Alternating Series Error Bound: Examples

The alternating series error bound says that if an alternating series passes the alternating series test, stopping after n terms gives an error no bigger than the first term you left out: $|S - S_n| \le b_{n+1}$ . The true sum S always lies between two consecutive partial sums.

Before this lesson: Alternating Series Test, Infinite Series and Partial Sums and Taylor Polynomials

Why the next term bounds the error

Take an alternating series $b_1 - b_2 + b_3 - b_4 + \cdots$ whose terms $b_n$ shrink to 0. Each partial sum adds or subtracts a smaller amount than the step before. So the partial sums zigzag above and below the true sum S, and the zigzags get smaller.

When you stop at $S_n$ , the next step, $b_{n+1}$ , would jump past S. So S is between $S_n$ and $S_{n+1}$ , and the gap from $S_n$ to S is less than $b_{n+1}$ .

Dot plot of the partial sums Sₙ of 1 − 1/4 + 1/9 − 1/16 + ⋯ for n = 1 to 10. The dots zigzag above and below a dashed orange line at the sum S ≈ 0.822, getting closer each time: 1, 0.75, 0.861, 0.799, 0.839 and so on. A bracket at n = 5 marks the gap from S₅ ≈ 0.839 to S, which is at most b₆ = 1/36. — The partial sums of an alternating series zigzag around the sum, so the error after n terms is less than the next term.

Alternating series error bound

Suppose $S = \sum_{n=1}^{\infty} (-1)^{n+1} b_n$ where every $b_n \gt 0$ , the $b_n$ decrease, and $\lim_{n \to \infty} b_n = 0$ . Then:

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

$S_n$ is the sum of the first n terms, and $b_{n+1}$ is the size of the first term left out.
The sign of $S - S_n$ matches the sign of the first term left out. If that term is negative, $S_n$ is too big; if it is positive, $S_n$ is too small.
This result is also called the alternating series estimation theorem.

How to use the error bound

Check the conditions: the terms alternate, $b_n$ decreases, and $b_n \to 0$ .
To bound the error of $S_n$ , find the first omitted term, $b_{n+1}$ .
To reach an error below a target E, find the first $b_k$ with $b_k \le E$ . Then add every term before it.

Alternating series error bound examples

Example 1: Bound the error of a partial sum

Estimate $S = \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n^2}$ with $S_5$ , and bound the error.

b_n = \dfrac{1}{n^2}

bₙ is positive, decreases and goes to 0, so the bound applies.

S_5 = 1 - \dfrac{1}{4} + \dfrac{1}{9} - \dfrac{1}{16} + \dfrac{1}{25}

Add the first five terms.

1 - 0.25 + 0.111111 - 0.0625 + 0.04 \approx 0.838611

Convert to decimals and combine.

|S - S_5| \le b_6 = \dfrac{1}{36} \approx 0.0278

The first term left out is −1/36.

0.838611 - 0.0278 \le S \le 0.838611

That term is negative, so S₅ is too big.

Answer: S ≈ 0.8386 with error at most 0.0278, and the sum is between about 0.8108 and 0.8386. (The exact sum is π²/12 ≈ 0.8225, which fits.)

Example 2: How many terms are needed?

How many terms of $\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n}$ guarantee an error of at most 0.01?

b_{n+1} = \dfrac{1}{n + 1} \le 0.01

The first omitted term must be at most 0.01.

n + 1 \ge 100

Take reciprocals of both sides; the inequality flips.

n \ge 99

Subtract 1 from both sides.

Answer: 99 terms. This series converges very slowly; its sum is ln 2 ≈ 0.693.

Example 3: A series with factorials

The series $\sum_{n=0}^{\infty} \dfrac{(-1)^n}{n!} = 1 - 1 + \dfrac{1}{2!} - \dfrac{1}{3!} + \cdots$ equals $\dfrac{1}{e}$ . Estimate 1/e with an error less than 0.001.

b_6 = \dfrac{1}{6!} = \dfrac{1}{720} \approx 0.00139

Too big: 0.00139 > 0.001.

b_7 = \dfrac{1}{7!} = \dfrac{1}{5040} \approx 0.000198

Small enough, so b₇ is the first term to leave out.

1 - 1 + \dfrac{1}{2} - \dfrac{1}{6} + \dfrac{1}{24} - \dfrac{1}{120} + \dfrac{1}{720}

Add the terms for n = 0 through 6 (2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720).

\dfrac{1}{2} - \dfrac{1}{6} + \dfrac{1}{24} - \dfrac{1}{120} + \dfrac{1}{720} = \dfrac{265}{720}

The first two terms cancel. Use the common denominator 720: 360 − 120 + 30 − 6 + 1 = 265.

\dfrac{265}{720} = \dfrac{53}{144} \approx 0.368056

Simplify and divide.

