If the difference is f(r) - f(r + 2) (a gap of two), then two terms survive at each end: f(1) + f(2) at the start and -f(n+1) - f(n+2) at the end. Always write out the first few and last few terms to see exactly what remains.

What are sign error in partial fractions?

Verify by recombining the partial fractions back to the original before summing.

Given 1r - 1r+1 = 1r(r+1), find _r=1^n 1r(r+1). [3 marks]

Deduce the sum to infinity of _r=1^∞ 1r(r+1). [2 marks]

SEAB Original-style practice: Express 1r(r+1) in partial fractions and hence find _r=1^n 1r(r+1). Deduce the sum to infinity.

Partial fractions: 1r(r+1) = 1r - 1r+1. The sum telescopes: _r=1^n(1r - 1r+1) = (1 - 12) + (12 - 13) + + (1n - 1n+1). All interior terms cancel, leaving 1 - 1n+1 = nn+1. As n → ∞, 1n+1 → 0, so the sum to infinity is 1. Markers reward the partial fractions, showing the cancellation explicitly, the closed form nn+1, and the limit.

SEAB Original-style practice: Given that 1r - 1r+2 = 2r(r+2), find _r=1^n 1r(r+2).

From the identity, 1r(r+2) = 12(1r - 1r+2). The sum is 12_r=1^n(1r - 1r+2). Because the gap is 2, two terms survive at each end: = 12[(1 + 12) - (1n+1 + 1n+2)] = 12[32 - 1n+1 - 1n+2]. So the sum is 34 - 12(n+1) - 12(n+2). Markers reward the factor (1)/(2), correctly identifying that two terms survive at each end (gap of two), and the closed form.

SingaporeMathsSyllabus dot point

How does a telescoping sum collapse, and how do we exploit it to evaluate a series?

Use the method of differences, including the use of partial fractions, to find the sum of a series whose terms telescope, and deduce the sum to infinity where it exists

A focused answer to the H2 Mathematics outcome on the method of differences. Writing a term as a difference (often via partial fractions), cancelling the telescoping sum, and deducing the sum to infinity.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
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What this dot point is asking

SEAB wants you to evaluate a series by the method of differences: write each term as the difference of consecutive (or near-consecutive) values of some function $\mathrm{f}(r)$ , so that when you sum, almost everything cancels (telescopes), leaving only the boundary terms. Partial fractions is the usual tool for producing the difference.

The answer

The telescoping idea

If each term can be written $u_r = \mathrm{f}(r) - \mathrm{f}(r + 1)$ , then

\sum_{r=1}^{n} u_r = \mathrm{f}(1) - \mathrm{f}(n + 1)

because every interior $\mathrm{f}(2), \mathrm{f}(3), \ldots, \mathrm{f}(n)$ appears once positive and once negative and cancels. Only the very first and very last survive.

Producing the difference with partial fractions

A term like $\dfrac{1}{r(r+1)}$ does not look like a difference, but partial fractions reveals one:

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

This is the standard route: factor the denominator, split into partial fractions, and recognise the difference of a function at consecutive arguments.

Wider gaps

If the difference is $\mathrm{f}(r) - \mathrm{f}(r + 2)$ (a gap of two), then two terms survive at each end: $\mathrm{f}(1) + \mathrm{f}(2)$ at the start and $-\mathrm{f}(n+1) - \mathrm{f}(n+2)$ at the end. Always write out the first few and last few terms to see exactly what remains.

Sum to infinity

Once you have the closed form, let $n \to \infty$ . If the surviving $n$ -dependent terms tend to zero, the sum to infinity is just the constant boundary part.

Worked example

Find $\displaystyle\sum_{r=1}^{n} \dfrac{1}{(2r-1)(2r+1)}$ and deduce its sum to infinity.

Step 1: Partial fractions

Write $\dfrac{1}{(2r-1)(2r+1)} = \dfrac{A}{2r-1} + \dfrac{B}{2r+1}$ . Then $1 = A(2r+1) + B(2r-1)$ . Setting $r = \tfrac{1}{2}$ : $1 = 2A$ , so $A = \tfrac{1}{2}$ . Setting $r = -\tfrac{1}{2}$ : $1 = -2B$ , so $B = -\tfrac{1}{2}$ .

