# Summation of Series

> CIE A-Level Further Mathematics · 9231 Further Pure 1
> Source: https://www.owlsprep.com/study/cie-9231-u1-summation-of-series/

This module covers standard summation formulae for integer powers, the method of differences for telescoping series, and partial fraction decomposition to sum rational functions, all core exam techniques for CIE FP1.

**Prerequisites:** [Working knowledge of partial fractions](https://www.owlsprep.com/study/cie-9231-u1-partial-fractions/); Basic understanding of algebraic series

## Learning objectives

- Recall and apply standard summation formulae for powers of integers
- Use the method of differences to sum telescoping series
- Decompose rational series using partial fractions for summation
- Adjust summation bounds correctly for non-starting-at-1 series

## Standard Summation Formulae

**Standard Power Summation Formulae** — Closed-form expressions for the sum of the first $n$ positive integer powers of $r$, proven by induction. These are used to sum polynomial expressions in $r$.

*Notation:* \sum_{r=1}^n r, \sum_{r=1}^n r^2, \sum_{r=1}^n r^3

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

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

1. Split the sum into separate standard sums using linearity:
2. $$\sum_{r=1}^{20} r^2 - 3 \sum_{r=1}^{20} r$$
3. Substitute the standard formulae with $n=20$:
4. $$\frac{n(n+1)(2n+1)}{6} - 3 \cdot \frac{n(n+1)}{2}
= \frac{20(21)(41)}{6} - 3 \cdot \frac{20(21)}{2}$$
5. Calculate each term and simplify:
6. $$2870 - 630 = 2240$$

> **tip**
>
> If your sum starts at $r=a$ where $a>1$, use the identity $
> \sum_{r=a}^n f(r) = \sum_{r=1}^n f(r) - \sum_{r=1}^{a-1} f(r)$ to adjust the bounds.

> **Exam tip:** Always check the upper and lower bounds of the sum before substituting formulae.

*Calculator:* allowed

## Method of Differences for Telescoping Series

**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.

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

1. Decompose the general term using partial fractions:
2. $$\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}$$
3. Write out the first three and last three terms to see cancellation:
4. $$\left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + ... + \left(\frac{1}{n-1} - \frac{1}{n}\right) + \left(\frac{1}{n} - \frac{1}{n+1}\right)$$
5. All intermediate terms cancel, leaving only the first term of the first expression and the last term of the last expression:
6. $$S_n = 1 - \frac{1}{n+1} = \frac{n}{n+1}$$

> **warning**
>
> If the gap between terms is greater than 1 (e.g. $f(r) - f(r+2)$), you will have more than one remaining term at the start and end.

*Calculator:* allowed

## Summing Rational Series with Partial Fractions

Most rational functions of $r$ 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.

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

1. Decompose into partial fractions:
2. $$\frac{1}{r(r+2)} = \frac{1}{2} \left( \frac{1}{r} - \frac{1}{r+2} \right)$$
3. Expand terms to see which cancel:
4. $$\frac{1}{2} \left[ \left(1 - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + ... + \left(\frac{1}{n-1} - \frac{1}{n+1}\right) + \left(\frac{1}{n} - \frac{1}{n+2}\right) \right]$$
5. Cancel intermediate terms, leaving the first two positive terms and last two negative terms:
6. $$S_n = \frac{1}{2} \left( 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2} \right) = \frac{3n^2 + 5n}{4(n+1)(n+2)}$$

**Check your understanding**

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}

   *Answer:* \frac{5}{6}

   *Why:* Correct: substituting $n=3$ 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.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting to adjust the sum when it starts at $r=a \neq 1$.
  - Why it fails: All standard formulae give the sum from $r=1$ to $n$, so starting at a higher bound gives an incorrect result.
  - Correct: 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:** Cancelling too many terms when the gap between terms is greater than 1.
  - Why it fails: For differences of the form $f(r) - f(r+k)$, the first $k$ and last $k$ terms remain uncancelled.
  - Correct: Always write out the first 3-4 and last 3-4 terms explicitly to check which terms are left.
- **Wrong:** Sign errors when splitting into partial fractions.
  - Why it fails: Incorrect signs mean the terms will not cancel correctly, leading to a wrong final expression.
  - Correct: Check your partial fraction decomposition by substituting a value of $r$ (e.g. $r=1$) before starting summation.
- **Wrong:** Using infinite series results for finite sum questions.
  - Why it fails: Confusion between summing the first $n$ terms and summing to infinity leads to missing the $n$ terms in the final answer.
  - Correct: Always read the question carefully to check if it asks for the sum of the first $n$ terms or the sum to infinity.

## Cheatsheet

| Sum Type | Formula |
| --- | --- |
| Sum of first $n$ $r$ | $\sum_{r=1}^n r = \frac{n(n+1)}{2}$ |
| Sum of first $n$ $r^2$ | $\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}$ |
| Sum of first $n$ $r^3$ | $\sum_{r=1}^n r^3 = \left(\frac{n(n+1)}{2}\right)^2$ |
| Telescoping $f(r)-f(r+1)$ | $S_n = f(1) - f(n+1)$ |
| Telescoping $f(r)-f(r+k)$ | $S_n = \sum_{i=1}^k f(i) - \sum_{i=n+1}^{n+k} f(i)$ |

## 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.

- [Matrices](https://www.owlsprep.com/study/cie-9231-u1-matrices/)
- [Polar coordinates](https://www.owlsprep.com/study/cie-9231-u1-polar-coordinates/)
- [Vectors](https://www.owlsprep.com/study/cie-9231-u1-vectors/)

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