\frac{1}{3} < e^{-1} < \frac{1}{3} + \frac{1}{24}.

Example 3: Arctangent series and “how many terms?”

You may see:

\arctan(x)=\sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{2n+1}, \quad |x|\le 1.

Approximate \arctan(1)=\frac{\pi}{4} within 10^{-4}.

• At x=1:

\frac{\pi}{4}=\sum_{n=0}^{\infty} (-1)^n\frac{1}{2n+1}.

• Here b_n=\frac{1}{2n+1}.
• Need b_{n+1}=\frac{1}{2(n+1)+1} = \frac{1}{2n+3} < 10^{-4}.
• Solve: 2n+3 > 10^4 \Rightarrow n > \frac{9997}{2} \Rightarrow n \ge 4999.

So using partial sum through n=4999 (that’s **5000 terms**) guarantees error under 10^{-4}.

Example 4: Using the bound to create a numeric interval

Let

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

Approximate using S_4 and give an interval containing S.

• b_n=\frac{1}{n^2} is decreasing to 0.
• Compute:

S_4=1-\frac{1}{4}+\frac{1}{9}-\frac{1}{16}.

• Error bound:

|R_4| \le b_5=\frac{1}{25}.

• Over/under: n=4 even, term 5 is positive, so R_4>0 and S_4 is an underestimate.

Interval:

S_4 < S < S_4 + \frac{1}{25}.

Common Mistakes & Traps

1. Using the error bound without checking the conditions

   • Wrong: Jumping straight to |R_n|\le b_{n+1} just because signs alternate.
   • Why wrong: The theorem needs b_n decreasing (at least eventually) and b_n\to 0.
   • Fix: Always state/check: b_n\ge 0, decreasing, limit 0.

2. Using b_n instead of b_{n+1}

   • Wrong: Saying |R_n|\le b_n.
   • Why wrong: The remainder after n terms is controlled by the next omitted term.
   • Fix: Write: “Error after n terms \le magnitude of term n+1.”

3. Index confusion (starting at 0 vs 1)

   • Wrong: Mixing up what “the nth term” means.
   • Why wrong: Taylor series often start at n=0, while many textbook series start at n=1.
   • Fix: Define your partial sum clearly (e.g., S_n=\sum_{k=0}^{n} \dots) and then use the next term accordingly.

4. Assuming “alternating” automatically means “decreasing”

   • Wrong: Not verifying b_{n+1} \le b_n.
   • Why wrong: Some alternating series have magnitudes that bounce around; AST (and the error bound) may fail.
   • Fix: Check monotonicity directly, or show it’s true for all n\ge N.

5. Forgetting that “eventually decreasing” is enough

   • Wrong: Rejecting AST/error bound because the first couple of terms don’t decrease.
   • Why wrong: The theorem requires decreasing for all sufficiently large n.
   • Fix: If needed, start applying the bound after the series becomes monotone.

6. Sign mistake on overestimate vs underestimate

   • Wrong: Claiming the wrong direction (especially with odd/even partial sums).
   • Why wrong: The remainder has the sign of term n+1.
   • Fix: Look at the next term’s sign. If R_n>0 then S_n

7. Solving the inequality but not rounding correctly

   • Wrong: Getting n>99 and choosing n=99.
   • Why wrong: You need an integer n that satisfies the strict inequality.
   • Fix: Always round up to the next integer that works.

8. Confusing alternating-series error with Taylor’s Lagrange error

   • Wrong: Using derivative-based Lagrange remainder when the problem is clearly about alternating series terms.
   • Why wrong: Different hypotheses, different bounds.
   • Fix: If the terms alternate and decrease to 0, the alternating bound is typically the fastest, simplest tool: |R_n|\le b_{n+1}.

Memory Aids & Quick Tricks

Quick sign check: determine the sign of term n+1. That’s the sign of R_n.

Quick Review Checklist

• [ ] Can you rewrite the series as \sum (-1)^n b_n with b_n\ge 0?
• [ ] Did you verify b_n is (eventually) decreasing and \lim b_n=0?
• [ ] Do you know the error bound: |S-S_n|=|R_n|\le b_{n+1}?
• [ ] For a desired error \varepsilon, did you solve b_{n+1}
• [ ] Can you state whether S_n is an overestimate or underestimate by checking the sign of term n+1?
• [ ] If needed, can you give an interval: S_n - b_{n+1} < S < S_n + b_{n+1} (and tighten it using the sign/odd-even rule)?

You’ve got this—if you can spot b_{n+1} quickly and track the sign of the next term, most AP questions on this become routine.