Pure Mathematics | Series

Welcome to the Chapter on Series!

Hello future mathematician! This chapter, Series, is one of the most elegant and satisfying topics in Further Pure Mathematics 1 (FP1).
At first glance, working with complicated sums might seem intimidating, but this unit provides powerful tools and clever tricks to simplify them dramatically.

What you will learn: You will master the use of summation notation and, crucially, learn the Method of Differences (also known as telescoping series) to find the exact sum of a finite series, even when it involves complex fractions.
Why it's important: These techniques are fundamental for understanding Calculus and higher-level Mathematics, providing rigorous ways to handle infinite processes and sums. Let's dive in!

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Section 1: Prerequisites – Summation Notation (\(\sum\))

1.1 Understanding Sigma Notation

The capital Greek letter sigma, \(\sum\), simply means "sum up" or "add together". It is a compact way of writing a long addition problem.

A general series is written as: \[ S_n = \sum_{r=1}^{n} u_r \]

• \(u_r\): This is the formula for the r-th term of the series.
• \(r=1\): This is the starting value for the index r (the lower limit).
• \(n\): This is the finishing value for the index r (the upper limit).

Example: If you see \(\sum_{r=1}^{4} (2r)\), this means:
(2(1)) + (2(2)) + (2(3)) + (2(4)) = 2 + 4 + 6 + 8 = 20.

1.2 Basic Rules of Summation

Series summation follows basic algebraic rules, which are essential when manipulating complex sums:

Rule 1: Constants
You can pull a constant (a number, k, that doesn't depend on r) outside the summation sign. \[ \sum_{r=1}^{n} k u_r = k \sum_{r=1}^{n} u_r \]

Rule 2: Addition/Subtraction
The sum of terms can be split into separate sums. \[ \sum_{r=1}^{n} (u_r + v_r) = \sum_{r=1}^{n} u_r + \sum_{r=1}^{n} v_r \]

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Section 2: The Standard Sums (Building Blocks)

In FP1, we often encounter sums that require you to substitute standard formulae. These formulae are typically provided in your examination booklet, but familiarity is key. They allow you to sum polynomials of r up to \(r^3\).

2.1 The Essential Standard Results

1. Sum of the first \(n\) integers (\(r\)): \[ \sum_{r=1}^{n} r = \frac{1}{2} n(n+1) \]

2. Sum of the first \(n\) squares (\(r^2\)): \[ \sum_{r=1}^{n} r^2 = \frac{1}{6} n(n+1)(2n+1) \]

3. Sum of the first \(n\) cubes (\(r^3\)): \[ \sum_{r=1}^{n} r^3 = \frac{1}{4} n^2(n+1)^2 \quad \text{or } \left( \sum_{r=1}^{n} r \right)^2 \]

2.2 Using Standard Sums to Solve Problems

When calculating a sum like \(\sum_{r=1}^{n} (2r^2 - r)\), we use the rules of summation to split it up, then apply the standard formulas.

Step-by-Step Example: Calculate \(S_n = \sum_{r=1}^{n} (2r^2 - r)\)

1. Split the Sum: \[ S_n = \sum_{r=1}^{n} 2r^2 - \sum_{r=1}^{n} r \]
2. Pull out Constants: \[ S_n = 2 \sum_{r=1}^{n} r^2 - \sum_{r=1}^{n} r \]
3. Substitute Formulas: \[ S_n = 2 \left[ \frac{1}{6} n(n+1)(2n+1) \right] - \left[ \frac{1}{2} n(n+1) \right] \]
4. Simplify (Factorisation is key!):

   This involves lots of algebraic simplification, often by factorising out the common factor \(\frac{1}{6} n(n+1)\) or \(\frac{1}{2} n(n+1)\).
   (In this case, factoring out \(\frac{1}{3} n(n+1)\) from the first term and \(\frac{1}{2} n(n+1)\) from the second helps.)

   If we use \(\frac{1}{2} n(n+1)\) as the common factor: \[ S_n = \frac{1}{2} n(n+1) \left[ \frac{2}{3}(2n+1) - 1 \right] \] \[ S_n = \frac{1}{2} n(n+1) \left[ \frac{4n+2}{3} - \frac{3}{3} \right] = \frac{1}{2} n(n+1) \left[ \frac{4n-1}{3} \right] \] \[ S_n = \frac{1}{6} n(n+1)(4n-1) \]

Memory Aid: Notice the lovely connection: \(\sum r^3 = (\sum r)^2\). A neat pattern!

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Section 3: The Method of Differences (Telescoping Series)

The standard sums work well for polynomials, but what if the general term \(u_r\) involves fractions or products? This is where the Method of Differences comes into play.

3.1 The Concept of Telescoping

Imagine you have an old brass telescope. When you collapse it, the sections slide inside each other and disappear, leaving only the front and back pieces.

The Method of Differences relies on expressing the general term \(u_r\) as the difference between two functions of \(r\): \[ u_r = f(r) - f(r+1) \]

When we sum this series, almost every intermediate term cancels out. This cancellation process is called telescoping.

