Why is "Silently deleting a $\frac{1}{n+1}$ term instead of stating it tends to $0$" a common mistake in Convergence of Series and Sum to Infinity?

Why it happens: Treating the limit as obvious and skipping the justification. How to avoid it: Always write 'as n , 1n+1 0' before giving S_ — the limit step earns its own mark. Source: 9231 Paper 1 Examiner Reports — sum-to-infinity questions

Why is "Quoting a 'sum to infinity' for a divergent series" a common mistake in Convergence of Series and Sum to Infinity?

Why it happens: Plugging n mechanically without checking whether S_n actually stays finite. How to avoid it: Check that S_n tends to a finite value first. If a term grows with n (e.g. \tfrac12 n(n+1)), the series diverges — say so. Source: 9231 Paper 1 mark schemes

Why is "Mishandling the limit of a single rational $S_n$ (e.g. $\tfrac{n}{2n+1} \to 1$)" a common mistake in Convergence of Series and Sum to Infinity?

Why it happens: Comparing leading terms loosely instead of dividing by the highest power of n. How to avoid it: Divide top and bottom by the highest power of n, then send \tfrac1n-terms to 0. n2n+1 = 12+1/n \tfrac12. Source: 9231 Paper 1 Examiner Reports

Why is "Losing one of two vanishing terms in a gap-2 result" a common mistake in Convergence of Series and Sum to Infinity?

Why it happens: Carrying over a sign error from the differences stage, or forgetting the second surviving end term. How to avoid it: Send EVERY fraction containing n to 0; the leftover constant is S_. Re-check the closed form from subtopic 3.1. Source: 9231 Paper 1 mark schemes

Why is "Concluding a series converges just because $u_r \to 0$" a common mistake in Convergence of Series and Sum to Infinity?

Why it happens: Treating the necessary condition as if it were sufficient. How to avoid it: u_r 0 is necessary, not sufficient (harmonic series diverges). Decide convergence from the closed form S_n. Source: 9231 Paper 1 Examiner Reports

Why is "Trying to telescope a term that is a SUM $f(r)+f(r+1)$, not a difference" a common mistake in Convergence of Series and Sum to Infinity?

Why it happens: Pattern-matching to 'two fractions' without checking the sign between them. How to avoid it: Only a DIFFERENCE f(r) - f(r+k) telescopes. A sum like \tfrac1r + 1r+1 does not collapse — handle it separately (often it diverges). Source: 9231 Paper 1 Examiner Reports

Cambridge International A Levels Further Mathematics (9231)

Convergence of Series and Sum to Infinity

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Detailed Study Notes

Detailed notes on Summation of Series for Cambridge International A Levels Further Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Summation of Series — Convergence and the Sum to Infinity (Cambridge International AS & A Level Further Mathematics 9231 Paper 1, 2026-2027 syllabus)

Once the method of differences has given you a closed form $S_n$ for the partial sum, let $n \to \infty$ . If $S_n$ tends to a finite limit the series converges, and that limit is the sum to infinity $S_\infty$ . This subtopic is all about taking limits of the closed forms you produced in subtopic 3.1.

At a glance

Prerequisite: first get a closed form $S_n$ (usually via the method of differences — subtopic 3.1).
Convergence: the series converges iff its partial sums $S_n$ tend to a finite limit as $n \to \infty$ .
Sum to infinity: $S_\infty = \displaystyle\lim_{n\to\infty} S_n$ — substitute the vanishing terms with $0$ .
Key limit fact: any $\dfrac{1}{n+k} \to 0$ as $n \to \infty$ , so those terms drop out; constants survive.
Beware: if a surviving term grows (e.g. a constant multiple of $n$ ), the series diverges — no sum to infinity exists.

What you’ll learn

Mapped to the Cambridge International A Level 9231 syllabus (2026-2027).

1.3d — Recognise, by considering the limit of the partial sum $S_n$ as $n \to \infty$ , when a series converges, and find its sum to infinity in such cases.
1.3e — Deduce the sum to infinity from a closed form obtained by the method of differences.
1.3f — Comment on convergence: identify which terms vanish and which (if any) prevent convergence.

What 'sum to infinity' means

Add more and more terms; if the running total settles on a finite value, that's the sum to infinity.

An infinite series $u_1 + u_2 + u_3 + \cdots$ is handled through its partial sums: $S_n = \sum_{r=1}^{n} u_r = u_1 + u_2 + \cdots + u_n.$

The definition. The series converges if the sequence of partial sums $S_n$ approaches a finite limit as $n$ gets larger and larger. That limit is the sum to infinity: $S_\infty = \sum_{r=1}^{\infty} u_r = \lim_{n\to\infty} S_n.$

If $S_n$ does not settle on a finite value — it grows without bound, or oscillates — the series diverges and has no sum to infinity.

