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

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 n=1(1)n·an or n=1(1)n+1·an

    • If the infinite series converges to a sum S

      • and if sn is the partial sum of the first n terms

    • Then |Ssn|an+1

      • 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) sn is less than the absolute value of the next term in the series, an+1

        • Note that an+1 is the absolute value of the (n+1)th term

    • and the sign of Ssn is the same as the sign of that next term

      • If the next term is negative then Ssn0    snS

      • or if the next term is positive then Ssn0    snS

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 two tables below for the first 9 terms of the alternating harmonic series n=1(1)n+1n=ln 2=0.693147...

    • The sn column (sum of the first n terms) tends towards ln 2=0.693147...

    • The Ssn column, the error between ln 2 and sn, tends towards zero

      • In the first table you can see how sn alternates between being above and below the value of the series sum

    • The absolute value of the numbers in the Ssn column are always less than the absolute value of the numbers in the (n+1)th term column, i.e. |Ssn|an+1

n

sn

S-sn

(n+1)th term

1

1

0.306852...

0.5

2

0.5

0.193147...

0.333333...

3

0.833333...

0.140186...

0.25

4

0.583333...

0.109813...

0.2

5

0.783333...

0.090186...

0.166666...

6

0.616666...

0.076480...

0.142857...

7

0.759523...

0.066376...

0.125

8

0.634523...

0.058623...

0.111111...

9

0.745634...

0.052487...

0.1

n

sn

|S-sn|

|(n+1)th term|

1

1

0.306852...

0.5

2

0.5

0.193147...

0.333333...

3

0.833333...

0.140186...

0.25

4

0.583333...

0.109813...

0.2

5

0.783333...

0.090186...

0.166666...

6

0.616666...

0.076480...

0.142857...

7

0.759523...

0.066376...

0.125

8

0.634523...

0.058623...

0.111111...

9

0.745634...

0.052487...

0.1

Worked Example

The series n=1(1)n+1n2 satisfies the conditions of the alternating series test, that is 112122132142... and limn1n2=0.

If n=1(1)n+1n2=S and sn is the nth partial sum, find the minimum value of n for which the alternating series error bound guarantees that |Ssn|0.001.

Answer:

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 |Ssn||an+1|=1(n+1)2

|Ssn|1(n+1)2 is guaranteed, so setting 0.001 to be greater than 1(n+1)2 will guarantee that |Ssn|<0.001

1(n+1)2<0.001

  |Ssn|1(n+1)2<0.001

  |Ssn|<0.001

Solve 1(n+1)2<0.001 for n

You can multiply both sides of the inequality by (n+1)2 as it is a positive quantity (so won't reverse the inequality sign)

1(n+1)2<0.00110.001<(n+1)21000<(n+1)21000<n+1

  n>10001=30.622776...

Remember that n must be an integer

  n31

A minimum value of n=31 guarantees that |Ssn|0.001

Examiner Tips and Tricks

Sometimes, the bound given using the alternating series error bound is less than the bound asked for in the question. In this case, you must state that the error bound is less than the given bound.

For example, you might be asked to show that |Lsn|<110, but an+1=115. So you should conclude that |Lsn|<115<110.

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Build on this topic

Roger B

Author: Roger B

Expertise: Development Editor

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

Reviewer: Mark Curtis

Expertise: Maths Content Creator

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.