Let's see the series \(S_n = \sum_{r=1}^{n} (f(r) - f(r+1))\):

• When \(r=1\): \(u_1 = f(1) - f(2)\)
• When \(r=2\): \(u_2 = f(2) - f(3)\)
• When \(r=3\): \(u_3 = f(3) - f(4)\)
• ...
• When \(r=n\): \(u_n = f(n) - f(n+1)\)

Adding them all together: \[ S_n = \begin{array}{l} (f(1) - \cancel{f(2)}) \\ + (\cancel{f(2)} - \cancel{f(3)}) \\ + (\cancel{f(3)} - \cancel{f(4)}) \\ + \ldots \\ + (\cancel{f(n)} - f(n+1)) \end{array} \]

The vast majority cancel, leaving only: \[ S_n = f(1) - f(n+1) \]

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Section 4: The Process – Using Partial Fractions

In FP1, the terms \(u_r\) that require the Method of Differences are usually rational functions (fractions) that must be split using Partial Fractions.

4.1 Step-by-Step Guide for Telescoping Series

Consider the series: \(S_n = \sum_{r=1}^{n} \frac{2}{r(r+2)}\)

Step 1: Decompose \(u_r\) using Partial Fractions

We must express the term as two separate, simpler fractions. \[ \frac{2}{r(r+2)} = \frac{A}{r} + \frac{B}{r+2} \] Multiplying by \(r(r+2)\) gives: \(2 = A(r+2) + Br\).

• Set \(r=0\): \(2 = A(2) \implies A=1\)
• Set \(r=-2\): \(2 = B(-2) \implies B=-1\)

So, the general term is: \[ u_r = \frac{1}{r} - \frac{1}{r+2} \] We now have a difference! (Here, \(f(r) = 1/r\))

Step 2: Write out the Terms and Identify Cancellation Pattern

Because the difference is \(f(r) - f(r+2)\) (a difference of 2 steps), the cancellation is slightly more complex than \(f(r) - f(r+1)\).

We strongly encourage students to write out at least the first three terms and the last three terms. Do not skip this!

\[ S_n = u_1 + u_2 + u_3 + \ldots + u_{n-1} + u_n \]

$$ \begin{array}{rcl} r=1: & u_1 = & \mathbf{1} - \cancel{\frac{1}{3}} \\ r=2: & u_2 = & \mathbf{\frac{1}{2}} - \cancel{\frac{1}{4}} \\ r=3: & u_3 = & \cancel{\frac{1}{3}} - \cancel{\frac{1}{5}} \\ r=4: & u_4 = & \cancel{\frac{1}{4}} - \cancel{\frac{1}{6}} \\ \vdots & & \vdots \\ r=n-1: & u_{n-1} = & \cancel{\frac{1}{n-1}} - \mathbf{\frac{1}{n+1}} \\ r=n: & u_n = & \cancel{\frac{1}{n}} - \mathbf{\frac{1}{n+2}} \end{array} $$

Step 3: Collect the Remaining Terms

After cancellation, the remaining (un-cancelled) terms are: \[ S_n = 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2} \]

Step 4: Combine into a Single Expression (Optional, but often required)

\[ S_n = \frac{3}{2} - \frac{1}{n+1} - \frac{1}{n+2} \] To combine the fractions involving \(n\), the common denominator is \((n+1)(n+2)\): \[ S_n = \frac{3}{2} - \left[ \frac{(n+2) + (n+1)}{(n+1)(n+2)} \right] \] \[ S_n = \frac{3}{2} - \frac{2n+3}{(n+1)(n+2)} \]

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Section 5: Dealing with Infinite Series and Other Structures

5.1 Summing to Infinity (\(S_{\infty}\))

A standard requirement in these problems is to find the sum of the series as \(n\) tends to infinity, \(S_{\infty}\). This only works if the terms of the series tend to zero (which they generally do in FP1 telescoping examples).

We simply take the limit of the finite sum \(S_n\) as \(n \to \infty\). \[ S_{\infty} = \lim_{n \to \infty} S_n \]

Returning to our example from Section 4: \[ S_n = \frac{3}{2} - \frac{1}{n+1} - \frac{1}{n+2} \]

As \(n \to \infty\), the fractions \(\frac{1}{n+1}\) and \(\frac{1}{n+2}\) both tend to zero. \[ S_{\infty} = \frac{3}{2} - 0 - 0 = \frac{3}{2} \]

5.2 Handling Starting Index Differences

If the series starts at \(r=k\) instead of \(r=1\), you cannot use the standard formula \(\sum_{r=1}^{n} u_r = f(1) - f(n+1)\) directly.

The standard method for non-\(r=1\) starts: \[ \sum_{r=k}^{n} u_r = \sum_{r=1}^{n} u_r - \sum_{r=1}^{k-1} u_r \]

This means you calculate the sum up to \(n\) starting from 1, and then subtract the terms you didn't want (the terms from 1 up to \(k-1\)).

Did you know? Telescoping series are not just theoretical! They are used heavily in probability and statistics, especially in analyzing sequences of events.

5.3 Summation by Differences with Products

Sometimes, \(u_r\) is a product of terms, not a fraction (e.g., \(u_r = r(r+1)\)). While these could be solved using standard sums (\(r^2+r\)), they can also sometimes be set up as a difference of products.

For example, for products, the key identity is often: \[ r(r+1) = \frac{1}{3} [r(r+1)(r+2) - (r-1)r(r+1)] \]

In this case, \(f(r) = r(r+1)(r+2)\), and \(u_r\) becomes \(\frac{1}{3} [f(r) - f(r-1)]\). This results in a telescoping series where the remaining terms are at \(r=n\) and \(r=0\) (if we include \(r-1\)).

Keep practicing those partial fractions and algebraic simplifications – they are the biggest hurdles in this chapter! You've got this!