As n increases the partial sums Sₙ climb but flatten out, getting ever closer to the horizontal limit S∞. The vanishing gap is the part that "tends to 0".

Cambridge tip. The exam almost never asks you to sum an infinite series from scratch. It hands you (or asks you to derive) the closed form $S_n$ first — your job is the limit. So master subtopic 3.1, then this becomes a one-line finish.

$S_n = \sum_{r=1}^{n} u_r$ is the partial sum.
Converges $\Leftrightarrow$ $S_n \to$ a finite limit.
$S_\infty = \lim_{n\to\infty} S_n$ when that limit exists.

See the full worked example for convergence of series and sum to infinity →

Taking the limit of a closed form

Replace every 1/(n+k) by 0; keep the constants. That's the sum to infinity.

Once you have a closed form $S_n$ , finding $S_\infty$ is mechanical. The one fact you need: $\frac{1}{n+k} \longrightarrow 0 \quad \text{as } n \to \infty \quad (\text{for any fixed } k).$

More generally, any fraction whose denominator grows without bound tends to $0$ : $\dfrac{1}{2n+1}$ , $\dfrac{3}{(n+1)(n+2)}$ , $\dfrac{1}{n^2}$ all vanish.

The routine:

Write the closed form $S_n$ .
Identify which pieces $\to 0$ as $n \to \infty$ (the fractions with $n$ in the denominator).
Set those pieces to $0$ ; what remains is $S_\infty$ .

Worked idea. From subtopic 3.1, $\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+1)} = 1 - \frac{1}{n+1}$ . As $n \to \infty$ , the $\dfrac{1}{n+1} \to 0$ , so: $\sum_{r=1}^{\infty}\frac{1}{r(r+1)} = \lim_{n\to\infty}\left(1 - \frac{1}{n+1}\right) = 1 - 0 = 1.$

Multiple vanishing terms. A gap-2 result like $\dfrac{3}{4} - \dfrac{1}{2(n+1)} - \dfrac{1}{2(n+2)}$ has two terms that vanish, leaving $S_\infty = \dfrac{3}{4}$ .

Cambridge tip. Show the limit step explicitly — write "as $n \to \infty$ , $\dfrac{1}{n+1} \to 0$ " — rather than just deleting the term. Examiners want to see you justify the limit, not silently drop it.

$\dfrac{1}{n+k} \to 0$ as $n \to \infty$ — the workhorse fact.
Set every vanishing fraction to $0$ ; the constants are $S_\infty$ .
State the limit step in words for full marks.

See the full worked example for convergence of series and sum to infinity →

When a series does NOT converge

If a surviving term grows with n, the partial sums blow up — the series diverges.

Not every series has a sum to infinity. Convergence fails when the closed form $S_n$ contains a term that grows without bound as $n \to \infty$ .

Tell-tale signs of divergence:

A term linear (or higher) in $n$ survives, e.g. $S_n = \tfrac12 n(n+1)$ — clearly $S_n \to \infty$ .
The general term $u_r$ itself does not tend to $0$ . (A necessary condition for convergence is $u_r \to 0$ ; if the terms don't shrink, the sum can't settle.)

Example. $\displaystyle\sum_{r=1}^{n} r = \tfrac12 n(n+1) \to \infty$ , so $\sum_{r=1}^{\infty} r$ diverges — no sum to infinity.

Contrast. $\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+1)} = 1 - \frac{1}{n+1} \to 1$ converges, because the general term $\dfrac{1}{r(r+1)} \to 0$ fast enough.

A subtle warning. $u_r \to 0$ is necessary but not sufficient: the harmonic series $\sum \tfrac1r$ has terms tending to $0$ yet diverges. In 9231 the convergence is always decided by the closed form you obtain, so always inspect $S_n$ , not just $u_r$ .

Cambridge tip. If asked to "comment on convergence", state clearly: "As $n \to \infty$ , $S_n \to L$ (a finite value), so the series converges and $S_\infty = L$ " — or, if it diverges, " $S_n \to \infty$ , so the series does not converge."

A term growing with $n$ in $S_n$ ⟹ divergence.
$u_r \to 0$ is necessary, not sufficient, for convergence.
Always decide from the closed form $S_n$ , not the term alone.

See the full worked example for convergence of series and sum to infinity →

Putting it together — differences then limit

Telescope to get S_n, then take n → ∞ in one clean line.

The most common Paper 1 question chains the two subtopics: method of differences to get $S_n$ , then limit to get $S_\infty$ .

The full template:

Express $u_r$ in partial fractions (subtopic 3.1).
Telescope to a closed form $S_n$ .
Combine into a single fraction (good practice).
Let $n \to \infty$ : drop the vanishing fractions.
State $S_\infty$ and, if asked, comment that the series converges.

