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AP®MathsCollege BoardCalculus BCStudy GuidesUnit 10: Infinite Sequences and SeriesTests for Divergence & ConvergenceAlternating Series Error Bound

Alternating Series Error Bound (College Board AP® Calculus BC)

Study Guide

Roger B

Written by: Roger B

Reviewed by: Mark Curtis

Updated on

Alternating series error bound

What is an alternating series error bound?

  • If an alternating series converges, the alternating series error bound places a bound on how far a partial sum (the sum of the first n terms) is from the value of the infinite series

  • For an alternating series of the form sum from n equals 1 to infinity of open parentheses negative 1 close parentheses to the power of n times a subscript n or sum from n equals 1 to infinity of open parentheses negative 1 close parentheses to the power of n plus 1 end exponent times a subscript n

    • If the infinite series converges to a sum S

      • and if s subscript n is the partial sum of the first n terms

    • Then open vertical bar S minus s subscript n close vertical bar less than a subscript n plus 1 end subscript

      • i.e. the absolute value of the difference between the sum of the infinite series, S, and the sum of the first n terms (the nth partial sum) s subscript n is less than the absolute value of the next term in the series, a subscript n plus 1 end subscript

        • Note that a subscript n plus 1 end subscript is the absolute value of the (n+1)th term

    • and the sign of S minus s subscript n is the same as the sign of that next term

      • i.e. if the next term is negative then S minus s subscript n less than 0 space space rightwards double arrow space space s subscript n greater than S

      • or if the next term is positive then S minus s subscript n greater than 0 space space rightwards double arrow space space s subscript n less than S

Examiner Tips and Tricks

Make sure that the alternating series converges first, before attempting to determine an error bound! You can do this using the alternating series test, or by showing the series converges absolutely.

  • As more and more terms are added to a converging alternating series, the sum of the terms will continually 'flip flop' above and below its infinite sum

    • However it will keep getting closer and closer to that infinite sum

    • An example is shown in the table below for the first 9 terms of the alternating harmonic series sum from n equals 1 to infinity of open parentheses negative 1 close parentheses to the power of n plus 1 end exponent over n equals ln space 2 equals 0.693147...

      • The s subscript n column (sum of the first n terms) tends towards ln space 2 equals 0.693147...

      • The S minus s subscript n column, the error between ln space 2 and s subscript n, tends towards zero

      • The absolute value of the numbers in the S minus s subscript n column are always less than the absolute value of the numbers in the (n+1)th term column, i.e. open vertical bar S minus s subscript n close vertical bar less than a subscript n plus 1 end subscript

n

sn

S-sn

(n+1)th term

1

1

negative 0.306852...

negative 0.5

2

0.5

0.193147...

0.333333...

3

0.833333...

negative 0.140186...

negative 0.25

4

0.583333...

0.109813...

0.2

5

0.783333...

negative 0.090186...

negative 0.166666...

6

0.616666...

0.076480...

0.142857...

7

0.759523...

negative 0.066376...

negative 0.125

8

0.634523...

0.058623...

0.111111...

9

0.745634...

negative 0.052487...

negative 0.1

Worked Example

The series sum from n equals 1 to infinity of open parentheses negative 1 close parentheses to the power of n plus 1 end exponent over n squared satisfies the hypotheses of the alternating series test, i.e. 1 over 1 squared greater or equal than 1 over 2 squared greater or equal than 1 over 3 squared greater or equal than 1 over 4 squared greater or equal than... and limit as n rightwards arrow infinity of 1 over n squared equals 0.

If sum from n equals 1 to infinity of open parentheses negative 1 close parentheses to the power of n plus 1 end exponent over n squared equals S and s subscript n is the nth partial sum, find the minimum value of n for which the alternating series error bound guarantees that open vertical bar S minus s subscript n close vertical bar less than 0.001.

We are told the conditions of the alternating series test are satisfied, which means that the series converges

Therefore the error bound is given by open vertical bar S minus s subscript n close vertical bar less than open vertical bar open parentheses n plus 1 close parentheses to the power of th space term close vertical bar equals 1 over open parentheses n plus 1 close parentheses squared

open vertical bar S minus s subscript n close vertical bar less than 1 over open parentheses n plus 1 close parentheses squared is guaranteed, so setting 0.001 to be greater than 1 over open parentheses n plus 1 close parentheses squared will guarantee that open vertical bar S minus s subscript n close vertical bar less than 0.001

table row cell 1 over open parentheses n plus 1 close parentheses squared end cell less than cell 0.001 end cell end table

table row blank rightwards double arrow cell space space open vertical bar S minus s subscript n close vertical bar less than 1 over open parentheses n plus 1 close parentheses squared less than 0.001 end cell end table

table row blank rightwards double arrow cell space space open vertical bar S minus s subscript n close vertical bar less than 0.001 end cell end table

Solve table row cell 1 over open parentheses n plus 1 close parentheses squared end cell less than cell 0.001 end cell end table for n

You can multiply both sides of the inequality by open parentheses n plus 1 close parentheses squared as it is a positive quantity (so won't reverse the inequality sign)

table row cell 1 over open parentheses n plus 1 close parentheses squared end cell less than cell 0.001 end cell row cell fraction numerator 1 over denominator 0.001 end fraction end cell less than cell open parentheses n plus 1 close parentheses squared end cell row 1000 less than cell open parentheses n plus 1 close parentheses squared end cell row cell square root of 1000 end cell less than cell n plus 1 end cell end table

rightwards double arrow space space n greater than square root of 1000 minus 1 equals 30.622776...

Remember that n must be an integer

rightwards double arrow space space n greater or equal than 31

A minimum value of n equals 31 guarantees that open vertical bar S minus s subscript n close vertical bar less than 0.001

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    Author
    Reviewer
    Roger B

    Author: Roger B

    Expertise: Maths

    Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

    Mark Curtis

    Author: Mark Curtis

    Expertise: Maths

    Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.