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 $n$th 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.