Worked idea. Suppose telescoping gives $\displaystyle S_n = \frac{n(n+3)}{4(n+1)(n+2)}$ . To take the limit, divide top and bottom by $n^2$ (the highest power): $S_n = \frac{1 + \tfrac{3}{n}}{4\left(1 + \tfrac{1}{n}\right)\left(1 + \tfrac{2}{n}\right)} \xrightarrow{n\to\infty} \frac{1 + 0}{4(1)(1)} = \frac{1}{4}.$

So $\displaystyle\sum_{r=1}^{\infty}\frac{1}{r(r+1)(r+2)} = \frac{1}{4}$ .

Two ways to take the limit:

If $S_n$ is left as constant minus vanishing fractions (e.g. $\tfrac14 - \tfrac{1}{2(n+1)(n+2)}$ ), just drop the fraction → $\tfrac14$ .
If $S_n$ is a single rational function, divide top and bottom by the highest power of $n$ , then send the $\tfrac1n$ terms to $0$ .

Cambridge tip. It is often cleaner not to fully combine into a single fraction before taking the limit — leaving " $\text{constant} - \text{vanishing}$ " makes the limit instant. Choose whichever form makes the limit obvious.

Telescope → $S_n$ → limit → $S_\infty$ in one chain.
Single rational $S_n$ : divide top & bottom by the highest power of $n$ .
'Constant − vanishing fraction' form makes the limit a one-liner.

See the full worked example for convergence of series and sum to infinity →

Quick recap

A series converges iff its partial sums $S_n$ tend to a finite limit; $S_\infty = \lim_{n\to\infty} S_n$ .
Any $\dfrac{1}{n+k} \to 0$ as $n \to \infty$ — drop those terms; constants survive.
From the method of differences: get $S_n$ , then take the limit for $S_\infty$ .
For a single rational $S_n$ , divide top and bottom by the highest power of $n$ .
A term growing with $n$ ⟹ divergence; $u_r \to 0$ is necessary but not sufficient.
State the limit step ('as $n \to \infty$ , ... $\to 0$ ') explicitly for full marks.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

$S_\infty = \lim_{n\to\infty} S_n$ — the sum to infinity is the limit of the partial sum.
$\dfrac{1}{n+k} \to 0$ and any fraction with $n \to \infty$ in the denominator vanishes.
Single rational $S_n$ : divide top and bottom by the highest power of $n$ to take the limit.
Converges $\Leftrightarrow$ $S_n$ tends to a finite value (decide from $S_n$ , not from $u_r$ alone).
Always state the limit step in words before writing $S_\infty$ .

How it’s examined

The sum to infinity is the natural follow-on to a method-of-differences question (subtopic 3.1) and appears in most Paper 1 papers as a final part worth 1-3 marks. A typical chain is: 'express in partial fractions [2] ... hence show that $\sum_{r=1}^{n}\ldots = S_n$ [3] ... deduce the sum to infinity [2]'. Marks are lost by (i) not stating the limit step (silently deleting a term), (ii) mishandling the limit of a single rational form, and (iii) claiming a sum to infinity for a divergent series. Keeping $S_n$ as 'constant − vanishing fraction' usually makes the final limit a one-liner.

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); 9231 Examiner Reports 2022-2024 (Paper 1); 9231/11, 9231/12, 9231/13 Further Pure Mathematics 1 — past papers and mark schemes; MF19 List of Formulae. Last reviewed 2026-06-23.

Take this whole topic with you

Step-by-step worked examples — Convergence of Series and Sum to Infinity

Step-by-step solutions to past-paper-style questions on convergence of series and sum to infinity, written exactly the way a tutor would explain them at the board.

Question type:

