✨ Finite Series: Comprehensive Study Notes (FP2 Pure Maths) ✨
Welcome to the chapter on Finite Series! Don't worry if the Greek letter Sigma (\(\Sigma\)) looks intimidating—this chapter is all about finding clever shortcuts to add up long lists of numbers following a pattern.
You already know how to sum Arithmetic Progressions (APs) and Geometric Progressions (GPs). This chapter extends that knowledge to much more complex patterns, using powerful algebraic techniques.
What will you learn? You will master standard formulas for sums of powers, and, most importantly, you will learn the fantastic Method of Differences (also called Telescoping Sums) to tackle almost any finite series.
1. The Building Blocks: Sums of Natural Numbers
In Further Maths, you need to be lightning-fast when calculating the sums of the first \(n\) natural numbers, the first \(n\) squares, and the first \(n\) cubes. These are your essential starting points (FP1.5 content).
1.1 Sum of the First \(n\) Integers (\(\sum r\))
This is the simplest sum, just adding \(1 + 2 + 3 + \dots + n\).
- The notation is: \(\sum_{r=1}^{n} r\)
- The formula is: \(\sum_{r=1}^{n} r = \frac{1}{2} n (n+1)\)
(Did you know? This formula is often attributed to the young mathematician Gauss!)
1.2 Sum of the First \(n\) Squares (\(\sum r^2\))
This is the sum of the squares: \(1^2 + 2^2 + 3^2 + \dots + n^2\).
- The notation is: \(\sum_{r=1}^{n} r^2\)
- The formula is: \(\sum_{r=1}^{n} r^2 = \frac{1}{6} n (n+1) (2n+1)\)
1.3 Sum of the First \(n\) Cubes (\(\sum r^3\))
This is the sum of the cubes: \(1^3 + 2^3 + 3^3 + \dots + n^3\).
- The notation is: \(\sum_{r=1}^{n} r^3\)
- The formula is: \(\sum_{r=1}^{n} r^3 = \frac{1}{4} n^2 (n+1)^2 = \left( \sum_{r=1}^{n} r \right)^2\)
🔑 Quick Review: Standard Sums
These three formulas must be known. They are often given in the formulae booklet, but knowing them speeds up your exam technique.
2. Summing Combined Series
Most series you encounter won't be pure \(\sum r\) or \(\sum r^2\). They will be polynomial expressions, for example: \(\sum_{r=1}^{n} (r^2 - 3r + 1)\).
2.1 The Rules of Summation (Linearity)
The beauty of the summation operator (\(\Sigma\)) is that it is linear. This means you can break a complicated series into simpler parts:
1. You can split the series across addition or subtraction:
\(\sum (u_r + v_r) = \sum u_r + \sum v_r\)
2. You can take constant multipliers outside the summation:
\(\sum c \cdot u_r = c \sum u_r\)
3. The sum of a constant \(c\) is just \(n\) times the constant (assuming the sum goes from \(r=1\) to \(n\)):
\(\sum_{r=1}^{n} c = nc\)
2.2 Step-by-Step Process for Polynomial Series
Example: Find a polynomial expression for \(S_n = \sum_{r=1}^{n} (r^2 - 3r + 1)\).
- Split the Series:
\(S_n = \sum r^2 - 3\sum r + \sum 1\)
- Apply Formulas: Substitute the standard results.
\(S_n = \frac{1}{6} n (n+1) (2n+1) - 3 \left[ \frac{1}{2} n (n+1) \right] + n\)
- Factorise and Simplify: This is the crucial step. Look for the largest common factor. Here, it’s \(\frac{1}{6}n\).
$$S_n = \frac{n}{6} [ (n+1)(2n+1) - 9(n+1) + 6 ]$$
Simplify the bracketed expression to get the final, concise polynomial form in terms of \(n\).
