10.10 - Alternating Series Error Bound

Prerequisites

Important: the alternating series must converge for error bounds to mean anything. If the series diverges, error bound analysis is pointless.

Understanding Alternating Series Behavior

Before getting into error bounds, we need to see how alternating series converge.

Example: The Alternating Harmonic Series

Consider

\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdots

This series converges by the Alternating Series Test.

Building a table of terms and partial sums:

$n$	Term $a_n$	Decimal	Partial Sum $S_n$	Decimal
1	$1$	$1.000$	$1$	$1.000$
2	$-\frac{1}{2}$	$-0.500$	$\frac{1}{2}$	$0.500$
3	$\frac{1}{3}$	$0.333\ldots$	$\frac{5}{6}$	$0.833\ldots$
4	$-\frac{1}{4}$	$-0.250$	$\frac{7}{12}$	$0.583\ldots$
5	$\frac{1}{5}$	$0.200$	$\frac{47}{60}$	$0.783\ldots$
6	$-\frac{1}{6}$	$-0.167$	$\frac{37}{60}$	$0.617\ldots$
7	$\frac{1}{7}$	$0.143$	$\frac{319}{420}$	$0.760\ldots$
8	$-\frac{1}{8}$	$-0.125$	$\frac{533}{840}$	$0.635\ldots$

Key observation: the partial sums oscillate around the true sum.

The true sum:

\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = \ln(2) \approx 0.6931

Calculating the Error (Remainder)

The error, also called the remainder, is the difference between the true sum and the partial sum.

R_n = |S - S_n|

where:

$S$ is the true sum of the infinite series

$S_n$ is the partial sum of the first $n$ terms

$R_n$ is the remainder, or error, after $n$ terms

Computing Remainders

For the alternating harmonic series, with

S = \ln(2) \approx 0.6931

$n$	$S_n$	Error $R_n = \lvert\ln(2) - S_n\rvert$
1	$1.000$	$0.307$
2	$0.500$	$0.193$
3	$0.833$	$0.140$
4	$0.583$	$0.110$
5	$0.783$	$0.090$
6	$0.617$	$0.076$
7	$0.760$	$0.066$
8	$0.635$	$0.059$

Notice: the errors shrink as

n

grows.

The Alternating Series Error Bound Theorem

For an alternating series

\sum_{n=1}^{\infty} (-1)^{n+1} a_n

that passes the Alternating Series Test (

a_n > 0

a_n

decreasing, and

a_n \to 0

\boxed{|S - S_n| \leq a_{n+1}}

In words: the error from approximating the sum with the first

n

terms is no larger than the absolute value of the next term.

The Relationship Between Error and Error Bound

R_n \leq |a_{n+1}|

where

R_n

is the actual error and

|a_{n+1}|

is the error bound, meaning the maximum possible error.

The error bound is always the size of the very next term you didn’t include.

Overestimate or Underestimate?

The partial sums take turns landing above and below

S

, so the true sum always sits between two partial sums in a row,

S_n

and

S_{n+1}

. That tells you which way the error goes:

If the next term (the first one you left out) is negative, adding it would pull the total down, so $S_n$ is an overestimate: $S < S_n$ .

If the next term is positive, $S_n$ is an underestimate: $S > S_n$ .

Check with the table:

S_3 = \frac{5}{6} \approx 0.833

, and the next term is

-\frac{1}{4}

(negative). So

S_3

is an overestimate, and sure enough,

0.833 > \ln(2) \approx 0.693

Verification with the Alternating Harmonic Series

Let’s confirm the theorem holds:

$n$	Error $R_n$	Next term $\lvert a_{n+1}\rvert$	Does $R_n \leq \lvert a_{n+1}\rvert$ ?
1	$0.307$	$0.5$	Yes, $0.307 \leq 0.5$
2	$0.193$	$0.333$	Yes, $0.193 \leq 0.333$
3	$0.140$	$0.25$	Yes, $0.140 \leq 0.25$
4	$0.110$	$0.2$	Yes, $0.110 \leq 0.2$
5	$0.090$	$0.167$	Yes, $0.090 \leq 0.167$

The theorem holds every time.

Key Definitions

Term $a_n$ : an individual value in the sequence.

Partial sum $S_n$ : the sum of the first

n

terms.

S_n = a_1 + a_2 + a_3 + \cdots + a_n

True sum $S$ : the limit of the partial sums.

S = \lim_{n \to \infty} S_n = \sum_{n=1}^{\infty} a_n

Error (remainder) $R_n$ : the difference between the true sum and the partial sum.

R_n = |S - S_n|

Error bound: the absolute value of the next term.

\text{Error bound} = |a_{n+1}|

Example 1: Finding How Many Terms You Need

How many terms of

\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n}

do you need so that the error is less than

0.001

We need

|S - S_n| \leq |a_{n+1}| < 0.001

Since

a_n = \frac{1}{n}

, we have

a_{n+1} = \frac{1}{n+1}

. Set up the inequality:

\frac{1}{n+1} < 0.001 \quad \Rightarrow \quad \frac{1}{n+1} < \frac{1}{1000} \quad \Rightarrow \quad 1000 < n+1 \quad \Rightarrow \quad n > 999

Answer: you need more than 999 terms, so at least 1000 terms to guarantee an error below

0.001

Example 2: Finding the Error Bound for a Given Partial Sum

Find the error bound when you use the first 10 terms to approximate

\sum_{n=1}^{\infty} \dfrac{(-1)^n}{5n + 2}

The error bound is the absolute value of the 11th term.

|a_{11}| = \left|\frac{(-1)^{11}}{5(11) + 2}\right| = \left|\frac{-1}{57}\right| = \frac{1}{57}

Answer: the error bound is

\frac{1}{57} \approx 0.0175

Interpretation: using the first 10 terms gives an approximation that’s off by at most

\frac{1}{57}

from the true sum.