Answer: 1/e ≈ 0.36806 with error less than 0.000198. (The true value is 0.36788.)

Example 4: The error of a Taylor polynomial

In the Taylor polynomials lesson, $\sin 0.5 \approx 0.5 - \frac{0.5^3}{3!} + \frac{0.5^5}{5!} \approx 0.4794271$ . The Maclaurin series of sin x continues $- \frac{x^7}{7!} + \frac{x^9}{9!} - \cdots$ . Bound the error.

\dfrac{0.5^7}{7!} = \dfrac{0.0078125}{5040}

At x = 0.5 the series alternates and its terms decrease, so the next term bounds the error. 0.5⁷ = 0.0078125.

\dfrac{0.0078125}{5040} \approx 0.00000155

Divide.

Answer: the error is less than 0.00000155, so sin 0.5 ≈ 0.479427 is correct to five decimal places. The next term is negative, so the estimate is a little too big. (The true value is 0.4794255.)

Example 5: A calculator chip (word problem)

A chip computes ln 1.5 with the alternating series $0.5 - \dfrac{0.5^2}{2} + \dfrac{0.5^3}{3} - \cdots$ , whose terms are $b_n = \dfrac{0.5^n}{n}$ . How many terms does it need for an error of at most 0.0005, and what is the estimate?

b_7 = \dfrac{0.5^7}{7} = \dfrac{0.0078125}{7} \approx 0.00112

Compute terms until one is at most 0.0005. This one is too big.

b_8 = \dfrac{0.5^8}{8} = \dfrac{0.00390625}{8} \approx 0.000488

Small enough: leave out b₈ and keep the 7 terms before it.

S_7 = 0.5 - 0.125 + 0.0416667 - 0.015625 + 0.00625 - 0.0026042 + 0.0011161

Write the first 7 terms as decimals.

S_7 \approx 0.4058036

Combine.

Answer: 7 terms, giving ln 1.5 ≈ 0.4058 with error at most 0.000488. (The true value is 0.4055.)

Common mistakes

Mistake	Example of the error	Fix
Using the last term kept	Bounding the error of S₅ by b₅ = 1/25	Use the first term left out: b₆ = 1/36.
Using it on a series that does not alternate	“The error of S₅ for $\sum \frac{1}{n^2}$ is at most 1/36”	For positive series the error is the whole tail, which is bigger than the next term. Use another method.
Skipping the conditions	Using the bound when $b_n$ does not decrease or does not go to 0	Check all the alternating series test conditions first.
Miscounting terms that start at n = 0	Saying “6 terms” for n = 0 through 6	n = 0 through 6 is 7 terms. Count carefully.
Thinking the error equals the next term	“The error is exactly 1/36”	The next term is an upper bound. The real error is smaller.
Rounding the bound down	Calling 0.000198 “about 0.0001”	Round a bound up, or keep enough digits.

Related methods

Lagrange error bound

When a Taylor series does not alternate, such as the series for $e^x$ at x = 0.5, the next term does not bound the error. Then use the Lagrange error bound, the next lesson.

Bounding the error of a positive series

For a positive series that passes the integral test, the error of $S_n$ is less than $\int_n^{\infty} f(x)\,dx$ . This uses an integral instead of a single term.

Practice problems

Estimate $\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n^3}$ with $S_4$ , and bound the error.
How many terms of $\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n^2}$ guarantee an error of at most 0.001?
How many terms of $\sum_{n=0}^{\infty} \dfrac{(-1)^n}{2n + 1}$ guarantee an error of at most 0.005?
Estimate $\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n \cdot 3^n}$ with $S_3$ , and bound the error.
Is $S_5$ of $\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n^2}$ larger or smaller than the sum S? Explain.
Use $1 - \dfrac{x^2}{2} + \dfrac{x^4}{24}$ to estimate cos 0.4, and bound the error. The series continues $- \frac{x^6}{720} + \cdots$ .
The series $\sum_{n=0}^{\infty} \dfrac{(-0.5)^n}{n!}$ equals $e^{-0.5}$ . How many terms guarantee an error of at most 0.0001?
Use $S_4$ and $S_5$ to give an interval that contains the sum of $\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n^2}$ .
Can you use the error bound on $\sum_{n=1}^{\infty} (-1)^{n+1} \dfrac{n}{n + 1}$ ? Explain.
A student uses $12\left(1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16} + \frac{1}{25} - \frac{1}{36}\right)$ to estimate π², since $\sum \frac{(-1)^{n+1}}{n^2} = \frac{\pi^2}{12}$ . Find the estimate and bound its error.

Show answers

S_4 = 1 - \dfrac{1}{8} + \dfrac{1}{27} - \dfrac{1}{64}

Add the first four terms; bₙ = 1/n³ decreases to 0.

