Summing Series Using Partial Fractions Revision Notes for AQA A-Level Further Maths

Summing Series Using Partial Fractions

Introduction to the method of differences

When you need to find the sum of a series, one powerful technique is the method of differences. This method works when the general term of a series can be written as the difference between two consecutive function values.

infoNote

Key principle: If the general term of a function can be expressed as $f(r+1) - f(r)$ , you can find the sum of the series using the method of differences.

The beauty of this method lies in the cancellation that occurs. When you write out the terms of the series and arrange them vertically, most terms cancel out, leaving only the first few and last few terms. This dramatic simplification transforms what might seem like a complex summation into a straightforward calculation.

Understanding partial fractions in series

To apply the method of differences effectively, you often need to express your series term using partial fractions. Partial fractions allow you to decompose a single complex rational expression into a sum or difference of simpler fractions.

chatImportant

Before attempting partial fraction decomposition, you must check that the degree of the numerator is smaller than the degree of the denominator. If this condition is not met, you need to perform polynomial division first.

The standard form for partial fractions with two linear factors is:

$\frac{px+q}{(x-a)(x-b)} = \frac{A}{(x-a)} + \frac{B}{(x-b)}$

To find the constants $A$ , $B$ , etc., you can use two main approaches:

Substitution method: Multiply through by the denominator, then substitute strategic values (usually values that make factors equal to zero) to eliminate variables and solve for each constant.
Comparing coefficients: Multiply through by the denominator, expand, and then compare the coefficients of like powers of the variable on both sides.

Often, a combination of both methods provides the most efficient solution.

The telescoping technique

Once you have expressed your series term as a difference using partial fractions, you can apply the method of differences. The key is to write the differences for successive terms vertically and identify which terms cancel.

When you sum a series where each term is of the form $f(r+1) - f(r)$ , the partial sums create a pattern where intermediate values appear with both positive and negative signs, causing them to cancel. This is called a telescoping series because, like a telescope, it collapses down to just a few terms.

infoNote

The terms that survive cancellation depend on the structure of your partial fractions. For example, if your partial fractions have factors differing by 2 (like $r$ and $r+2$ ), then terms two rows apart will cancel.

Worked example 1: Proving the sum formula

lightbulbExample

Worked Example: Proving the Natural Number Sum Formula

Let's prove that $\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$ using the method of differences.

Given identity: $2r \equiv r(r+1) - r(r-1)$

Step 1: Start with the identity and sum both sides:

$\sum_{r=1}^{n} 2r = \sum_{r=1}^{n} [r(r+1) - r(r-1)]$

Step 2: Write out the differences vertically to see the cancellation pattern:

$2\sum_{r=1}^{n} r = [2 - 0] + [6 - 2] + [12 - 6] + \ldots + [(n-1)n - (n-1)(n-2)] + [n(n+1) - n(n-1)]$

Arranging vertically:

Term 1: $2 - 0$
Term 2: $6 - 2$
Term 3: $12 - 6$
...
Last term: $n(n+1) - n(n-1)$

Step 3: Notice that all middle terms cancel. The $2$ from term 1 cancels with the $-2$ from term 2, the $6$ from term 2 cancels with the $-6$ from term 3, and so on.

After cancellation: $2\sum_{r=1}^{n} r = n(n+1) - 0$

Step 4: Solve for the sum:

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

This result can also be verified using the formula for an arithmetic progression, demonstrating the consistency of the method.

Worked example 2: Summing with two factors

lightbulbExample

Worked Example: Summing a Series with Two Factors

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

Part (a): Find the partial fractions

Let $\frac{1}{r(r+2)} = \frac{A}{r} + \frac{B}{r+2}$

Multiply through by the denominator $r(r+2)$ :

$1 \equiv A(r+2) + Br$

Finding constant A: Substitute $r = 0$ :

$1 = A(2) + B(0)$

$1 = 2A$

$A = \frac{1}{2}$

Finding constant B: Substitute $r = -2$ :

$1 = A(0) + B(-2)$

$1 = -2B$

$B = -\frac{1}{2}$

Therefore: $\frac{1}{r(r+2)} = \frac{1}{2r} - \frac{1}{2(r+2)} = \frac{1}{2}\left(\frac{1}{r} - \frac{1}{r+2}\right)$

Part (b): Use the method of differences

$\sum_{r=1}^{n} \frac{1}{r(r+2)} = \frac{1}{2}\sum_{r=1}^{n}\left(\frac{1}{r} - \frac{1}{r+2}\right)$

Write the differences vertically:

$r=1$ : $\frac{1}{1} - \frac{1}{3}$
$r=2$ : $\frac{1}{2} - \frac{1}{4}$
$r=3$ : $\frac{1}{3} - \frac{1}{5}$
...
$r=n-1$ : $\frac{1}{n-1} - \frac{1}{n+1}$
$r=n$ : $\frac{1}{n} - \frac{1}{n+2}$

Notice that the terms that cancel are two rows apart. The $\frac{1}{3}$ from row 1 cancels with the $\frac{1}{3}$ from row 3, and so on.

