Paper 1 · Further Pure 1 ✦

Summation of Series

Three standard results, one telescoping trick, and a limit. Between them they close every FP1 series question — and the marks are almost all in the presentation.

Syllabus 1.3Σr, Σr², Σr³ given by MF19Method of differences

What the examiner expects you to do

Use the standard results for $\sum r, \sum r^{2}, \sum r^{3}$ to find related sums.
Use partial fractions to put a general term into a suitable form.
Use the method of differences to sum a finite series.
Recognise — by direct consideration of the sum to $n$ terms — when a series converges, and find its sum to infinity.

§1 The three standard results

These three are printed in MF19 — you are not asked to memorise them, only to use them fluently and to factorise the answer at the end.

◆ Given on the formula sheet

\sum_{r = 1}^{n} r = \frac{1}{2} n (n + 1), \sum_{r = 1}^{n} r^{2} = \frac{1}{6} n (n + 1) (2 n + 1), \sum_{r = 1}^{n} r^{3} = \frac{1}{4} n^{2} (n + 1)^{2} .

The fourth one you do have to remember, because it looks too obvious to write down:

\sum_{r = 1}^{n} c = c n (a constant term summed n times — not c).

◇ Linearity — the only rule you need

Summation is linear, so you may split a sum term by term and pull constants out:

\sum_{r = 1}^{n} (a r^{2} + b r + c) = a \sum_{r = 1}^{n} r^{2} + b \sum_{r = 1}^{n} r + c n .

It is not multiplicative: $\sum r (r + 2) \neq (\sum r) (\sum (r + 2))$ . Always expand the bracket first.

✎ Worked example — expand, then factorise

Find $\sum_{r = 1}^{n} r (r + 2)$ in a fully factorised form.

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

Take out the common factor $\frac{1}{6} n (n + 1)$ — this is the step that earns the final mark:

= \frac{1}{6} n (n + 1) [(2 n + 1) + 6] = \frac{1}{6} n (n + 1) (2 n + 7) .

Check with $n = 2$ : the series is $1 (3) + 2 (4) = 11$ , and $\frac{1}{6} (2) (3) (11) = 11$ . ✓

✎ Worked example — a sum that does not start at

r = 1

Find $\sum_{r = n + 1}^{2 n} r^{2}$ . The standard results only run from $r = 1$ , so subtract the missing head:

\sum_{r = n + 1}^{2 n} r^{2} = \sum_{r = 1}^{2 n} r^{2} - \sum_{r = 1}^{n} r^{2} = \frac{1}{6} (2 n) (2 n + 1) (4 n + 1) - \frac{1}{6} n (n + 1) (2 n + 1) .

Common factor $\frac{1}{6} n (2 n + 1)$ :

= \frac{1}{6} n (2 n + 1) [2 (4 n + 1) - (n + 1)] = \frac{1}{6} n (2 n + 1) (7 n + 1) .

Check with $n = 2$ : the series is $3^{2} + 4^{2} = 25$ , and $\frac{1}{6} (2) (5) (15) = 25$ . ✓

★ A* tip — the curiosity worth knowing

Comparing the first and third results, $\sum_{r = 1}^{n} r^{3} = (\sum_{r = 1}^{n} r)^{2}$ . It is a genuine identity, it is a favourite induction exercise, and it is a fast sanity check: $1 + 8 + 27 = 36 = 6^{2}$ . ✓

§2 The method of differences

When the general term is a fraction, the standard results are useless. Instead force the term into the shape something minus the next something — then almost everything cancels down the middle. This is telescoping.

◆ The principle

If you can write $u_{r} = f (r) - f (r + 1)$ , then

\sum_{r = 1}^{n} u_{r} = [f (1) - f (2)] + [f (2) - f (3)] + \dots + [f (n) - f (n + 1)] = f (1) - f (n + 1),

because every interior $f$ appears once positive and once negative. Only the ends survive.

✎ Worked example — find the partial fractions yourself

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

Partial fractions: $\frac{1}{r (r + 1)} = \frac{1}{r} - \frac{1}{r + 1}$ , which is already $f (r) - f (r + 1)$ with $f (r) = \frac{1}{r}$ . Writing out the terms:

(\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n + 1}) = 1 - \frac{1}{n + 1} = \frac{n}{n + 1} .