1Limit of a simple closed form
Getting startedDirect calculation• sum-to-infinity, limit
Question
A series has partial sum $S_n = 1 - \dfrac{1}{n+1}$ . Find the sum to infinity.
Step-by-step solution
1. Step 1
  Identify the vanishing term. As $n \to \infty$ , the denominator $n+1 \to \infty$ , so $\dfrac{1}{n+1} \to 0$ .
  $\frac{1}{n+1} \to 0 \quad \text{as } n \to \infty$
2. Step 2
  Take the limit of $S_n$ , dropping the vanishing term.
  $S_\infty = \lim_{n\to\infty}\left(1 - \frac{1}{n+1}\right) = 1 - 0 = 1$
Answer
$S_\infty = 1$ .
Examiner tip
State 'as $n \to \infty$ , $\tfrac{1}{n+1} \to 0$ ' explicitly — examiners want the limit justified, not the term silently deleted.
2Limit of a single rational form
Getting startedDirect calculation• sum-to-infinity, limit
Question
Find $\displaystyle\lim_{n\to\infty}\frac{n}{n+1}$ , and hence the sum to infinity of a series with $S_n = \dfrac{n}{n+1}$ .
Step-by-step solution
1. Step 1
  Divide top and bottom by $n$ (the highest power of $n$ ).
  $\frac{n}{n+1} = \frac{1}{1 + \tfrac{1}{n}}$
2. Step 2
  Send $\tfrac{1}{n} \to 0$ .
  $\frac{1}{1 + \tfrac{1}{n}} \to \frac{1}{1 + 0} = 1$
Answer
$\displaystyle\lim_{n\to\infty}\frac{n}{n+1} = 1$ , so $S_\infty = 1$ .
Examiner tip
Dividing by the highest power of $n$ is the safe way to handle a single rational $S_n$ . Here $\tfrac{n}{n+1}$ and $1 - \tfrac{1}{n+1}$ are the same series — both give $S_\infty = 1$ .
3Sum to infinity from a telescoped result
Building confidenceMulti-step problem• sum-to-infinity, method-of-differences
Question
It is given that $\displaystyle\sum_{r=1}^{n}\frac{2}{(2r-1)(2r+1)} = \frac{2n}{2n+1}$ . Determine whether the series converges and, if so, state its sum to infinity.
Step-by-step solution
1. Step 1
  Examine the partial sum as $n \to \infty$ . Divide top and bottom by $n$ .
  $\frac{2n}{2n+1} = \frac{2}{2 + \tfrac{1}{n}}$
2. Step 2
  Take the limit: $\tfrac{1}{n} \to 0$ , giving a finite value.
  $\frac{2}{2 + \tfrac{1}{n}} \to \frac{2}{2} = 1$
3. Step 3
  Conclude. $S_n \to 1$ , a finite limit, so the series converges.
  $\sum_{r=1}^{\infty}\frac{2}{(2r-1)(2r+1)} = 1$
Answer
The series converges, with sum to infinity $1$ .
Examiner tip
A 'determine whether it converges' instruction needs the explicit statement that $S_n$ tends to a finite limit. State both the convergence conclusion and the value.
4Limit with two vanishing terms (gap-2)
Building confidenceMulti-step problem• sum-to-infinity, gap-2
Question
Given that $\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+2)} = \frac{3}{4} - \frac{1}{2(n+1)} - \frac{1}{2(n+2)}$ , find the sum to infinity.
Step-by-step solution
1. Step 1
  Spot the vanishing terms. Both $\dfrac{1}{2(n+1)}$ and $\dfrac{1}{2(n+2)} \to 0$ as $n \to \infty$ .
  $\frac{1}{2(n+1)} \to 0, \qquad \frac{1}{2(n+2)} \to 0$
2. Step 2
  Only the constant $\tfrac34$ survives.
  $S_\infty = \frac{3}{4} - 0 - 0 = \frac{3}{4}$
Answer
$\displaystyle\sum_{r=1}^{\infty}\frac{1}{r(r+2)} = \frac{3}{4}$ .
Examiner tip
Keeping $S_n$ as 'constant minus vanishing fractions' makes the limit instant — both fractions drop, leaving $\tfrac34$ . Don't lose one of the two vanishing terms.
5Commenting on a divergent series
StretchShow that / prove• convergence, divergence
Question
Explain why the series $\displaystyle\sum_{r=1}^{\infty} r$ does not have a sum to infinity, but $\displaystyle\sum_{r=1}^{\infty}\frac{1}{r(r+1)}$ does.
Step-by-step solution
1. Step 1
  Closed form of the first series (standard result). It grows without bound.
  $S_n = \sum_{r=1}^{n} r = \tfrac{1}{2}n(n+1) \to \infty \;\text{as}\; n \to \infty$
2. Step 2
  Since $S_n \to \infty$ , the partial sums do not approach a finite limit — the series diverges, so no sum to infinity exists.
3. Step 3
  Closed form of the second series (telescoping). It tends to a finite limit.
  $S_n = \sum_{r=1}^{n}\frac{1}{r(r+1)} = 1 - \frac{1}{n+1} \to 1$
4. Step 4
  Since $S_n \to 1$ (finite), this series converges with sum to infinity $1$ . Note the general term $\tfrac{1}{r(r+1)} \to 0$ , whereas $r \not\to 0$ .
Answer
$\sum r$ diverges ( $S_n = \tfrac12 n(n+1) \to \infty$ ); $\sum \tfrac{1}{r(r+1)}$ converges to $1$ ( $S_n = 1 - \tfrac{1}{n+1} \to 1$ ).
Examiner tip
Convergence is decided by the behaviour of $S_n$ , not by inspection of the term alone. The contrast — terms that don't shrink vs terms that vanish — is the key teaching point.
6Differences then limit — the full chain
StretchMulti-step problem• sum-to-infinity, method-of-differences
Question
Given $\dfrac{1}{r(r+1)(r+2)} = \dfrac{1}{2}\left[\dfrac{1}{r(r+1)} - \dfrac{1}{(r+1)(r+2)}\right]$ , find $\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+1)(r+2)}$ and deduce the sum to infinity.
Step-by-step solution
1. Step 1
  Telescope with $f(r) = \dfrac{1}{r(r+1)}$ , so the term is $\tfrac12\big[f(r) - f(r+1)\big]$ , giving $\tfrac12\big[f(1) - f(n+1)\big]$ .
  $S_n = \frac{1}{2}\left[\frac{1}{1\cdot 2} - \frac{1}{(n+1)(n+2)}\right] = \frac{1}{4} - \frac{1}{2(n+1)(n+2)}$
2. Step 2
  Take the limit. The fraction $\dfrac{1}{2(n+1)(n+2)} \to 0$ as $n \to \infty$ (denominator grows without bound).
  $\frac{1}{2(n+1)(n+2)} \to 0$
3. Step 3
  Only the constant survives — the series converges.
  $S_\infty = \frac{1}{4} - 0 = \frac{1}{4}$
Answer
$S_n = \dfrac{1}{4} - \dfrac{1}{2(n+1)(n+2)}$ ; the series converges with $\displaystyle\sum_{r=1}^{\infty}\frac{1}{r(r+1)(r+2)} = \frac{1}{4}$ .
Examiner tip
Leaving $S_n$ in the 'constant − vanishing fraction' form (rather than the combined $\tfrac{n(n+3)}{4(n+1)(n+2)}$ ) makes the final limit a one-liner. Either form earns the marks, but this one is faster.

