Further Mathematics (9665) | Series

Welcome to FP1 Series: Unlocking the Power of Summation!

Hello! This chapter, "Series," is all about finding clever and efficient ways to add up long lists of numbers that follow a specific pattern. Instead of adding term-by-term (which would take forever!), we use powerful formulas and techniques.

Mastering series is essential because it builds the foundation for more advanced calculus and mathematical modelling. Don't worry if the notation looks heavy; we will break down everything step-by-step!

1. Sums of Powers of Natural Numbers

This section deals with finding the sum of the first \(n\) natural numbers, their squares, and their cubes. Natural numbers are simply \(1, 2, 3, \dots\).

1.1 Standard Formulae (The Building Blocks)

You should be familiar with the summation notation, \(\sum_{r=1}^{n} f(r)\), which means adding up the expression \(f(r)\) for every integer value of \(r\) from 1 up to \(n\).

The core knowledge required here includes the formulas for the sum of the first \(n\) integers, squares, and cubes. These formulas are usually provided in your formulae booklet, but knowing how to use them is key!

A. Sum of the first \(n\) natural numbers (\(\sum r\))

This is often considered prerequisite knowledge, but it's crucial for context:
$$\sum_{r=1}^{n} r = 1 + 2 + 3 + \dots + n = \frac{1}{2} n (n+1)$$

B. Sum of the squares of the first \(n\) natural numbers (\(\sum r^2\))

This is the first major formula you must apply correctly in FP1:
$$\sum_{r=1}^{n} r^2 = 1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{1}{6} n (n+1) (2n+1)$$

C. Sum of the cubes of the first \(n\) natural numbers (\(\sum r^3\))

This formula looks complex, but notice the satisfying pattern—it's just the square of the \(\sum r\) formula!
$$\sum_{r=1}^{n} r^3 = 1^3 + 2^3 + 3^3 + \dots + n^3 = \frac{1}{4} n^2 (n+1)^2$$
Memory Aid: If you know the formula for \(\sum r\), you automatically know \(\sum r^3\) by squaring the entire expression!

1.2 Applying the Summation Rules

When dealing with series that are polynomial expressions (like \(\sum (r^2 + 2r)\)), we use the rules of linearity:

• The Sum Rule: We can split a sum into separate sums:
  $$\sum (a_r + b_r) = \sum a_r + \sum b_r$$
• The Constant Multiple Rule: We can pull constants outside the summation sign:
  $$\sum c \cdot a_r = c \sum a_r$$

Step-by-Step Example Process:

1. Expand: Write the expression in terms of powers of \(r\). For example, \(\sum_{r=1}^{n} r(r+2)\) becomes \(\sum_{r=1}^{n} (r^2 + 2r)\).
2. Split: Use the rules to split the sum: \(\sum r^2 + 2 \sum r\).
3. Substitute: Replace \(\sum r^2\) and \(\sum r\) with their respective formulas (using \(n\)).
4. Simplify: Factorise the resulting polynomial to reach the simplest answer. This often involves taking out common factors like \(\frac{1}{6} n(n+1)\).

2. Summation by the Method of Differences (Telescoping Series)

The Method of Differences (MoD) is an incredibly powerful technique used to sum finite series where the terms don't follow a simple polynomial pattern like \(r^2\) or \(r^3\).

2.1 The Concept of Telescoping

Imagine you have an old telescope. When you collapse it, the sections slide into each other and cancel out, leaving only the ends visible. This is exactly what happens in a telescoping series!

The essential idea is to rewrite the general term, \(u_r\), as the difference between two consecutive terms of another function, \(f(r)\):
$$u_r = f(r) - f(r-1) \quad \text{or} \quad u_r = f(r+1) - f(r)$$

When you sum the series, \(\sum u_r\), most of the internal terms cancel out:

Let \(S_n = \sum_{r=1}^{n} (f(r) - f(r-1))\):
$$S_n = (f(1) - f(0))$$ $$+ (f(2) - f(1))$$ $$+ (f(3) - f(2))$$ $$\dots$$ $$+ (f(n) - f(n-1))$$

Notice how the \(+f(1)\) cancels the \(-f(1)\), the \(+f(2)\) cancels the \(-f(2)\), and so on.