Key Takeaway: Always manipulate the expression into standard \(\sum r^k\) forms before substituting the formulas. Don't stop until you have fully factored and simplified your final polynomial expression for \(S_n\).
3. The Method of Differences (Telescoping Sums)
When the term \(u_r\) is a product or a rational function (a fraction), the standard formulas don't work. This is where the Method of Differences (FP1.5 & FP2.5) comes to the rescue. This technique is often examined.
3.1 The Concept: The Telescoping Analogy
Imagine a series where, when you write out the terms, almost everything cancels out. This is a Telescoping Sum. Think of an old handheld telescope—when you push it together, the middle sections disappear, and only the ends remain.
The Goal: To express the general term, \(u_r\), as the difference between two consecutive functions of \(r\):
$$u_r = f(r) - f(r+1) \quad \text{ or } \quad u_r = f(r+1) - f(r)$$
When you sum this series, \(S_n\), all the intermediate terms cancel:
\(S_n = \sum_{r=1}^{n} (f(r) - f(r+1))\)
\(S_n = [f(1) - f(2)] + [f(2) - f(3)] + [f(3) - f(4)] + \dots + [f(n) - f(n+1)]\)
All the inner terms (like \(-f(2) + f(2)\)) cancel, leaving only the first part of the first term and the last part of the last term:
$$S_n = f(1) - f(n+1)$$
3.2 Technique 1: Using Partial Fractions (For Rational Functions)
This method is essential for rational expressions (fractions) like \(\frac{1}{r(r+1)}\) or \(\frac{2}{(2r-1)(2r+1)}\).
Step 1: Express \(u_r\) using Partial Fractions.
Example: Sum the series \(S_n = \sum_{r=1}^{n} \frac{1}{r(r+1)}\)
We decompose the general term \(u_r\):
$$\frac{1}{r(r+1)} \equiv \frac{A}{r} + \frac{B}{r+1}$$
By substitution or algebraic manipulation, you find \(A=1\) and \(B=-1\).
$$u_r = \frac{1}{r} - \frac{1}{r+1}$$
Notice how \(u_r\) is now in the required difference form \(f(r) - f(r+1)\), where \(f(r) = \frac{1}{r}\).
Step 2: Write out the Partial Sum (\(S_n\)) and Observe the Cancellation.
Substitute the values of \(r\) from 1 to \(n\):
- \(r=1\): \(\frac{1}{1} - \frac{1}{2}\)
- \(r=2\): \(\frac{1}{2} - \frac{1}{3}\)
- \(r=3\): \(\frac{1}{3} - \frac{1}{4}\)
- \(\dots\)
- \(r=n\): \(\frac{1}{n} - \frac{1}{n+1}\)
When you add them up, the terms cancel diagonally:
$$S_n = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \dots + \left( \frac{1}{n} - \frac{1}{n+1} \right)$$
Step 3: State the Result for the Partial Sum (\(S_n\)).
The only remaining terms are the first part of the first bracket and the second part of the last bracket:
$$S_n = 1 - \frac{1}{n+1}$$
⚠️ Common Mistake Alert!
If your partial fractions result in a separation where terms jump (e.g., \(u_r = f(r) - f(r+2)\)), you will have two uncancelled terms at the start and two at the end. For example, if \(u_r = f(r) - f(r+2)\), then:
$$S_n = [f(1) + f(2)] - [f(n+1) + f(n+2)]$$
Always write out the first few and last few terms to check the pattern of cancellation!
3.3 Technique 2: Using Products (For Factorials and Polynomials)
Sometimes, the expression \(u_r\) involves factorials or products that can be cleverly rewritten as a difference. This is often necessary for series involving \(r!\) (FP1.5 example: \(\sum r \cdot r!\)).
Example: Sum the series \(S_n = \sum_{r=1}^{n} r \cdot r!\)
Step 1: Express \(u_r\) as a difference.
We know that \((r+1)! = (r+1) \times r!\).