Model Answers — Convergence of Series and Sum to Infinity

High-scoring sample answers for convergence of series and sum to infinity on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
3 marks
A series has $n$ th partial sum $S_n = 2 - \dfrac{3}{n+2}$ . State, with reason, whether the series converges, and give its sum to infinity if it does.
Model answer
As $n \to \infty$ , the denominator $n+2 \to \infty$ , so $\frac{3}{n+2} \to 0.$

Therefore the partial sums tend to a finite limit: $S_\infty = \lim_{n\to\infty}\left(2 - \frac{3}{n+2}\right) = 2 - 0 = 2.$

Since $S_n$ approaches the finite value $2$ , the series converges, with sum to infinity $2$ .
Why this scores
M1 for stating $\tfrac{3}{n+2} \to 0$ , A1 for $S_\infty = 2$ , B1 for the explicit 'converges because $S_n$ tends to a finite limit' statement. The reason must be given, not just the value.
Question 2
3 marks
Find $\displaystyle\lim_{n\to\infty}\frac{3n^2 + 1}{6n^2 + 5n}$ , showing your method.
Model answer
Divide every term, top and bottom, by the highest power of $n$ , which is $n^2$ : $\frac{3n^2 + 1}{6n^2 + 5n} = \frac{3 + \dfrac{1}{n^2}}{6 + \dfrac{5}{n}}.$

As $n \to \infty$ , both $\dfrac{1}{n^2} \to 0$ and $\dfrac{5}{n} \to 0$ , so $\lim_{n\to\infty}\frac{3n^2 + 1}{6n^2 + 5n} = \frac{3 + 0}{6 + 0} = \frac{1}{2}.$
Why this scores
M1 for dividing by $n^2$ , M1 for sending the $\tfrac1n$ -type terms to $0$ , A1 for $\tfrac12$ . The limit equals the ratio of leading coefficients $\tfrac{3}{6}$ — a useful sanity check.
Question 3
5 marks
(a) Given that $\dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1}$ , find $\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+1)}$ . (b) Deduce $\displaystyle\sum_{r=1}^{\infty}\frac{1}{r(r+1)}$ .
Model answer
(a) With $f(r) = \dfrac{1}{r}$ , each term is $f(r) - f(r+1)$ . Writing the sum out: $\sum_{r=1}^{n}\frac{1}{r(r+1)} = \left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{n}-\frac{1}{n+1}\right).$

The interior terms cancel, leaving $S_n = 1 - \frac{1}{n+1} = \frac{n}{n+1}.$

(b) As $n \to \infty$ , $\dfrac{1}{n+1} \to 0$ , so the partial sums tend to a finite limit: $\sum_{r=1}^{\infty}\frac{1}{r(r+1)} = \lim_{n\to\infty}\left(1 - \frac{1}{n+1}\right) = 1.$