Example 3: Approximating a Series with an Error Bound

Use the first 5 terms to approximate

\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1} \cdot 4}{n^2}

and find the error bound.

Step 1: Calculate

S_5

with technology.

S_5 = \sum_{n=1}^{5} \frac{(-1)^{n+1} \cdot 4}{n^2}

Using a calculator with summation notation:


sum((-1)^(n+1) * 4/n^2, n, 1, 5)

Result:

S_5 \approx 3.3544

Step 2: Find the error bound, which is

|a_6|

|a_6| = \left|\frac{(-1)^{6+1} \cdot 4}{6^2}\right| = \frac{4}{36} = \frac{1}{9} \approx 0.1111

Step 3: Write the approximation with its error bound.

The true sum

S

satisfies

S_5 - |a_6| \leq S \leq S_5 + |a_6|

3.3544 - 0.1111 \leq S \leq 3.3544 + 0.1111

3.2433 \leq S \leq 3.4655

Answer: the sum is about

3.3544

, with an error bound of

\frac{1}{9}

, so the true sum lies between

3.243

and

3.466

Example 4: Proving an Approximation Meets an Error Requirement

Given

f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots

show that

1 - \frac{1}{3!}

approximates

f(1)

with an error less than

0.01

Step 1: Evaluate at

x = 1

f(1) = 1 - \frac{1^3}{3!} + \frac{1^5}{5!} - \frac{1^7}{7!} + \cdots = 1 - \frac{1}{6} + \frac{1}{120} - \frac{1}{5040} + \cdots

Step 2: Identify the approximation.

S_2 = 1 - \frac{1}{3!} = 1 - \frac{1}{6}

This uses the first 2 terms,

a_1 = 1

and

a_2 = -\frac{1}{3!}

Step 3: Find the error bound, which is

|a_3|

|a_3| = \left|\frac{(-1)^2 \cdot 1^5}{5!}\right| = \frac{1}{120}

Step 4: Compare to the required error.

We need the error to be less than

0.01 = \frac{1}{100}

, and

\frac{1}{120} < \frac{1}{100} \quad \checkmark

Conclusion: since

\frac{1}{120} < \frac{1}{100}

, the approximation

1 - \frac{1}{3!}

has an error less than

0.01

Finding Intervals of Convergence with Error Bounds

For power series with a parameter

x

, we can find where the approximations meet an error requirement.

Example 5: Error Bound with a Variable (Advanced)

The series

\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1} x^n}{n^2}

converges for

0 < x \leq 1

by the AST. For which of these

x

values can you guarantee that the partial sum

S_3

approximates the sum with an error less than

0.01

Step 1: Check the AST conditions. For

0 < x \leq 1

, the terms

a_n = \frac{x^n}{n^2}

are positive, they decrease (the numerator never grows and the denominator does), and they approach

0

Step 2: Write

S_3

and the first term left out.

S_3 = x - \frac{x^2}{4} + \frac{x^3}{9}, \qquad |a_4| = \frac{x^4}{16}

Step 3: Make the error bound less than

0.01

\frac{x^4}{16} < 0.01 \quad \Rightarrow \quad x^4 < 0.16 \quad \Rightarrow \quad x < (0.16)^{1/4} \approx 0.632

Answer: for

0 < x < 0.632

S_3

is guaranteed to be within

0.01

of the true sum. Closer to

x = 1

the terms are bigger, so you’d need more terms.

Important Notes and Common Applications

When Can You Use the Error Bound Theorem?

Requirements:

The series has to be alternating

The series has to converge

The terms have to be decreasing in absolute value

The terms have to approach zero

These are the same requirements as the Alternating Series Test.

Why This Matters

Practical use: you can approximate infinite sums with finite calculations and still guarantee your accuracy.

On the AP exam: error bound problems show up regularly on BC free response questions.

Strategy for Error Bound Problems

Type 1: “How many terms do I need for error less than $\epsilon$ ?”

Set up $|a_{n+1}| < \epsilon$

Solve for $n$

Answer with the smallest integer $n$ that satisfies the inequality

Type 2: “What’s the error bound using $n$ terms?”

The error bound is $|a_{n+1}|$

Calculate the $(n+1)$ st term

Take the absolute value

Type 3: “Approximate the sum with an error bound”

Calculate $S_n$ with technology

Find $|a_{n+1}|$ , the error bound

Report $S_n - |a_{n+1}| \leq S \leq S_n + |a_{n+1}|$

Type 4: “Prove the approximation meets the error requirement”

Identify which partial sum is being used

Calculate the next term, which is the error bound

Compare it to the required error

State your conclusion

Key Takeaways

Error bound theorem: for an alternating series that passes the AST, $|S - S_n| \leq |a_{n+1}|$

The error bound is the next term, meaning the size of the very next term you left out

It has to converge first. Error bounds only apply to convergent alternating series

Partial sums oscillate, landing alternately above and below the true sum

Errors shrink as $n$ grows

It’s the next term, not the next partial sum. The bound is $|a_{n+1}|$ , not $S_{n+1}$

Absolute value matters. Take $|a_{n+1}|$ even when the term is negative

Technology helps. Use your calculator’s summation notation

Common AP problem. It shows up often on the BC exam

Practical application: it lets you approximate with a guaranteed accuracy

Four problem types: terms needed, find the bound, approximate with a bound, and prove a requirement

Visual understanding: the graph shows oscillation around the true sum with shrinking amplitude

Back Next