FP2 Chapter 2: Series — Method of Differences

FP2 Lecture Notes: Method of Differences

The method of differences is a powerful technique for finding sums of sequences, particularly those involving rational expressions. We begin with a fascinating problem that motivated the development of this method: proving that the sum of reciprocal squares converges.

Motivating Problem: Basel Problem

Consider the infinite series:

$\sum_{n=1}^{\infty} \frac{1}{n^2}$

How can we prove this sum is finite? And what is its value?

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This famous problem, known as the Basel Problem, was solved by Euler in 1735. He proved that:

$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$

While Euler’s proof used deep analysis, we can prove the sum is finite using the method of differences.

To show that the sum is finite, we use the comparison test. Notice that:

$\frac{1}{n^2} < \frac{1}{n(n-1)} \quad \text{for } n \geq 2$

This implies:

$\sum_{n=2}^{\infty} \frac{1}{n^2} < \sum_{n=2}^{\infty} \frac{1}{n(n-1)}$

The series on the right can be simplified using partial fractions:

$\frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}$

This results in a telescoping series:

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

The terms cancel, leading to a finite limit. Thus, since $\sum_{n=2}^{\infty} \frac{1}{n(n-1)}$ converges, we conclude that $\sum_{n=1}^{\infty} \frac{1}{n^2}$ is also finite.

Module 1: Basic Method of Differences

Definition and Key Concept

Definition — Method of Differences: The method of differences is used to find sums of sequences by:

1. First expressing the term as a difference of two or more terms
2. Then using cancellation to find the sum

Key Steps for Partial Fractions and Method of Differences

Standard Approach

1. For expressions like $\dfrac{cn+d}{(n+a)(n+b)}$:

• Split into partial fractions: $\dfrac{A}{n+a} + \dfrac{B}{n+b}$
• Find $A$ and $B$ by multiplying both sides by $(n+a)(n+b)$

2. For sums using method of differences:

• Identify the non-cancelling terms at start and end
• Write out first 3 terms and last 2 terms to spot the pattern
• Put over a common denominator to simplify

Example 1

Express $\dfrac{1}{(r+3)(r+5)}$ in partial fractions. Hence find $\displaystyle\sum_{r=1}^n \frac{1}{(r+3)(r+5)}$ using the method of differences.

Solution:

1. Let $\dfrac{1}{(r+3)(r+5)} = \dfrac{A}{r+3} + \dfrac{B}{r+5}$

2. Multiply both sides by $(r+3)(r+5)$:

$1 = A(r+5) + B(r+3)$

3. $A = \_\_\_\_$ and $B = \_\_\_\_$

4. Therefore:

$\frac{1}{(r+3)(r+5)} = \_\_\_\_$

5. Write out first 3 terms and last 2 terms:

$\_\_\_\_ + \_\_\_\_ + \_\_\_\_ + \cdots + \_\_\_\_ + \_\_\_\_$

6. Non-cancelling terms are:

$\_\_\_\_ + \_\_\_\_ - \_\_\_\_ - \_\_\_\_$

7. Put over common denominator and simplify:

$= \_\_\_\_$

In-class Exercise

Exercise

Express $\dfrac{2}{(r-1)(r+1)}$ in partial fractions and hence find:

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

Module 2: Three-Term Partial Fractions

Three-Term Partial Fractions Decomposition

Example 2 — Three-Term Partial Fractions

Express $\dfrac{1}{(2r-1)(2r+1)(2r+3)}$ in partial fractions. Hence find $\displaystyle\sum_{r=1}^n \frac{1}{(2r-1)(2r+1)(2r+3)}$ using the method of differences.

Solution:

1. Let $\dfrac{1}{(2r-1)(2r+1)(2r+3)} = \dfrac{A}{2r-1} + \dfrac{B}{2r+1} + \dfrac{C}{2r+3}$

2. Multiply throughout by $(2r-1)(2r+1)(2r+3)$:

$1 = A(2r+1)(2r+3) + B(2r-1)(2r+3) + C(2r-1)(2r+1)$

3. Put $r = \_\_\_\_$, $r = \_\_\_\_$, $r = \_\_\_\_$ to find $A$, $B$, $C$:

$A = \_\_\_\_, \quad B = \_\_\_\_, \quad C = \_\_\_\_$

4. Therefore:

$\frac{1}{(2r-1)(2r+1)(2r+3)} = \_\_\_\_$

5. Write out first three terms and the last two terms:

$\_\_\_\_ + \_\_\_\_ + \_\_\_\_ + \cdots + \_\_\_\_ + \_\_\_\_$

6. Identify non-cancelling terms:

$\_\_\_\_ - \_\_\_\_ + \_\_\_\_ - \_\_\_\_$

7. Put over common denominator:

$= \_\_\_\_$

Key Points for Three-Term Problems

1. When finding partial fractions:

• Write $\dfrac{A}{ax+b} + \dfrac{B}{cx+d} + \dfrac{C}{ex+f}$
• Use strategic values of $x$ to find coefficients
• Check your work by expanding back

