What is the method of differences (telescoping)?

If a term can be written as a difference u_r = f(r) - f(r+1), then the sum collapses:

What is using partial fractions to telescope?

Many fractional terms become telescoping after partial fractions. For example 1r(r+1) = 1r - 1r+1 has exactly the difference form. For 1r(r+2) the gap is two, so the cancellation leaves the first two and last two terms.

What is not displaying the cancellation?

The proof structure (write the first and last brackets) is itself worth marks.

Write 1r(r+1) as a difference of two fractions. [1 mark]

Use the standard results to find _r=1^n (r^2 + r). [2 marks]

State the sum to infinity of _r=1^∞(1r - 1r+1). [1 mark]

SEAB Original-style practice: Using the method of differences, find _r=1^n (1)/(r(r+1)) in terms of n, and deduce the sum to infinity.

Express the term in partial fractions: 1r(r+1) = 1r - 1r+1. Then the sum telescopes: $_r=1^n((1)/(r) - (1)/(r+1)) = ((1)/(1) - (1)/(2)) + ((1)/(2) - (1)/(3)) + + ((1)/(n) - (1)/(n+1)).$ All intermediate terms cancel, leaving $= 1 - (1)/(n+1) = (n)/(n+1).$ As n → ∞, 1n+1 → 0, so the sum to infinity is 1. Markers reward the partial fractions, displaying the telescoping cancellation, the closed form nn+1, and the limit 1.

SEAB Original-style practice: Find _r=1^n (2r - 1)^2 in terms of n, simplifying your answer.

Expand: (2r-1)^2 = 4r^2 - 4r + 1. So $_r=1^n(2r-1)^2 = 4_r=1^n r^2 - 4_r=1^n r + _r=1^n 1.$ Use the standard results sum r^2 = 16n(n+1)(2n+1), sum r = 12n(n+1), sum 1 = n: $= 4·16n(n+1)(2n+1) - 4·12n(n+1) + n.$ $= 23n(n+1)(2n+1) - 2n(n+1) + n.$ Taking out 13n and simplifying gives 13n(2n-1)(2n+1) = 13n(4n^2 - 1). Markers reward expanding the square, splitting into standard sums, substituting the standard results, and a fully simplified factorised answer.

SingaporeFurther MathsSyllabus dot point

How do we sum a finite series in closed form using the method of differences and the standard power sums?

Sum finite series using the method of differences, standard results for powers of integers, and partial fractions

A focused answer to the H2 Further Mathematics outcome on summing series. The method of differences (telescoping), the standard sums of powers of integers, partial fractions to create a telescoping form, and recovering sums 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

SEAB wants you to find closed-form expressions for finite series. The two main tools are the method of differences (telescoping), often set up with partial fractions, and the standard sums of the first, second and third powers of the integers. You should also be able to take the limit to recover a sum to infinity where it exists.

The answer

The standard power sums

These results, themselves proved by induction, are the building blocks:

\sum_{r=1}^{n} 1 = n, \qquad \sum_{r=1}^{n} r = \frac{1}{2}n(n+1),

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

Any polynomial in $r$ can be summed by expanding and applying these termwise, using the linearity of $\sum$ .

The method of differences (telescoping)

If a term can be written as a difference $u_r = \mathrm{f}(r) - \mathrm{f}(r+1)$ , then the sum collapses:

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

because every interior term cancels with its neighbour. Writing out the first two and last two terms makes the cancellation visible and is expected in a proof.

Using partial fractions to telescope

Many fractional terms become telescoping after partial fractions. For example $\dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1}$ has exactly the difference form. For $\dfrac{1}{r(r+2)}$ the gap is two, so the cancellation leaves the first two and last two terms.

Recovering the sum to infinity

A finite closed form often has a clear limit as $n \to \infty$ . If the leftover terms tend to zero, the series converges and its sum to infinity is that limit. For $\sum \dfrac{1}{r(r+1)} = \dfrac{n}{n+1}$ , the limit is $1$ .

Worked example

Find $\displaystyle\sum_{r=1}^{n} \frac{2}{(2r-1)(2r+1)}$ in closed form and state the sum to infinity.

Step 1: Split into partial fractions

Write $\dfrac{2}{(2r-1)(2r+1)} = \dfrac{A}{2r-1} + \dfrac{B}{2r+1}$ , so $2 = A(2r+1) + B(2r-1)$ . Setting $r = \tfrac{1}{2}$ gives $2 = 2A$ , $A = 1$ ; setting $r = -\tfrac{1}{2}$ gives $2 = -2B$ , $B = -1$ . Thus the term is $\dfrac{1}{2r-1} - \dfrac{1}{2r+1}$ .

Step 2: Write out the telescoping sum

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

Step 3: Cancel the interior terms

Every $-\dfrac{1}{2r+1}$ cancels the $+\dfrac{1}{2(r+1)-1}$ of the next bracket, leaving

= 1 - \frac{1}{2n+1}.

Step 4: Simplify and take the limit

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

As $n \to \infty$ , $\dfrac{1}{2n+1} \to 0$ , so the sum to infinity is $1$ .

Examples in context

Example 1. Probability over a discrete distribution. Verifying that the probabilities of a distribution sum to one often requires a telescoping or geometric series. The method of differences turns an apparently hard infinite sum into a single surviving term.

Example 2. Bounding a partial sum. Comparing $\sum \dfrac{1}{r^2}$ with the telescoping $\sum \dfrac{1}{r(r+1)} = 1 - \dfrac{1}{n+1}$ shows the partial sums are bounded above, a first taste of the convergence arguments used with series and improper integrals.

Try this

Q1. Write $\dfrac{1}{r(r+1)}$ as a difference of two fractions. [1 mark]

Cue. $\dfrac{1}{r} - \dfrac{1}{r+1}$ .

Q2. Use the standard results to find $\sum_{r=1}^{n} (r^2 + r)$ . [2 marks]

Cue. $\tfrac{1}{6}n(n+1)(2n+1) + \tfrac{1}{2}n(n+1) = \tfrac{1}{3}n(n+1)(n+2)$ .

Q3. State the sum to infinity of $\sum_{r=1}^{\infty}\left(\dfrac{1}{r} - \dfrac{1}{r+1}\right)$ . [1 mark]

Cue. $1$ , since the partial sum is $1 - \dfrac{1}{n+1} \to 1$ .

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 marksUsing the method of differences, find

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

in terms of

n

, and deduce the sum to infinity.

Show worked answer →

Express the term in partial fractions: $\dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1}$ .

Then the sum telescopes:

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

All intermediate terms cancel, leaving

= 1 - \frac{1}{n+1} = \frac{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, displaying the telescoping cancellation, the closed form $\tfrac{n}{n+1}$ , and the limit $1$ .

Original5 marksFind

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

in terms of

n

, simplifying your answer.

Show worked answer →

Expand: $(2r-1)^2 = 4r^2 - 4r + 1$ . So

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

Use the standard results

\sum r^2 = \tfrac{1}{6}n(n+1)(2n+1)

\sum r = \tfrac{1}{2}n(n+1)

\sum 1 = n

= 4\cdot\tfrac{1}{6}n(n+1)(2n+1) - 4\cdot\tfrac{1}{2}n(n+1) + n.

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

Taking out

\tfrac{1}{3}n

and simplifying gives

\tfrac{1}{3}n(2n-1)(2n+1) = \tfrac{1}{3}n(4n^2 - 1)

Markers reward expanding the square, splitting into standard sums, substituting the standard results, and a fully simplified factorised answer.