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 $n$th 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 $n$th 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 $S_n<S$; if $R_n<0$ then $S_n>S$.

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 $b_{n+1}<\varepsilon$ and round up?
• [ ] 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.