The series therefore converges, with sum to infinity $1$ .
Why this scores
Part (a): M1 for writing out the terms, M1 for the cancellation, A1 for $1 - \tfrac{1}{n+1}$ . Part (b): M1 for the limit step, A1 for $1$ . 'Deduce' means use part (a) — do not start a fresh method.
Question 4
6 marks
(a) Show that $\dfrac{1}{(2r-1)(2r+1)} = \dfrac{1}{2}\left(\dfrac{1}{2r-1} - \dfrac{1}{2r+1}\right)$ . (b) Hence find $\displaystyle\sum_{r=1}^{n}\frac{1}{(2r-1)(2r+1)}$ in terms of $n$ . (c) Write down the sum to infinity.
Model answer
(a) Combine the right-hand side over a common denominator: $\frac{1}{2}\left(\frac{1}{2r-1} - \frac{1}{2r+1}\right) = \frac{1}{2}\cdot\frac{(2r+1) - (2r-1)}{(2r-1)(2r+1)} = \frac{1}{2}\cdot\frac{2}{(2r-1)(2r+1)} = \frac{1}{(2r-1)(2r+1)}. \;\checkmark$

(b) With $f(r) = \dfrac{1}{2r-1}$ and the $\tfrac12$ pulled outside, the term is $\tfrac12\big[f(r) - f(r+1)\big]$ (since $2(r+1)-1 = 2r+1$ ). Telescoping: $\sum_{r=1}^{n}\frac{1}{(2r-1)(2r+1)} = \frac{1}{2}\left(\frac{1}{1} - \frac{1}{2n+1}\right) = \frac{1}{2}\cdot\frac{2n}{2n+1} = \frac{n}{2n+1}.$

(c) As $n \to \infty$ , divide top and bottom by $n$ : $\dfrac{n}{2n+1} = \dfrac{1}{2 + \tfrac1n} \to \dfrac{1}{2}$ . $\sum_{r=1}^{\infty}\frac{1}{(2r-1)(2r+1)} = \frac{1}{2}.$
Why this scores
Part (a): M1 A1 for the common-denominator verification ('Show that' — numerator $2$ must appear). Part (b): M1 for the telescope, M1 for keeping the $\tfrac12$ factor, A1 for $\tfrac{n}{2n+1}$ . Part (c): B1 for $\tfrac12$ . Forgetting the outside $\tfrac12$ is the classic slip.
Question 5
9 marks
(a) Express $\dfrac{4r+6}{r(r+1)(r+2)}$ in partial fractions. (b) Hence find $\displaystyle\sum_{r=1}^{n}\frac{4r+6}{r(r+1)(r+2)}$ . (c) State the sum to infinity, justifying convergence.
Model answer
(a) Write $\dfrac{4r+6}{r(r+1)(r+2)} = \dfrac{A}{r} + \dfrac{B}{r+1} + \dfrac{C}{r+2}$ , so $4r+6 = A(r+1)(r+2) + Br(r+2) + Cr(r+1)$ .

$r=0$ : $6 = A(1)(2) = 2A \Rightarrow A = 3$ .

$r=-1$ : $2 = B(-1)(1) = -B \Rightarrow B = -2$ .

$r=-2$ : $-2 = C(-2)(-1) = 2C \Rightarrow C = -1$ .

$\frac{4r+6}{r(r+1)(r+2)} = \frac{3}{r} - \frac{2}{r+1} - \frac{1}{r+2}.$

(b) Regroup so the sum telescopes. Split $\dfrac{3}{r} - \dfrac{2}{r+1} - \dfrac{1}{r+2}$ as two difference pairs: $= \left(\frac{2}{r} - \frac{2}{r+1}\right) + \left(\frac{1}{r} - \frac{1}{r+2}\right) = 2\left(\frac{1}{r} - \frac{1}{r+1}\right) + \left(\frac{1}{r} - \frac{1}{r+2}\right).$ (Check: $\tfrac2r - \tfrac{2}{r+1} + \tfrac1r - \tfrac{1}{r+2} = \tfrac3r - \tfrac{2}{r+1} - \tfrac{1}{r+2}$ . ✓)

The first piece telescopes (gap 1) to $2\left(1 - \dfrac{1}{n+1}\right)$ . The second telescopes (gap 2) to $\left(\dfrac{1}{1} + \dfrac{1}{2} - \dfrac{1}{n+1} - \dfrac{1}{n+2}\right)$ . Adding: $S_n = 2 - \frac{2}{n+1} + \frac{3}{2} - \frac{1}{n+1} - \frac{1}{n+2} = \frac{7}{2} - \frac{3}{n+1} - \frac{1}{n+2}.$