1 - 0.125 + 0.037037 - 0.015625 \approx 0.896412

Convert to decimals and combine.

|S - S_4| \le b_5 = \dfrac{1}{125} = 0.008

The first term left out is 1/125.

Answer: S ≈ 0.8964 with error at most 0.008

\dfrac{1}{(n + 1)^2} \le 0.001

The first omitted term must be at most 0.001.

(n + 1)^2 \ge 1000

Take reciprocals; the inequality flips.

n + 1 \ge \sqrt{1000} \approx 31.62

Take the square root of both sides.

n + 1 \ge 32

n + 1 is a whole number.

Answer: n = 31 terms

\dfrac{1}{2(N + 1) + 1} = \dfrac{1}{2N + 3} \le 0.005

Using terms n = 0 to N, the first omitted term is n = N + 1.

2N + 3 \ge 200

Take reciprocals; the inequality flips.

N \ge 98.5, \text{ so } N = 99

Subtract 3, divide by 2, and round up.

Answer: n = 0 through 99, which is 100 terms

S_3 = \dfrac{1}{3} - \dfrac{1}{18} + \dfrac{1}{81}

Terms: 1/(1 · 3), 1/(2 · 9), 1/(3 · 27).

0.333333 - 0.055556 + 0.012346 \approx 0.290123

Convert to decimals and combine.

|S - S_3| \le b_4 = \dfrac{1}{4 \cdot 81} = \dfrac{1}{324} \approx 0.00309

The first term left out is 1/(4 · 3⁴).

Answer: S ≈ 0.2901 with error at most 0.00309

-\dfrac{1}{36}

The first term left out after S₅ is −1/36, which is negative.

S - S_5 \lt 0

S − S₅ has the same sign as that term.

Answer: S₅ is larger than S (it overestimates)

1 - \dfrac{0.16}{2} + \dfrac{0.0256}{24}

Substitute x = 0.4: 0.4² = 0.16 and 0.4⁴ = 0.0256.

1 - 0.08 + 0.0010667 \approx 0.9210667

Divide, then combine.

\dfrac{0.4^6}{720} = \dfrac{0.004096}{720} \approx 0.0000057

The terms alternate and decrease, so the next term bounds the error.

Answer: cos 0.4 ≈ 0.921067 with error less than 0.0000057

b_5 = \dfrac{0.5^5}{5!} = \dfrac{0.03125}{120} \approx 0.00026

Compute terms until one is at most 0.0001. Too big.

b_6 = \dfrac{0.5^6}{6!} = \dfrac{0.015625}{720} \approx 0.0000217

Small enough: leave out the n = 6 term.

Answer: keep n = 0 through 5, which is 6 terms

S_4 = 1 - \dfrac{1}{4} + \dfrac{1}{9} - \dfrac{1}{16} \approx 0.798611

Add four terms.

S_5 = S_4 + \dfrac{1}{25} \approx 0.838611

Add the fifth term.

Answer: 0.7986 < S < 0.8386

\lim_{n \to \infty} \dfrac{n}{n + 1} = 1

The terms do not go to 0.

Answer: no. The conditions fail; in fact the series diverges by the divergence test, so there is no sum to estimate

10.

1 - 0.25 + 0.111111 - 0.0625 + 0.04 - 0.027778 \approx 0.810833

Add the six terms as decimals.

12(0.810833) \approx 9.73

Multiply by 12.

12 \cdot \dfrac{1}{49} \approx 0.245

The first term left out is 1/49, and the factor of 12 scales the error too.

Answer: π² ≈ 9.73 with error at most 0.245 (the true value is 9.8696)

Alternating series error bound FAQ

What is the alternating series estimation theorem?

If an alternating series passes the alternating series test, the error from stopping after n terms is at most the first term left out: $|S - S_n| \le b_{n+1}$ .

How do you find how many terms you need?

Set the first omitted term less than or equal to the allowed error, $b_{n+1} \le E$ , and solve for n. For factorial terms, list the terms until one is small enough.

Is the partial sum bigger or smaller than the actual sum?

Look at the first term left out. If it is negative, the partial sum is too big. If it is positive, the partial sum is too small.

When can’t you use the alternating series error bound?

You can’t use it when the signs don’t alternate, or when the terms don’t decrease to 0. For a Taylor polynomial that doesn’t alternate, use the Lagrange error bound.

Summary

For an alternating series with $b_n$ decreasing to 0, $|S - S_n| \le b_{n+1}$ .
The bound is the first term left out, not the last term kept.
The sign of the first omitted term tells whether $S_n$ is too big or too small.
The sum is always between two consecutive partial sums.
It also bounds the error of alternating Taylor polynomials, like those for sin x and cos x.

Previous lesson: Taylor Polynomials

Next lesson: Lagrange Error Bound

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