Study Guide

Summation of Series

CIE A-Level Further MathematicsΒ· 30 min read

1. Standard Summation Formulaeβ˜…β˜…β˜†β˜†β˜†β± 15 min

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πŸ“˜ Definition

Standard Power Summation Formulae

βˆ‘r=1nr,βˆ‘r=1nr2,βˆ‘r=1nr3\sum_{r=1}^n r, \sum_{r=1}^n r^2, \sum_{r=1}^n r^3

Closed-form expressions for the sum of the first positive integer powers of , proven by induction. These are used to sum polynomial expressions in .

Example:

Used to find $ \sum_{r=1}^n (2r^2 - 5r + 1)$ by splitting into individual sums.

πŸ“ Worked Example

Find the value of $ \sum_{r=1}^{20} (r^2 - 3r)$

  1. 1

    Split the sum into separate standard sums using linearity:

  2. 2
    βˆ‘r=120r2βˆ’3βˆ‘r=120r\sum_{r=1}^{20} r^2 - 3 \sum_{r=1}^{20} r
  3. 3

    Substitute the standard formulae with :

  4. 4
    n(n+1)(2n+1)6βˆ’3β‹…n(n+1)2=20(21)(41)6βˆ’3β‹…20(21)2\frac{n(n+1)(2n+1)}{6} - 3 \cdot \frac{n(n+1)}{2} = \frac{20(21)(41)}{6} - 3 \cdot \frac{20(21)}{2}
  5. 5

    Calculate each term and simplify:

  6. 6
    2870βˆ’630=22402870 - 630 = 2240

Exam tip:

Always check the upper and lower bounds of the sum before substituting formulae.

2. Method of Differences for Telescoping Seriesβ˜…β˜…β˜…β˜†β˜†β± 20 min

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πŸ“˜ Definition

Telescoping Series

A series where intermediate terms cancel out when expanded fully, leaving only a small number of terms from the start and end of the sequence.

Example:

Any series of the form $ \sum_{r=1}^n [f(r) - f(r+1)]$ is telescoping.

πŸ“ Worked Example

Find $ \sum_{r=1}^n \frac{1}{r(r+1)}$ by writing the general term as a difference.

  1. 1

    Decompose the general term using partial fractions:

  2. 2
    1r(r+1)=1rβˆ’1r+1\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}
  3. 3

    Write out the first three and last three terms to see cancellation:

  4. 4
    (1βˆ’12)+(12βˆ’13)+(13βˆ’14)+...+(1nβˆ’1βˆ’1n)+(1nβˆ’1n+1)\left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + ... + \left(\frac{1}{n-1} - \frac{1}{n}\right) + \left(\frac{1}{n} - \frac{1}{n+1}\right)
  5. 5

    All intermediate terms cancel, leaving only the first term of the first expression and the last term of the last expression:

  6. 6
    Sn=1βˆ’1n+1=nn+1S_n = 1 - \frac{1}{n+1} = \frac{n}{n+1}

3. Summing Rational Series with Partial Fractionsβ˜…β˜…β˜…β˜…β˜†β± 25 min

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Most rational functions of can be decomposed into partial fractions, then written as a difference to form a telescoping series. This is one of the most common exam questions on this topic.

πŸ“ Worked Example

Find $ \sum_{r=1}^n \frac{1}{r(r+2)}$

  1. 1

    Decompose into partial fractions:

  2. 2
    1r(r+2)=12(1rβˆ’1r+2)\frac{1}{r(r+2)} = \frac{1}{2} \left( \frac{1}{r} - \frac{1}{r+2} \right)
  3. 3

    Expand terms to see which cancel:

  4. 4
    12[(1βˆ’13)+(12βˆ’14)+(13βˆ’15)+...+(1nβˆ’1βˆ’1n+1)+(1nβˆ’1n+2)]\frac{1}{2} \left[ \left(1 - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + ... + \left(\frac{1}{n-1} - \frac{1}{n+1}\right) + \left(\frac{1}{n} - \frac{1}{n+2}\right) \right]
  5. 5

    Cancel intermediate terms, leaving the first two positive terms and last two negative terms:

  6. 6
    Sn=12(1+12βˆ’1n+1βˆ’1n+2)=3n2+5n4(n+1)(n+2)S_n = \frac{1}{2} \left( 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2} \right) = \frac{3n^2 + 5n}{4(n+1)(n+2)}
βœ“ Quick check

What is the value of $ \sum_{r=1}^3 \frac{1}{r(r+2)}$ using the formula above?

  1. Select the correct value:

    • \frac{2}{3}

    • \frac{5}{6}

    • \frac{3}{4}

    • \frac{11}{12}

    Reveal answer
    1 β€”

    Correct: substituting gives $ \frac{3(9) + 5(3)}{4(4)(5)} = \frac{42}{80}? No, wait: $ \frac{1}{2}(1 + 1/2 - 1/4 - 1/5) = 1/2 ( (10 + 5 - 2.5 - 1)/10 ) = 5/6. Yes.

4. Common Pitfalls

Wrong move:

Forgetting to adjust the sum when it starts at .

Why:

All standard formulae give the sum from to , so starting at a higher bound gives an incorrect result.

Correct move:

Use $ \sum_{r=a}^n f(r) = \sum_{r=1}^n f(r) - \sum_{r=1}^{a-1} f(r)$ to adjust the bounds before calculating.

Wrong move:

Cancelling too many terms when the gap between terms is greater than 1.

Why:

For differences of the form , the first and last terms remain uncancelled.

Correct move:

Always write out the first 3-4 and last 3-4 terms explicitly to check which terms are left.

Wrong move:

Sign errors when splitting into partial fractions.

Why:

Incorrect signs mean the terms will not cancel correctly, leading to a wrong final expression.

Correct move:

Check your partial fraction decomposition by substituting a value of (e.g. ) before starting summation.

Wrong move:

Using infinite series results for finite sum questions.

Why:

Confusion between summing the first terms and summing to infinity leads to missing the terms in the final answer.

Correct move:

Always read the question carefully to check if it asks for the sum of the first terms or the sum to infinity.

5. Quick Reference Cheatsheet

Sum Type

Formula

Sum of first

Sum of first

Sum of first

Telescoping

Telescoping

When this came up on past exams

AI-estimated based on syllabus patterns β€” cross-check with official past papers for accuracy. Use only as revision-focus signals.

  • 2023 Β· 1

    Telescoping series sum question

  • 2022 Β· 1

    Standard formulae summation

  • 2021 Β· 1

    Partial fraction summation

Going deeper

What's Next

Summation of series is a foundational technique for almost all further pure topics. You will use these methods again when proving results for series by induction, solving recurrence relations, and studying infinite series convergence in further pure 2. Exam questions often combine this topic with partial fractions and induction to create longer, multi-mark questions. Mastery of telescoping cancellation and standard formulae will save you valuable time in the exam and help you tackle more complex topics later in the course.