Further Pure | Topic 3: Summation of Series

Curriculum Objectives:

• use the standard results for $\sum r $, $\sum r^2 $, $\sum r^3 $ to find related sums;
• use the method of differences to obtain the sum of a finite series, e.g. by expressing the general term in partial fractions;
• recognise, by direct consideration of a sum to n terms, when a series is convergent, and find the sum to infinity in such cases.

'$\sum $' notation and properties thereof:

A very useful shorthand for the sum of series containing a large number of terms is the sigma notation. The symbol $\sum\limits_{k = 1}^n {{a_k}} $ denotes the sum ${a_1} + {a_2} + ... + {a_n}$. The a's are terms of the sum: ${a_1}$ is the initial term, ${a_2}$ is the second term, ${a_k}$ is the kth term, and ${a_n}$ is the nth and last term of the sum. The variable k is called the index of summation. The values of k include all values from 1 to n.

Consider the following summations:

$\sum\limits_{k = 1}^5 k  = 1 + 2 + 3 + 4 + 5 = 15$
$\sum\limits_{k = 1}^3 {{{( - 1)}^k}k}  = {( - 1)^1}(1) + {( - 1)^2}(2) + {( - 1)^3}(3) =  - 2$

You will be required to have a firm grasp of the alegbra with these sums. Summarised below are some properties of this notation.

1. Sum Rule:

$\sum\limits_{k = 1}^n {({a_k} + {b_k})}  = \sum\limits_{k = 1}^n {{a_k}}  + \sum\limits_{k = 1}^n {{b_k}} $

2. Difference Law

$\sum\limits_{k = 1}^n {({a_k} - {b_k})}  = \sum\limits_{k = 1}^n {{a_k}}  - \sum\limits_{k = 1}^n {{b_k}} $

3. Constant Multiple Rule

$\sum\limits_{k = 1}^n {c{a_k}}  = c\cdot\sum\limits_{k = 1}^n {{a_k}} $ (for any number c)

4. Constant Value Rule

$\sum\limits_{k = 1}^n c  = n \cdot c$    (c is any constant value)

You may also need to use your knowledge of arithmetic and geometric series, to solve these questions. Formulae pertaining to those series, are thusly summarised:
For an arithmetic series:

${a_n} = {a_1} + (n - 1)d$, where $a_n$ is the nth term, $a_1$ is the first term, and d is the common difference.

${a_n} = {a_m} + (n - m)d$, for any general term, m.

The sum is given by:

${S_n} = \frac{n}{2}\left( {2{a_1} + (n - 1)d} \right)$

For a geometric progression:

${a_n} = {a_1}{r^{n - 1}}$, for the general term n, where $a_n$ is the nth term, $a_1$ the first term, and r is the common ratio.

The sum of a geometric progression, say $a{r^0} + a{r^1} + a{r^2} + ... + a{r^{n - 1}}$ is given by:

${S_n} = \frac{{a(1 - {r^n})}}{{1 - r}}$

The following general results are quoted in your MF10 formula sheet, and you should be aware of them.

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

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

$\sum\limits_{r = 1}^n {{r^3}}  = \frac{1}{4}{n^2}{(n + 1)^2}$

The following examples highlight the level of algebra you are expected to have mastered.

Ex. Find the following sums:

$a)\sum\limits_{r = 7}^{20} {{r^2}} $
$b)\sum\limits_{r = 12}^{25} {{r^3}} $

Sol:

The initial value here is not the usual 1 (or 0). To evaluate these summations, we evaluate two different summations, one from 1 to one less than the initial value and the other from 1 to the final value, and then subtract one from the other to arrive at the answer.

$\sum\limits_{r = 7}^{20} {{r^2}}  = \sum\limits_{r = 1}^{20} {{r^2}}  - \sum\limits_{r = 1}^6 {{r^2}}  = \frac{{20}}{6}(21)(41) - \frac{6}{6}(7)(13) = 2870 - 91 = 2779$

$\sum\limits_{r = 12}^{25} {{r^3}}  = \sum\limits_{r = 1}^{25} {{r^3}}  - \sum\limits_{r = 1}^{11} {{r^3}}  = \frac{{{{25}^2}}}{4}({26^2}) - \frac{{{{11}^2}}}{4}({12^2}) = 105625 - 4356 = 101269$

Ex. Prove that

$\sum\limits_{r = 1}^n {r(r + 1) \equiv } \frac{1}{3}n(n + 1)(n + 2)$

Sol.