2. When using method of differences:

• Write out enough terms to establish the pattern
• Pay attention to coefficients when identifying non-cancelling terms
• Combine terms carefully over a common denominator

In-class Exercise

Exercise

Express $\dfrac{r+4}{r(r+1)(r+2)}$ in partial fractions and hence find:

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

Module 3: Advanced Applications

Polynomial Differences

Guided Exercise 1

Use the binomial expansion to show that:

$n^5 - (n-1)^5 = 5n^4 - 10n^3 + 10n^2 - 5n + 1$

Hence, show that:

$\sum_{r=1}^n r^4 = \frac{1}{30}n(n+1)(2n+1)(3n^2 + 3n - 1)$

Guidance:

1. Rearrange to express $r^4$ in terms of the other expressions.
2. Write out the sum $\sum_{r=1}^n r^4$ using this expression.
3. Use these known formulas:
   • $\displaystyle\sum_{r=1}^n r = \frac{n(n+1)}{2}$
   • $\displaystyle\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}$
   • $\displaystyle\sum_{r=1}^n r^3 = \left(\frac{n(n+1)}{2}\right)^2$
4. Substitute and simplify to get the required form.

In-class Exercise

Exercise

Use the method of differences to prove that:

$\sum_{r=1}^n r^3 = \left(\frac{n(n+1)}{2}\right)^2$

Hint: Consider the difference $(r+1)^4 - r^4$.

Sums with Square Roots

Example 3 — Sum with Square Root

Show that for $r \geq 1$:

$\frac{r}{\sqrt{r(r+1)} + \sqrt{r(r-1)}} = A\left(\sqrt{r(r+1)} - \sqrt{r(r-1)}\right)$

where $A$ is a constant to be determined.

Solution:

1. Let the left side be $L$ and the right side be $R$:

$L = \frac{r}{\sqrt{r(r+1)} + \sqrt{r(r-1)}}, \quad R = A\left(\sqrt{r(r+1)} - \sqrt{r(r-1)}\right)$

2. To find $A$, multiply $L$ by $\dfrac{\sqrt{r(r+1)} - \sqrt{r(r-1)}}{\sqrt{r(r+1)} - \sqrt{r(r-1)}}$ to rationalise the denominator:

$L = \frac{r\left(\sqrt{r(r+1)} - \sqrt{r(r-1)}\right)}{\_\_\_\_}$

3. Therefore:

$L = \_\_\_\_ \left(\sqrt{r(r+1)} - \sqrt{r(r-1)}\right)$

4. Comparing with $R$, we find $A = \_\_\_\_$.

Example 4 — Method of Differences with Square Roots

Using the previous result, find:

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

Solution:

1. From the previous example:

$\frac{r}{\sqrt{r(r+1)} + \sqrt{r(r-1)}} = \_\_\_\_ \left(\sqrt{r(r+1)} - \sqrt{r(r-1)}\right)$

2. Write out first three terms and last two terms, then identify the telescoping pattern.

3. Therefore:

$\sum_{r=1}^n \frac{r}{\sqrt{r(r+1)} + \sqrt{r(r-1)}} = \_\_\_\_$

Example 5 — Extension

Find the constant $k$ such that:

$\sum_{r=1}^n \frac{kr}{\sqrt{r(r+1)} + \sqrt{r(r-1)}} = \sqrt{\sum_{r=1}^n r}$

Solution:

1. From the previous example:

$\sum_{r=1}^n \frac{kr}{\sqrt{r(r+1)} + \sqrt{r(r-1)}} = \_\_\_\_$

2. The right side is:

$\sqrt{\sum_{r=1}^n r} = \_\_\_\_$

3. Equating these and solving: $k = \_\_\_\_$.

Trigonometric Forms*

Example 6 — Trigonometric Identity

Using the formula

$\cos(A+B) = \cos A \cos B - \sin A \sin B$

show that:

$2\sin\left(\frac{1}{2}\right)\sin(n) = \cos\left(n - \frac{1}{2}\right) - \cos\left(n + \frac{1}{2}\right)$

Solution:

1. Use the given formula:

$\cos(A-B) - \cos(A+B) = \cos A \cos B + \sin A \sin B - (\cos A \cos B - \sin A \sin B) = 2 \sin A \sin B$

2. Let $A = n$ and $B = \dfrac{1}{2}$:

$2\sin\left(\frac{1}{2}\right)\sin(n) = \cos\left(n - \frac{1}{2}\right) - \cos\left(n + \frac{1}{2}\right)$

Example 7 — Method of Differences with Trigonometric Terms

Using the previous result, show that:

$\sum_{n=1}^N \sin(n) = \frac{1}{2}\csc\left(\frac{1}{2}\right)\left[\cos\left(\frac{1}{2}\right) - \cos\left(\frac{2N+1}{2}\right)\right]$

Solution:

1. From the previous example:

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

2. Write out first few terms:

$\frac{1}{2}\csc\left(\frac{1}{2}\right)\left[\cos\left(\frac{1}{2}\right) - \cos\left(\frac{3}{2}\right)\right] + \frac{1}{2}\csc\left(\frac{1}{2}\right)\left[\cos\left(\frac{3}{2}\right) - \cos\left(\frac{5}{2}\right)\right] + \cdots$

$+ \frac{1}{2}\csc\left(\frac{1}{2}\right)\left[\cos\left(\frac{2N-1}{2}\right) - \cos\left(\frac{2N+1}{2}\right)\right]$

3. Notice the telescoping pattern:

   • First term contains $\cos\left(\frac{1}{2}\right)$
   • Last term contains $-\cos\left(\frac{2N+1}{2}\right)$
   • All other terms cancel in pairs

4. Therefore:

$\sum_{n=1}^N \sin(n) = \frac{1}{2}\csc\left(\frac{1}{2}\right)\left[\cos\left(\frac{1}{2}\right) - \cos\left(\frac{2N+1}{2}\right)\right]$

Summary

Key Points

1. Basic Method:

• Look for terms that telescope
• Use partial fractions to split complex fractions
• The sum equals first term minus last term

2. Advanced Applications:

• Three-term partial fractions often require two steps
• Trigonometric terms can be treated as constants
• Square roots may require algebraic manipulation

Common Pitfalls

• Don’t forget to check if terms actually telescope
• Be careful with signs when splitting fractions
• Watch for opportunities to factor before using partial fractions
• Remember that the method works only if you can find a difference form

Homework

E.O.C.1

1. Express

$\frac{3}{(3r-1)(3r+2)}$

in partial fractions.

2. Using your answer to part 1 and the method of differences, show that

$\sum_{r=1}^n \frac{3}{(3r-1)(3r+2)} = \frac{3n}{2(3n+2)}$

3. Evaluate

$\sum_{r=100}^{1000} \frac{3}{(3r-1)(3r+2)}$

giving your answer to 3 significant figures.

E.O.C.2

Given that

$(2r+1)^3 = Ar^3 + Br^2 + Cr + 1$

where $A$, $B$ and $C$ are constants to be determined:

1. Find $A$, $B$ and $C$.

2. Show that

$(2r+1)^3 - (2r-1)^3 = 24r^2 + 2$

3. Using the result in part 2 and the method of differences, show that

$\sum_{r=1}^n r^2 = \frac{1}{6}n(n+1)(2n+1)$

Challenge Question (Optional): Method of Differences with Inverse Trigonometric Functions

Challenge Problem

Given that $A > B > 0$, and letting $x = \arctan A$ and $y = \arctan B$:

(a) Prove that:

$\arctan A - \arctan B = \arctan\left(\frac{A-B}{1+AB}\right)$

(b) Show that when $A = r + 2$ and $B = r$:

$\frac{A-B}{1+AB} = \frac{2}{(1+r)^2}$

(c) Hence, using the method of differences, show that:

$\sum_{r=1}^n \arctan\left(\frac{2}{(1+r)^2}\right) = \arctan(n+p) + \arctan(n+q) - \arctan 2 - \frac{\pi}{4}$

where $p$ and $q$ are integers to be determined.

(d) Hence, making your reasoning clear, determine:

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

giving the answer in the form $k\pi - \arctan 2$, where $k$ is a constant.