Check with $n = 3$ : $\frac{1}{2} + \frac{1}{6} + \frac{1}{12} = \frac{3}{4}$ , and $\frac{3}{3 + 1} = \frac{3}{4}$ . ✓

✎ Worked example — when the gap is 2, two terms survive at each end

Find $\sum_{r = 1}^{n} \frac{1}{r (r + 2)}$ . Here $\frac{1}{r (r + 2)} = \frac{1}{2} (\frac{1}{r} - \frac{1}{r + 2})$ , so each negative piece cancels the positive piece two rows later, not one:

2 \sum_{r = 1}^{n} \frac{1}{r (r + 2)} = \underset{start survives}{\underset{⏟}{\frac{1}{1} + \frac{1}{2}}} - \underset{end survives}{\underset{⏟}{\frac{1}{n + 1} - \frac{1}{n + 2}}} .

Hence, halving and putting it over a common denominator,

\sum_{r = 1}^{n} \frac{1}{r (r + 2)} = \frac{3}{4} - \frac{2 n + 3}{2 (n + 1) (n + 2)} = \frac{n (3 n + 5)}{4 (n + 1) (n + 2)} .

Check with $n = 2$ : $\frac{1}{3} + \frac{1}{8} = \frac{11}{24}$ , and $\frac{2 (11)}{4 (3) (4)} = \frac{22}{48} = \frac{11}{24}$ . ✓

✎ Worked example — the "show that … hence" format

A very common CAIE wording. (i) Show that $\frac{1}{2 r - 1} - \frac{1}{2 r + 1} = \frac{2}{(2 r - 1) (2 r + 1)}$ . (ii) Hence find $\sum_{r = 1}^{n} \frac{1}{(2 r - 1) (2 r + 1)}$ .

(i) Over a common denominator,

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

(ii) So the term is $\frac{1}{2} (\frac{1}{2 r - 1} - \frac{1}{2 r + 1})$ with $f (r) = \frac{1}{2 r - 1}$ , and the sum telescopes:

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

Check with $n = 2$ : $\frac{1}{3} + \frac{1}{15} = \frac{2}{5}$ , and $\frac{2}{2 (2) + 1} = \frac{2}{5}$ . ✓

★ A* tip — never "guess" what cancels

Write out the first three brackets and the last two in full, with the dots between them. That display is worth method marks on its own, and it is the only reliable way to see how many terms survive at each end.

§3 Convergence and the sum to infinity

The syllabus is precise about the route: you decide convergence by direct consideration of the sum to $n$ terms. So there is only ever one method.

◆ The method — two lines, always

Find the closed form $S_{n}$ (standard results, or method of differences).
Let $n \to \infty$ . If $S_{n} \to L$ , a finite limit, the series converges and $S_{\infty} = L$ . If $S_{n}$ grows without bound, it diverges and there is no sum to infinity.

✎ Worked example — three verdicts

From the sums found above:

S_{n} = \frac{n}{n + 1} = \frac{1}{1 + \frac{1}{n}} ⟶ 1, S_{n} = \frac{n}{2 n + 1} = \frac{1}{2 + \frac{1}{n}} ⟶ \frac{1}{2},

so $\sum_{r = 1}^{\infty} \frac{1}{r (r + 1)} = 1$ and $\sum_{r = 1}^{\infty} \frac{1}{(2 r - 1) (2 r + 1)} = \frac{1}{2}$ . But

S_{n} = \frac{1}{6} n (n + 1) (2 n + 1) ⟶ \infty,

so $\sum r^{2}$ diverges — it has no sum to infinity.

★ A* tip — how to take the limit cleanly

Divide numerator and denominator by the highest power of $n$ present, then use $\frac{1}{n} \to 0$ . For $\frac{n (3 n + 5)}{4 (n + 1) (n + 2)}$ , divide top and bottom by $n^{2}$ :

\frac{3 + \frac{5}{n}}{4 (1 + \frac{1}{n}) (1 + \frac{2}{n})} ⟶ \frac{3}{4} .

Writing " $\to \frac{3}{4}$ as $n \to \infty$ " explicitly is what scores; a bare answer usually does not.

⚠ Trap — a shrinking term is not enough

Terms getting smaller does not guarantee convergence, and the syllabus does not want you to argue from the terms at all. Find $S_{n}$ first, then take the limit of $S_{n}$ . That is the whole examinable method.

§4 Interactive: watch a series telescope and converge

Pick a series, then slide $n$ . The panel adds the terms up one by one and compares that running total with the closed form from the notes above — so you can see for yourself that the formula is right. The strip underneath shows exactly which pieces cancel.