(c) As $n \to \infty$ , both $\dfrac{3}{n+1} \to 0$ and $\dfrac{1}{n+2} \to 0$ , so $S_n$ tends to the finite limit $S_\infty = \frac{7}{2} - 0 - 0 = \frac{7}{2}.$ Since $S_n$ approaches a finite value, the series converges, with sum to infinity $\dfrac{7}{2}$ .
Why this scores
Part (a): M1 setup, A1 A1 for $A=3, B=-2, C=-1$ . Part (b): M1 for a telescoping regrouping, M1 for the gap-1 piece, M1 for the gap-2 piece, A1 for $\tfrac72 - \tfrac{3}{n+1} - \tfrac{1}{n+2}$ . Part (c): M1 for the vanishing-term reasoning, A1 for $\tfrac72$ with a convergence statement. The regrouping into two clean difference pairs is the discriminating step.
Question 6
8 marks
The $r$ th term of a series is $u_r = \dfrac{2r+1}{r(r+1)}$ . (a) Show that $u_r = \dfrac{1}{r} + \dfrac{1}{r+1}$ . (b) By finding $\displaystyle\sum_{r=1}^{n} u_r$ , determine, with justification, whether the series converges.
Model answer
(a) Combine the proposed sum over a common denominator: $\frac{1}{r} + \frac{1}{r+1} = \frac{(r+1) + r}{r(r+1)} = \frac{2r+1}{r(r+1)} = u_r. \;\checkmark$

(b) This is a sum, not a difference, so it does not telescope. Split it instead: $\sum_{r=1}^{n} u_r = \sum_{r=1}^{n}\frac{1}{r} + \sum_{r=1}^{n}\frac{1}{r+1} = \left(1 + \frac{1}{2} + \cdots + \frac{1}{n}\right) + \left(\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n+1}\right).$

Both bracketed sums are partial sums of the harmonic series, which grows without bound as $n \to \infty$ . Equivalently, every term $u_r = \dfrac{1}{r} + \dfrac{1}{r+1} > \dfrac{1}{r}$ , and $\sum \dfrac{1}{r}$ already diverges, so by comparison the partial sums increase without limit: $S_n \to \infty \quad \text{as } n \to \infty.$

Since $S_n$ does not approach a finite limit, the series diverges and has no sum to infinity.
Why this scores
Part (a): M1 A1 for the common-denominator verification ('Show that'). Part (b): M1 for recognising the term is a SUM (so no telescoping), M1 for relating to the harmonic series, A1 for the divergence conclusion, plus marks for the comparison reasoning. The trap is blindly telescoping a $f(r) + f(r+1)$ structure — only a DIFFERENCE collapses.

Key Formulae — Convergence of Series and Sum to Infinity

The formulae you need to memorise for convergence of series and sum to infinity on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Sum to infinity
$S_\infty = \sum_{r=1}^{\infty} u_r = \lim_{n\to\infty} S_n$
$S_n$
the $n$ th partial sum $\sum_{r=1}^{n} u_r$
$S_\infty$
the sum to infinity, when the limit exists
When to use
Once a closed form for $S_n$ is known. Defined only when the limit is finite (the series converges).
Standard vanishing limit
$\lim_{n\to\infty}\frac{1}{n+k} = 0 \quad (k \text{ fixed})$
$k$
any fixed constant
When to use
Any fraction whose denominator grows without bound tends to $0$ — drop those terms when taking the limit.
Limit of a rational function (divide by highest power)
$\lim_{n\to\infty}\frac{a_pn^p + \cdots}{b_qn^q + \cdots} = \frac{a_p}{b_q} \text{ if } p=q, \;=0 \text{ if } p<q$
$p, q$
degrees of numerator and denominator
When to use
When $S_n$ is a single rational function. Divide top and bottom by the highest power of $n$ , then send $\tfrac1n$ -terms to $0$ .
Necessary condition for convergence
$\text{converges} \Rightarrow u_r \to 0 \text{ as } r \to \infty$
$u_r$
the general term of the series
When to use
If the terms do NOT tend to $0$ , the series diverges. (The converse fails — e.g. the harmonic series.)

Key Definitions and Keywords — Convergence of Series and Sum to Infinity

Definitions to memorise and the exact keywords mark schemes credit for convergence of series and sum to infinity answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Partial sum $S_n$
Examiner keyword
The sum of the first $n$ terms of a series, $S_n = \sum_{r=1}^{n} u_r$ . Convergence is decided entirely by the behaviour of $S_n$ as $n \to \infty$ .
Convergence
Examiner keyword
A series converges if its partial sums $S_n$ approach a finite limit as $n \to \infty$ . That limit is the sum to infinity.
Divergence
Examiner keyword
A series diverges if its partial sums do not approach a finite limit — they grow without bound or oscillate. No sum to infinity exists.
Sum to infinity $S_\infty$
Examiner keyword
$S_\infty = \lim_{n\to\infty} S_n$ , the finite value the partial sums settle on for a convergent series.
Limit as $n \to \infty$
Examiner keyword
The fixed value an expression approaches as $n$ grows without bound. Fractions with $n$ in the denominator tend to $0$ ; the surviving constants give the limit.
Necessary condition $u_r \to 0$
For a series to converge the terms must tend to $0$ . This is necessary but NOT sufficient — the harmonic series has $u_r \to 0$ yet diverges.