We begin with the following consideration:

$\sum\limits_{r = 1}^n {r(r + 1) = \sum\limits_{r = 1}^n {({r^2} + r)} }  = \sum\limits_1^n {{r^2}}  + \sum\limits_1^n r $
$ = \frac{1}{6}n(n + 1)(2n + 1) + \frac{1}{2}n(n + 1)$
$ = \frac{1}{6}n(n + 1)((2n + 1) + 3))$
$ = \frac{1}{6}n(n + 1)(2n + 4)$
$ = \frac{1}{3}n(n + 1)(n + 2)$

The Method of Differences

While there does not exist, a single method to sum a finite series, the method of differences is a particularly elegant one for some. The general approach is to find a function, f(r), such that the rth term, $u_r$, of a series can be expressed as

${u_r} = f(r + 1) - f(r)$

Then we add all the terms, ${u_1},{u_2},...,{u_n}$, making cancellations as we go along to arrive at the final answer. The following example elucidates this:

Say we wanted to prove the sum 1 + 2 + 3 + ... + n, i.e.,

$\sum\limits_{r = 1}^n r $

Consider the identity:

$2r \equiv r(r + 1) - (r - 1)r$

By taking consecutive terms, 1, 2,..., n for r we get:

$2(1) = 1(2) - (0)1$
$2(2) = 2(3) - (1)(2)$
$2(3) = 3(4) - (2)(3)$
.
.
$2(n - 1) = (n - 1)(n) - (n - 2)(n - 1)$
$2n = n(n + 1) - (n - 1)(n)$

Now if we were to sum every term, we find that almost all terms cancel out leaving:

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

In an examination, you will either be given the function, or information, wherewith you maybe derive it yourself.

Ex. (May/June 2013 Paper 12 Q5.)
Use the method of differences to show that $\sum\limits_{r = 1}^N {\frac{1}{{(2r + 1)(2r + 3)}}}  = \frac{1}{6} - \frac{1}{{2(2N + 3)}}$.

Sol.

We use partial decomposition:

$\frac{1}{{(2r + 1)(2r + 3)}} = \frac{A}{{(2r + 1)}} + \frac{B}{{(2r + 3)}}$

Multiplying across by (2r + 1) and then putting $r =  - \frac{1}{2}$, we find A is equal to (1/2) and by multiplying across by (2r + 3) and then putting in $r =  - \frac{3}{2}$, to find B is equal to (-1/2). Thus:

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

For different values of r we have:

r = 1

$\frac{1}{{2(3)}} - \frac{1}{{2(5)}}$

r = 2

$\frac{1}{{2(5)}} - \frac{1}{{2(7)}}$

r = N - 1

$\frac{1}{{2(2N - 1)}} - \frac{1}{{2(2N + 1)}}$

r = N

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

Adding all the terms, the cancellations are apparent and we are left with the following result:

$\sum\limits_{r = 1}^N {\frac{1}{{(2r + 1)(2r + 3)}}}  = \frac{1}{6} - \frac{1}{{2(2N + 3)}}$

While the question doesn't ask for it, consider what would happen if n approached infinity. The fraction, would become smaller and smaller to the point that it no longer matters to the calculation at hand to any good approximation. Therefore:

$\sum\limits_{r = 1}^\infty  {\frac{1}{{(2r + 1)(2r + 3)}}}  = \frac{1}{6}$

 Deduce then $\sum\limits_{r = N + 1}^{2N} {\frac{1}{{(2r + 1)(2r + 3)}}}  < \frac{1}{{8N}}$

Sol.

$\sum\limits_{r = N + 1}^{2N} {\frac{1}{{(2r + 1)(2r + 3)}}}  = \sum\limits_{r = 1}^{2N} {\frac{1}{{(2r + 1)(2r + 3)}}}  - \sum\limits_{r = 1}^N {\frac{1}{{(2r + 1)(2r + 3)}}} $ 
$ = \frac{1}{6} - \frac{1}{{2(4N + 3)}} - \left( {\frac{1}{6} - \frac{1}{{2(2N + 3)}}} \right)$
$ = \frac{1}{6} - \frac{1}{{2(4N + 3)}} - \frac{1}{6} + \frac{1}{{2(2N + 3)}}$
$ = \frac{{ - 2N - 3 + 4N + 3}}{{2(4N + 3)(2N + 3)}}$
$ = \frac{{2N}}{{2(4N + 3)(2N + 3)}}$
$ = \frac{N}{{8{N^2} + 12N + 6N + 9}}$
$ = \frac{N}{{8{N^2} + 18N + 9}}$
$ = \frac{1}{{8N + 18 + \frac{9}{N}}}$

As N is greater than or equal to 1:

$18 + \frac{9}{N} > 0$
$\frac{1}{{8N + 18 + \frac{9}{N}}} < \frac{1}{{8N}}$

Hence proved.

Ex.(October/November 2012 Paper 12 Q4)

Let $f(r) = r(r + 1)(r + 2)$. Show that:

$f(r) - f(r - 1) = 3r(r + 1)$

Sol.