We want to rewrite \(r \cdot r!\). We can use the identity:
$$r \cdot r! = (r+1 - 1) r!$$
$$r \cdot r! = (r+1)r! - 1 \cdot r!$$
$$r \cdot r! = (r+1)! - r!$$
We have successfully expressed \(u_r\) in the form \(f(r+1) - f(r)\), where \(f(r) = r!\).
Step 2: Apply the Telescoping Sum.
The sum becomes:
- \(r=1\): \(2! - 1!\)
- \(r=2\): \(3! - 2!\)
- \(r=3\): \(4! - 3!\)
- \(\dots\)
- \(r=n\): \((n+1)! - n!\)
Cancellation occurs: \((2!-2!) + (3!-3!) + \dots\)
Step 3: State the Result.
$$S_n = (n+1)! - 1!$$
$$S_n = (n+1)! - 1$$
💡 Memory Aid: The Difference Cheat
If you have an expression \(u_r = P(r) \cdot Q(r)\) where \(P(r)\) is a polynomial and \(Q(r)\) is a simple function like a factorial or a product of linear terms (e.g., \(r(r+1)\)), try to introduce a term that increases the index in \(Q(r)\) by one, as shown in the factorial example above.
4. Extension to Infinite Series
Sometimes, we are asked to find the sum of a series that continues forever—an Infinite Series (FP1.5 content).
4.1 Partial Sums and Convergence
The sum of an infinite series, denoted \(\sum_{r=1}^{\infty} u_r\), is defined as the Limit of the Partial Sum (\(S_n\)) as \(n\) tends to infinity:
$$\sum_{r=1}^{\infty} u_r = \lim_{n \to \infty} S_n$$
A series only has a finite sum if this limit exists. If the limit exists, the series is said to converge. If the limit does not exist (e.g., it tends to infinity), the series diverges.
4.2 Calculating the Sum to Infinity
To find the sum to infinity for a series solved using the Method of Differences, you must take the expression for \(S_n\) you derived and apply the limit.
Example: Find the sum to infinity of the series used in Section 3.2: \(\sum_{r=1}^{\infty} \frac{1}{r(r+1)}\)
Step 1: Use the Partial Sum \(S_n\).
We already found the partial sum:
$$S_n = 1 - \frac{1}{n+1}$$
Step 2: Apply the limit as \(n \to \infty\).
$$\sum_{r=1}^{\infty} u_r = \lim_{n \to \infty} \left( 1 - \frac{1}{n+1} \right)$$
As \(n\) gets very large, the term \(\frac{1}{n+1}\) tends towards zero:
$$\lim_{n \to \infty} \frac{1}{n+1} = 0$$
Step 3: State the Sum to Infinity.
$$\sum_{r=1}^{\infty} \frac{1}{r(r+1)} = 1 - 0 = 1$$
The series converges to 1.
Example of Divergence (Non-Examinable Series, but useful concept):
If you tried to find the sum to infinity of the sums of squares (Section 1.2), the result for \(S_n\) was a cubic in \(n\). As \(n \to \infty\), that cubic tends to \(\infty\). Therefore, the series of natural numbers and the series of squares diverge.
🎯 Chapter Summary: Key Takeaways
- Standard Sums: Use \(\sum r\), \(\sum r^2\), and \(\sum r^3\) for any polynomial series. Remember the linearity rules (splitting the sum and taking constants out).
- Method of Differences: This is your main tool for complex non-polynomial series. The key step is rewriting \(u_r\) as a difference \(f(r) - f(r+k)\) using algebraic manipulation or partial fractions.
- Telescoping: Ensure you correctly identify which terms are left uncancelled at the start and end of the series.
- Infinite Sums: The sum to infinity is calculated by finding \(\lim_{n \to \infty} S_n\). If this limit is a finite number, the series converges.
Keep practicing the algebraic manipulation and factorisation—it’s crucial for achieving the final, required polynomial form! You've got this!