Number of terms

n

the middle always cancels ✎

§5 Examiner traps & A* checklist

⚠ The six most common mark-losers

Writing $\sum_{r = 1}^{n} c = c$ instead of $c n$ — the single most frequent slip in this topic.
Leaving the answer unfactorised. CAIE mark schemes quote a factorised form; take out $\frac{1}{6} n (n + 1)$ or whatever is common.
Starting the sum at the wrong place. For $\sum_{r = k}^{m}$ , compute $\sum_{r = 1}^{m} - \sum_{r = 1}^{k - 1}$ — note $k - 1$ , not $k$ .
Assuming exactly one term survives at each end. With a gap of 2 (or 3) it is two (or three) at each end.
Not writing out enough terms to justify the cancellation — the display is worth method marks.
Claiming a sum to infinity without producing $S_{n}$ and taking the limit.

★ Before you turn the page

Did I expand every bracket before summing?
Is my answer factorised, and does it give the right value at $n = 1$ and $n = 2$ ?
If the question said "hence", did I actually use the previous part?

§6 Video explainers (curated)

▶The standard results Σr, Σr², Σr³
Curated search — using and factorising related sums.↗ ▶Method of differences
Curated search — partial fractions and telescoping, worked.↗ ▶Convergence & sum to infinity
Curated search — taking the limit of S_n properly.↗

§7 Check yourself

Score 0 / 5

Q1 · linearity

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

Solution

$2 \sum r + \sum 1 = 2 \cdot \frac{1}{2} n (n + 1) + n = n^{2} + 2 n = n (n + 2)$ . The distractor $(n + 1)^{2}$ is what you get if you sum $\sum 1$ as $n + 1$ — i.e. if you start at $r = 0$ . Check $n = 2$ : $3 + 5 = 8 = 2 (4)$ . ✓

Q2 · related sum

$\sum_{r = 1}^{n} r (r + 1)$ equals:

Solution

$\sum r^{2} + \sum r = \frac{1}{6} n (n + 1) (2 n + 1) + \frac{1}{2} n (n + 1) = \frac{1}{6} n (n + 1) [(2 n + 1) + 3] = \frac{1}{3} n (n + 1) (n + 2)$ . Check $n = 2$ : $2 + 6 = 8 = \frac{1}{3} (2) (3) (4)$ . ✓

Q3 · telescoping

Given $u_{r} = \frac{1}{r} - \frac{1}{r + 1}$ , the sum $\sum_{r = 1}^{n} u_{r}$ is:

Solution

Only the ends survive: $f (1) - f (n + 1) = 1 - \frac{1}{n + 1} = \frac{n}{n + 1}$ .

Q4 · a gap of two

For $n \geq 2$ , $\sum_{r = 1}^{n} (\frac{1}{r} - \frac{1}{r + 2})$ equals:

Solution

Each negative piece cancels the positive piece two rows later, so two survive at each end: $(\frac{1}{1} + \frac{1}{2}) - (\frac{1}{n + 1} + \frac{1}{n + 2})$ . Check $n = 2$ : $(1 - \frac{1}{3}) + (\frac{1}{2} - \frac{1}{4}) = \frac{11}{12}$ , and $\frac{3}{2} - \frac{1}{3} - \frac{1}{4} = \frac{11}{12}$ . ✓

Q5 · full method — try it on paper first

(i) Show that $\frac{1}{r (r + 1)} - \frac{1}{(r + 1) (r + 2)} = \frac{2}{r (r + 1) (r + 2)}$ .

(ii) Hence find $\sum_{r = 1}^{n} \frac{1}{r (r + 1) (r + 2)}$ , and deduce the sum to infinity.

Solution

(i) Common denominator $r (r + 1) (r + 2)$ :

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

(ii) So with $f (r) = \frac{1}{r (r + 1)}$ the general term is $\frac{1}{2} [f (r) - f (r + 1)]$ , and the sum telescopes to the two ends:

\sum_{r = 1}^{n} \frac{1}{r (r + 1) (r + 2)} = \frac{1}{2} [\frac{1}{1 \cdot 2} - \frac{1}{(n + 1) (n + 2)}] = \frac{1}{4} - \frac{1}{2 (n + 1) (n + 2)} .

Over a common denominator this is the fully factorised form

\frac{(n + 1) (n + 2) - 2}{4 (n + 1) (n + 2)} = \frac{n (n + 3)}{4 (n + 1) (n + 2)} .

Sum to infinity: as $n \to \infty$ , $\frac{1}{2 (n + 1) (n + 2)} \to 0$ , so $S_{n} \to \frac{1}{4}$ . Hence $\sum_{r = 1}^{\infty} \frac{1}{r (r + 1) (r + 2)} = \frac{1}{4}$ .

Check with $n = 2$ : $\frac{1}{6} + \frac{1}{24} = \frac{5}{24}$ , and $\frac{2 (5)}{4 (3) (4)} = \frac{10}{48} = \frac{5}{24}$ . ✓ $■$

FP1 · Topic 3 · Summation of Series