9231. Further Pure 1. Summation of Series

A series is the sum of all the terms in a sequence (the sequence may be finite or infinite).

You have already met arithmetic and geometric series and applied the formulae for their series:

We will build on and extend this work, by looking at convergent series and series of squares and cubes of numbers.

Summation formulae: Σr, Σr², Σr³

Let’s think about language first.

$\sum\limits_{r=1}^{n} k$ means that k is added together n times, i.e. this is equal to kn.

However, $\sum\limits_{r=1}^{n} r$ is quite different and means the sum of the digits from 1 to n, i.e. an arithmetic sequence with first term 1 and difference 1, which from the formula above has a sum of $\frac{n(n+1)}{2}$

In summary, $\sum\limits_{r=1}^{n} r =\frac{n(n+1)}{2}$

Worked Example

Find $\sum\limits_{r=1}^{n} (4r+1)$ and hence calculate $\sum\limits_{17}^{48} (4r+1)$

Note

We are not restricted to “n” as our upper limit, for instance $\sum\limits_{r=1}^{2n} r$ would be $\frac{2n(2n+1)}{2}=2n^2+n$

We can use this to deduce $\sum\limits_{n+1}^{2n} r$ as this is the sum of the first 2n terms minus the first n terms, so is $\sum\limits_{1}^{2n} r - \sum\limits_{1}^{n}r$ , which is $2n^2 + n - \frac{n(n+1)}{2} = \frac{3}{2}n^2 + \frac{1}{2}n$

Worked example 2

Find an expression for the first n terms of: 1 + 3 + 5 + 7 + …

Σr²

Let’s work with the binomial expansion (r+1)³=r³+3r²+3r+1

If we substitute 1,2,3, …,n into this equation we get:

2³ = 1³ + 3×1² + 3×1 + 1

3³ = 2³ + 3×2² + 3×2 + 1

4³ = 3³ + 3×3² + 3×3 +1

…

(n+1)³ = n³ + 3xn² +3xn +1

If we add all of the terms on the right hand side, we have $\sum\limits_{r=1}^{n} r^3 + 3 \sum\limits_{r=1}^{n}r^2 + 3 \sum\limits_{r=1}^{n}r +n$

We can then add 1 (i.e. 1³) to both aides of the equation to give us:

$\sum\limits_{r=1}^{n+1} r^3=\sum\limits_{r=1}^{n} r^3 + 3 \sum\limits_{r=1}^{n}r^2 + 3 \sum\limits_{r=1}^{n}r +n+1$

We notice that both sides of the equation include $\sum\limits_{r=1}^{n} r^3$ so we can subtract this leaving $(n+1)^3= 3 \sum\limits_{r=1}^{n}r^2 + 3 \sum\limits_{r=1}^{n}r +n+1$ which we can then solve to give a formula for $\sum\limits_{r=1}^{n}r^2$

Worked Example

Find an expression in terms of n for $\sum\limits_{r=1}^{n}(3r^2-4r+2)$

Σr³

Similar to our last example, let’s start with the binomial expansion (r+1)⁴=r⁴+4r³+6r²+4r+1, and substitute in successive natural numbers, giving:

2⁴ = 1⁴ + 4×1³ + 6×1² + 4×1 + 1

3⁴ = 2⁴ + 4×2³ + 6×2² + 4×2 + 1

4⁴ = 3⁴ + 4×3³ + 6×3² + 4×3 +1

…

(n+1)⁴ = n⁴ + 4xn³ + 6xn² +4xn +1.

Adding 1 to both sides and summing give:

$\sum\limits_{r=1}^{n+1} r^4=\sum\limits_{r=1}^{n} r^4 + 4 \sum\limits_{r=1}^{n}r^3 + 6 \sum\limits_{r=1}^{n}r^2 + 4 \sum\limits_{r=1}^{n}r +n+1$

Following the same procedures as above, we can rearrange this to find $\sum\limits_{r=1}^{n} r^3$

Standard Summations Reminder (in formula book, but also good to memorise)

Worked Example

Find an expression in terms of n for $\sum\limits_{r=1}^{n} r^2(r-1)$

Exercise 1

Answers to Exercise 1

1,2. Proof; 3. -n, 4. Proof; 5. $\frac{1}{2}n(6n^2-9n-5)$ 6. $\frac{1}{3}n(2+9n+7n^2)$ , 7. 328,350, 8. Proof, 32,760, 9. 204,020,000

Worked Solutions to Exercise 1

Converging Series

We will now look at infinite series with a finite sum. In order to identify the finite sum, we will have to spot patterns within the series.

Let’s consider first $\sum\limits_{r=1}^{n} \frac{1}{r(r+1)}$ .

If we look at the first few terms, $\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20}$ it is hard to spot a pattern.

So let’s try a different methods. Using the partial fractions method, let’s rewrite $\frac{1}{r(r+1)}$ (do you remember how to do this?)

So we can now look at $\sum\limits_{r=1}^{n} (\frac{1}{r}-\frac{1}{r(r+1)})$ .

If we look at the first few terms of this: $(\frac{1}{1}-\frac{1}{2}) + (\frac{1}{2}-\frac{1}{3}) + (\frac{1}{3}-\frac{1}{4}) + ... + (\frac{1}{n-1}-\frac{1}{n}) + (\frac{1}{n} - \frac{1}{n+1})$ we see that all of the terms in the middle cancel out, leaving us with just $1 - \frac{1}{n+1}$ , i.e. $\frac{n}{n+1}$

When we cancel terms in this way we are using what is called the method of differences.

Worked examples

Find in terms of n, the sum of
- $\sum\limits_{r=1}^{n} (\frac{1}{(r+1)(r+2)})$ ;
- $\sum\limits_{r=1}^{n} (\frac{1}{r(r+2)})$ ;
- $\sum\limits_{r=1}^{n} (\frac{1}{(r-1)(r+1)})$ ;
- $\sum\limits_{r=2}^{n} (\frac{5}{r(r-1)})$ .