After cancellation, we're left with:

$= \frac{1}{2}\left[1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2}\right]$

$= \frac{1}{2}\left[\frac{3}{2} - \frac{1}{n+1} - \frac{1}{n+2}\right]$

This can be simplified by factorising:

$= \frac{1}{2}\left[\frac{3}{2} - \frac{(n+2) + (n+1)}{(n+1)(n+2)}\right]$

$= \frac{1}{2}\left[\frac{3}{2} - \frac{2n+3}{(n+1)(n+2)}\right]$

$= \frac{1}{2} \times \frac{3(n+1)(n+2) - 2(2n+3)}{2(n+1)(n+2)}$

$= \frac{3n^2 + 9n + 6 - 4n - 6}{4(n+1)(n+2)}$

Final result:

$= \frac{3n^2 + 5n}{4(n+1)(n+2)} = \frac{n(3n+5)}{4(n+1)(n+2)}$

infoNote

Exam tip: When terms cancel two rows apart, be systematic in your working to ensure you identify the correct surviving terms.

Worked example 3: Infinite series with three factors

lightbulbExample

Worked Example: Infinite Series with Three Factors

Express $\frac{16}{r(r+2)(r+4)}$ in partial fractions, then show that $\sum_{r=1}^{\infty} \frac{16}{r(r+2)(r+4)} = \frac{11}{6}$

Step 1: Set up the partial fractions

$\frac{16}{r(r+2)(r+4)} = \frac{A}{r} + \frac{B}{r+2} + \frac{C}{r+4}$

Multiply through by the denominator:

$16 \equiv A(r+2)(r+4) + Br(r+4) + Cr(r+2)$

Step 2: Find the constants

Finding A: Substitute $r = 0$ :

$16 = A(2)(4) = 8A$

$A = 2$

Finding B: Substitute $r = -2$ :

$16 = B(-2)(2) = -4B$

$B = -4$

Finding C: Substitute $r = -4$ :

$16 = C(-4)(-2) = 8C$

$C = 2$

Therefore: $\frac{16}{r(r+2)(r+4)} = \frac{2}{r} - \frac{4}{r+2} + \frac{2}{r+4}$

Step 3: Apply the method of differences

$\sum_{r=1}^{n} \frac{16}{r(r+2)(r+4)} = \sum_{r=1}^{n}\left(\frac{2}{r} - \frac{4}{r+2} + \frac{2}{r+4}\right)$

Writing the terms vertically and identifying cancellations (this time terms cancel several rows apart due to the three-factor structure):

After systematic cancellation:

$= \frac{2}{1} - \frac{4}{3} + \frac{2}{2} - \frac{4}{4} + \frac{2}{3} - \frac{4}{4} + \frac{2}{n+1} + \frac{2}{n+2} - \frac{4}{n+2} + \frac{2}{n+4}$

Collecting surviving terms:

$= \frac{11}{6} - \frac{2}{n+1} - \frac{2}{n+2} - \frac{2}{n+3} - \frac{2}{n+4}$

Step 4: Consider the limit as n → ∞

For an infinite series, we need to consider what happens as $n \to \infty$ :

As $n \to \infty$ , each of $\frac{2}{n+1}$ , $\frac{2}{n+2}$ , $\frac{2}{n+3}$ , and $\frac{2}{n+4}$ approaches $0$ .

Therefore:

$\sum_{r=1}^{\infty} \frac{16}{r(r+2)(r+4)} = \frac{11}{6}$

infoNote

Exam tip: When dealing with infinite series, always explicitly state that you're taking the limit as $n \to \infty$ and show that the remaining terms vanish.

Key exam guidance

Checking before you start:

Always verify that the degree of the numerator is less than the degree of the denominator before attempting partial fractions.
If the degrees are equal or the numerator has higher degree, perform polynomial division first.

Finding constants efficiently:

Use the substitution method for speed - choose values that make factors zero.
For checking your answer, you can use a different substitution value or compare coefficients.

The cancellation pattern:

Write terms vertically to clearly see the pattern.
The spacing of cancellation depends on the difference between factors in the denominator.
For factors like $r$ and $r+2$ , terms two rows apart cancel.
For factors like $r$ and $r+1$ , adjacent terms cancel.

chatImportant

Common mistakes to avoid:

Forgetting to multiply through by the denominator when setting up partial fractions.
Arithmetic errors when simplifying the final expression - always factorise carefully.
Missing the limit consideration for infinite series.
Not showing enough working for the cancellation process in exams.

Remember!

bookmarkSummary

Key Points to Remember:

The method of differences works when the general term can be written as $f(r+1) - f(r)$ , causing a telescoping effect where middle terms cancel.
Always check the degree of numerator and denominator before attempting partial fractions - numerator degree must be smaller.
Use the substitution method to find constants in partial fractions by choosing values that make factors equal to zero.
Write the series terms vertically to clearly identify which terms cancel in the telescoping process.
For infinite series, take the limit as $n \to \infty$ and show that the remaining terms approach zero to find the final sum.

Summing Series Using Partial Fractions (AQA A-Level Further Maths): Revision Notes