Common Mistakes and Misconceptions — Convergence of Series and Sum to Infinity

The traps other students keep falling into on convergence of series and sum to infinity questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Silently deleting a $\frac{1}{n+1}$ term instead of stating it tends to $0$
9231 Paper 1 Examiner Reports — sum-to-infinity questions
Why it happens
Treating the limit as obvious and skipping the justification.
How to avoid it
Always write 'as $n \to \infty$ , $\tfrac{1}{n+1} \to 0$ ' before giving $S_\infty$ — the limit step earns its own mark.
✕Quoting a 'sum to infinity' for a divergent series
9231 Paper 1 mark schemes
Why it happens
Plugging $n \to \infty$ mechanically without checking whether $S_n$ actually stays finite.
How to avoid it
Check that $S_n$ tends to a finite value first. If a term grows with $n$ (e.g. $\tfrac12 n(n+1)$ ), the series diverges — say so.
✕Mishandling the limit of a single rational $S_n$ (e.g. $\tfrac{n}{2n+1} \to 1$ )
9231 Paper 1 Examiner Reports
Why it happens
Comparing leading terms loosely instead of dividing by the highest power of $n$ .
How to avoid it
Divide top and bottom by the highest power of $n$ , then send $\tfrac1n$ -terms to $0$ . $\tfrac{n}{2n+1} = \tfrac{1}{2+1/n} \to \tfrac12$ .
✕Losing one of two vanishing terms in a gap-2 result
9231 Paper 1 mark schemes
Why it happens
Carrying over a sign error from the differences stage, or forgetting the second surviving end term.
How to avoid it
Send EVERY fraction containing $n$ to $0$ ; the leftover constant is $S_\infty$ . Re-check the closed form from subtopic 3.1.
✕Concluding a series converges just because $u_r \to 0$
9231 Paper 1 Examiner Reports
Why it happens
Treating the necessary condition as if it were sufficient.
How to avoid it
$u_r \to 0$ is necessary, not sufficient (harmonic series diverges). Decide convergence from the closed form $S_n$ .
✕Trying to telescope a term that is a SUM $f(r)+f(r+1)$ , not a difference
9231 Paper 1 Examiner Reports
Why it happens
Pattern-matching to 'two fractions' without checking the sign between them.
How to avoid it
Only a DIFFERENCE $f(r) - f(r+k)$ telescopes. A sum like $\tfrac1r + \tfrac{1}{r+1}$ does not collapse — handle it separately (often it diverges).

Convergence of Series and Sum to Infinity

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Limit of a simple closed form

2Limit of a single rational form

3Sum to infinity from a telescoped result

4Limit with two vanishing terms (gap-2)

5Commenting on a divergent series

6Differences then limit — the full chain

Sum to infinity

Standard vanishing limit

Limit of a rational function (divide by highest power)

Necessary condition for convergence

Partial sum SnS_nSn​

Convergence

Divergence

Sum to infinity S∞S_\inftyS∞​

Limit as n→∞n \to \inftyn→∞

Necessary condition ur→0u_r \to 0ur​→0

✕Silently deleting a 1n+1\frac{1}{n+1}n+11​ term instead of stating it tends to 000

✕Quoting a 'sum to infinity' for a divergent series

✕Mishandling the limit of a single rational SnS_nSn​ (e.g. n2n+1→1\tfrac{n}{2n+1} \to 12n+1n​→1)

✕Losing one of two vanishing terms in a gap-2 result

✕Concluding a series converges just because ur→0u_r \to 0ur​→0

✕Trying to telescope a term that is a SUM f(r)+f(r+1)f(r)+f(r+1)f(r)+f(r+1), not a difference

Partial sum $S_n$

Sum to infinity $S_\infty$

Limit as $n \to \infty$

Necessary condition $u_r \to 0$

✕Silently deleting a $\frac{1}{n+1}$ term instead of stating it tends to $0$

✕Mishandling the limit of a single rational $S_n$ (e.g. $\tfrac{n}{2n+1} \to 1$ )

✕Concluding a series converges just because $u_r \to 0$

✕Trying to telescope a term that is a SUM $f(r)+f(r+1)$ , not a difference