All the middle terms disappear, leaving only:
$$\mathbf{S_n = f(n) - f(0)}$$

2.2 Step-by-Step Procedure for MoD

In the FP1 syllabus, you will be expected to use MoD on series like \(\sum r \cdot r!\) or similar structures where the difference formula is either provided or easily constructed (often involving factorials or rational functions that resemble the examples in FP2's partial fractions).

Process:

1. Identify the difference: Ensure the term \(u_r\) is already written in the form \(f(r) - f(r-k)\).
   Example from the syllabus: You might be given the expression: $$r \cdot r! = (r+1)! - r!$$ In this case, \(u_r = r \cdot r!\) and \(f(r) = r!\). The difference form is \(f(r+1) - f(r)\).
2. List the terms: Write out the first few and last few terms of the summation, making sure to show the cancellation.
   • For \(r=1\): \(f(2) - f(1)\)
   • For \(r=2\): \(f(3) - f(2)\)
   • For \(r=3\): \(f(4) - f(3)\)
   • ...
   • For \(r=n-1\): \(f(n) - f(n-1)\)
   • For \(r=n\): \(f(n+1) - f(n)\)
3. Cancel: Observe that the second term of one line cancels the first term of the next line (the telescoping action).
4. State the Partial Sum (\(S_n\)): Collect the remaining terms—the "ends" of the telescope.
   In the \(f(r+1) - f(r)\) example above, \(S_n\) is: $$\mathbf{S_n = f(n+1) - f(1)}$$

Common Mistake to Avoid: Always check the index of summation (the starting value, often \(r=1\)). If the summation starts at \(r=k\) instead of \(r=1\), your final expression will be \(S_n = f(n+1) - f(k)\).

3. Extension to Infinite Series (Convergence)

Sometimes, we don't just want to find the sum up to \(n\) terms (\(S_n\)), but the sum of the entire infinite series, \(S_\infty\). This is only possible if the series converges.

3.1 Convergence and Partial Sums

An infinite series converges (has a finite sum) if and only if the limit of its partial sum \(S_n\) as \(n\) tends to infinity exists and is a finite number.

If the limit exists, we write:
$$\mathbf{S_{\infty} = \lim_{n \to \infty} S_n}$$

If \(\lim_{n \to \infty} S_n\) tends to infinity or does not settle on a single value, the series diverges and has no sum.

3.2 The Process for Finding \(S_{\infty}\)

To find the sum to infinity, you must first master the Method of Differences (Section 2), as this is the primary way you will find \(S_n\) for non-geometric series in FP1.

Step-by-Step Procedure:

1. Find \(S_n\): Use the Method of Differences to find a closed expression for the partial sum \(S_n\) (in terms of \(n\)).
2. Apply the Limit: Calculate the limit as \(n \to \infty\).
   $$\mathbf{S_{\infty} = \lim_{n \to \infty} (\text{Your expression for } S_n)}$$
3. Evaluate the terms: Focus on the terms involving \(n\). If \(n\) is in the denominator (e.g., \(\frac{1}{n}\) or \(\frac{1}{n+1}\)), that term tends to 0 as \(n \to \infty\).
   Example: If you found \(S_n = 5 - \frac{1}{n+1}\), then $$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left(5 - \frac{1}{n+1}\right) = 5 - 0 = 5$$ The series converges to 5.

Key Point: When dealing with infinite sums resulting from MoD, only the terms that contain \(n\) in the denominator will vanish. The constant terms and any terms from the starting point (e.g., \(f(1)\) or \(f(0)\)) will define the final sum.