$f(r) - f(r - 1) = r(r+1)(r+2)-(r-1)r(r+1)$
$= r(r + 1)(r + 2 - (r - 1))$
$= r(r + 1)(r + 2 - r + 1)$
$= 3r(r + 1)$

Hence show that:

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

Using the identity proven above we have:

For r = 1

$1(2)(3) - (0)(1)(2)$

r = 2

$2(3)(4) - (1)(2)(3)$

r = n - 1

$(n - 1)(n)(n + 1) - (n - 2)(n - 1)(n)$

r = n

$n(n + 1)(n + 2) - (n - 1)(n)(n + 1)$

Adding and cancelling terms, we get:

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

Using the standard result for the $\sum r $, deduce the general formula for $\sum r^2 $

Sol.

We know that:

$\sum\limits_{r = 1}^n {r(r + 1)}  = \frac{1}{3}n(n + 1)(n + 2)$
$\sum\limits_{r = 1}^n {{r^2} + r}  = \frac{1}{3}n(n + 1)(n + 2)$
$\sum\limits_{r = 1}^n {{r^2}}  = \frac{1}{3}n(n + 1)(n + 2) - \sum\limits_{r = 1}^n r $
$\sum\limits_{r = 1}^n {{r^2}}  = \frac{1}{3}n(n + 1)(n + 2) - \frac{1}{2}n(n + 1)$
$\sum\limits_{r = 1}^n {{r^2}}  = \frac{1}{6}n(n + 1)(2(n + 2) - 3)$
$\sum\limits_{r = 1}^n {{r^2}}  = \frac{1}{6}n(n + 1)(2n + 1)$

Find the sum of the series:

${1^2} + 2 \times {2^2} + {3^2} + 2 \times {4^2} + {5^2} + 2 \times {6^2} + ... + 2{(n - 1)^2} + {n^2}$

where n is odd

Sol.

For this question we rearrange the sequence thusly:

$({1^2} + {2^2} + {3^2} + {4^2} + {6^2} + ... + {(n - 1)^2} + {n^2}) + ({2^2} + {4^2} + {5^2} + {6^2} + ... + {(n - 1)^2})$

The first series is simply the sum of the squares of natural numbers from one to the odd number n, i.e.

$\sum\limits_1^n {{r^2}}  = \frac{1}{6}n(n + 1)(2n + 1)$

The second series is the sum of the square of even numbers up to the even number (n -1).

An even number is simply twice a natural number, therefore substituting 2n into $\sum\limits_1^n r^2 $, should give us an expression for the sum of even numbers i.e.

$\sum\limits_{r = 1}^{2n} {{{\left( {2r} \right)}^2} = 4\sum\limits_{r = 1}^{2n} {{r^2}} } $

Another important thing to be taken note of is that this expression would give us the sum of the squares of even numbers from one to 2n. We however require that up to n - 1. For this we substitute $\frac{{n - 1}}{2}$ into our expression of the sum of the squares of the even numbers, i.e.

$4\sum\limits_{r = 1}^{2\left( {\frac{{n - 1}}{2}} \right)} {{r^2}}  = 4\left( {\frac{1}{6}} \right)\left( {\frac{{n - 1}}{2}} \right)\left( {\frac{{n - 1}}{2} + 1} \right)\left( {2\left( {\frac{{n - 1}}{2}} \right) + 1} \right)$

$4\sum\limits_{r = 1}^{n - 1} {{r^2}}  = 4\left( {\frac{1}{6}} \right)\left( {\frac{{n - 1}}{2}} \right)\left( {\frac{{n + 1}}{2}} \right)\left( n \right)$

Adding these two expressions:

$\sum\limits_{r = 1}^n {{r^2}}  + 4\sum\limits_{r = 1}^{n - 1} {{r^2}}  = \frac{1}{6}n(n + 1)(2n + 1) + 4\left( {\frac{1}{6}} \right)\left( {\frac{{n - 1}}{2}} \right)\left( {\frac{{n + 1}}{2}} \right)n = \frac{1}{2}{n^2}(n + 1)$ 

(Optional) Double Summations

While not necessarily required for the course you may be interested in the use of double or even more summations performed concurrently. The following examples exhibit a few double summations for the reader to mull over.

$\sum\limits_{j = 1}^2 {\sum\limits_{i = 1}^3 {{a_{ij}}} }  = \sum\limits_{j = 1}^2 {\left( {\sum\limits_{i = 1}^3 {{a_{ij}}} } \right) = \sum\limits_{j = 1}^2 {\left( {{a_{1j}} + {a_{2j}} + {a_{3j}}} \right) = \sum\limits_{j = 1}^2 {{a_{1j}} + \sum\limits_{j = 1}^2 {{a_{2j}} + \sum\limits_{j = 1}^2 {{a_{3j}}} } } } }  = {a_{11}} + {a_{12}} + {a_{21}} + {a_{22}} + {a_{31}} + {a_{32}}$