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

Step 2: Write the telescoping sum

\frac{1}{2}\sum_{r=1}^{n}\left(\frac{1}{2r-1} - \frac{1}{2r+1}\right) = \frac{1}{2}\left[\left(1 - \tfrac{1}{3}\right) + \left(\tfrac{1}{3} - \tfrac{1}{5}\right) + \cdots + \left(\tfrac{1}{2n-1} - \tfrac{1}{2n+1}\right)\right]

Step 3: Cancel interior terms

Only the first and last survive: $\dfrac{1}{2}\left(1 - \dfrac{1}{2n+1}\right)$ .

Step 4: Simplify

= \frac{1}{2} \cdot \frac{2n}{2n+1} = \frac{n}{2n+1}

Step 5: Sum to infinity

As $n \to \infty$ , $\dfrac{n}{2n+1} \to \dfrac{1}{2}$ , so the sum to infinity is $\dfrac{1}{2}$ .

Examples in context

Example 1. A converging resistance ladder. A series $\sum \dfrac{1}{r(r+1)} = 1 - \dfrac{1}{n+1}$ models cumulative contributions that approach a finite total of $1$ , a clean physical analogue of a quantity saturating as more stages are added.

Example 2. Checking a closed form. Having found $\sum_{r=1}^{n} \dfrac{1}{(2r-1)(2r+1)} = \dfrac{n}{2n+1}$ , substituting $n = 1$ gives $\dfrac{1}{3}$ , which matches the single term $\dfrac{1}{1 \times 3}$ , a quick verification of the algebra.

Try this

Q1. Given $\dfrac{1}{r} - \dfrac{1}{r+1} = \dfrac{1}{r(r+1)}$ , find $\displaystyle\sum_{r=1}^{n} \dfrac{1}{r(r+1)}$ . [3 marks]

Cue. Telescopes to $1 - \dfrac{1}{n+1} = \dfrac{n}{n+1}$ .

Q2. State what is meant by a telescoping sum and why it simplifies. [2 marks]

Cue. Each term is a difference of consecutive values, so interior terms cancel in pairs, leaving only the boundary terms.

Q3. Deduce the sum to infinity of $\displaystyle\sum_{r=1}^{\infty} \dfrac{1}{r(r+1)}$ . [2 marks]

Cue. $\dfrac{n}{n+1} \to 1$ as $n \to \infty$ , so the sum to infinity is $1$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original6 marksExpress

\dfrac{1}{r(r+1)}

in partial fractions and hence find

\displaystyle\sum_{r=1}^{n} \dfrac{1}{r(r+1)}

. Deduce the sum to infinity.

Show worked answer →

Partial fractions: $\dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1}$ .

The sum telescopes:
$\sum_{r=1}^{n}\left(\dfrac{1}{r} - \dfrac{1}{r+1}\right) = \left(1 - \tfrac{1}{2}\right) + \left(\tfrac{1}{2} - \tfrac{1}{3}\right) + \cdots + \left(\tfrac{1}{n} - \tfrac{1}{n+1}\right)$ .

All interior terms cancel, leaving $1 - \dfrac{1}{n+1} = \dfrac{n}{n+1}$ .

As $n \to \infty$ , $\dfrac{1}{n+1} \to 0$ , so the sum to infinity is $1$ .

Markers reward the partial fractions, showing the cancellation explicitly, the closed form $\dfrac{n}{n+1}$ , and the limit.

Original5 marksGiven that

\dfrac{1}{r} - \dfrac{1}{r+2} = \dfrac{2}{r(r+2)}

, find

\displaystyle\sum_{r=1}^{n} \dfrac{1}{r(r+2)}

Show worked answer →

From the identity, $\dfrac{1}{r(r+2)} = \dfrac{1}{2}\left(\dfrac{1}{r} - \dfrac{1}{r+2}\right)$ .

The sum is $\dfrac{1}{2}\sum_{r=1}^{n}\left(\dfrac{1}{r} - \dfrac{1}{r+2}\right)$ . Because the gap is $2$ , two terms survive at each end:

$= \dfrac{1}{2}\left[\left(1 + \tfrac{1}{2}\right) - \left(\tfrac{1}{n+1} + \tfrac{1}{n+2}\right)\right] = \dfrac{1}{2}\left[\dfrac{3}{2} - \dfrac{1}{n+1} - \dfrac{1}{n+2}\right]$ .

So the sum is $\dfrac{3}{4} - \dfrac{1}{2(n+1)} - \dfrac{1}{2(n+2)}$ .

Markers reward the factor $\frac{1}{2}$ , correctly identifying that two terms survive at each end (gap of two), and the closed form.