Method of Differences

The method of differences — also called telescoping — is the trick for summing a series whose terms cancel in cascade. If you can rewrite the general term $u_r$ur​ as a difference of consecutive values of some function, say $u_r = f(r) - f(r+1)$ur​=f(r)−f(r+1), then when you add up the terms almost everything cancels and only a handful of boundary pieces survive. It is the natural partner to the standard-results algebra of the previous lesson: where standard results handle polynomial summands, the method of differences handles rational and factorial summands that no standard result can touch.

This lesson develops the technique rigorously: the telescoping identity, how partial fractions generate the required difference form, how to write out and cancel terms reliably, and how to read off limiting ($n \to \infty$n→∞) values.

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Where this sits in AQA 7367

This is the method of differences strand of the compulsory pure content examined on Paper 1 and Paper 2 (each 2 hours, 100 marks, 33⅓% of the A-Level). It draws on partial fractions (an A-Level Maths and Further-Maths technique) and complements the summation standard results. The mechanical decompose-cancel-simplify routine is AO1; the "show that the sum equals …, and hence find its limit as $n\to\infty$n→∞" framing carries the AO2 reasoning marks, and the cancellation argument is itself a small proof.

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Core theory: why telescoping works

Suppose the general term of a series can be written as the difference of consecutive values of a function $f$f:

$u_r = f(r) - f(r+1).$ur​=f(r)−f(r+1).

Then the sum from $r = 1$r=1 to $n$n collapses:

$\sum_{r=1}^{n} u_r = \big[f(1) - f(2)\big] + \big[f(2) - f(3)\big] + \cdots + \big[f(n) - f(n+1)\big].$∑r=1n​ur​=[f(1)−f(2)]+[f(2)−f(3)]+⋯+[f(n)−f(n+1)].

Every interior value $f(2), f(3), \ldots, f(n)$f(2),f(3),…,f(n) appears once with a $+$+ sign and once with a $-$− sign, so they all cancel, leaving only the first and last survivors:

$\boxed{\;\sum_{r=1}^{n}\big[f(r) - f(r+1)\big] = f(1) - f(n+1).\;}$r=1∑n​[f(r)−f(r+1)]=f(1)−f(n+1).​

The picture is a collapsing telescope — hence the name. The reverse form is just as common:

$\sum_{r=1}^{n}\big[f(r+1) - f(r)\big] = f(n+1) - f(1).$∑r=1n​[f(r+1)−f(r)]=f(n+1)−f(1).

The art of the topic is producing the difference form $u_r = f(r) - f(r+1)$ur​=f(r)−f(r+1). For rational $u_r$ur​ the engine is partial fractions; for factorial or product terms it is an algebraic identity such as $r\cdot r! = (r+1)! - r!$r⋅r!=(r+1)!−r!.

It is worth pausing on why the cancellation is total and not merely partial. Lay the terms out in a grid: the first bracket contributes $+f(1)$+f(1) and $-f(2)$−f(2); the second contributes $+f(2)$+f(2) and $-f(3)$−f(3); and so on. Reading down the $+$+ column you see $f(1), f(2), f(3), \ldots, f(n)$f(1),f(2),f(3),…,f(n); reading down the $-$− column you see $f(2), f(3), \ldots, f(n), f(n+1)$f(2),f(3),…,f(n),f(n+1). The two columns share every entry except $f(1)$f(1) (only in the $+$+ column) and $f(n+1)$f(n+1) (only in the $-$− column). That is the entire content of the method: a sum of differences is a difference of the extreme values. Everything else is finding the right $f$f.

A second observation guides which decomposition to aim for. A telescoping difference is recognised by its shape: the two pieces are the same function evaluated at shifted arguments. So when you decompose a rational summand, do not stop at "some partial fractions" — push until the pieces visibly read as $f(r)$f(r) and $f(r+1)$f(r+1) (or $f(r+k)$f(r+k)). For $\dfrac{1}{r(r+1)}$r(r+1)1​ the pieces $\dfrac1r$r1​ and $\dfrac{1}{r+1}$r+11​ are $f(r)$f(r) and $f(r+1)$f(r+1) with $f(r) = \tfrac1r$f(r)=r1​; for $\dfrac{2}{r(r+1)(r+2)}$r(r+1)(r+2)2​ the right grouping is $\dfrac{1}{r(r+1)}$r(r+1)1​ and $\dfrac{1}{(r+1)(r+2)}$(r+1)(r+2)1​, which are $f(r)$f(r) and $f(r+1)$f(r+1) with $f(r) = \tfrac{1}{r(r+1)}$f(r)=r(r+1)1​. Spotting the shifted-function structure is the single most valuable skill in the topic, because it tells you in advance what the surviving terms will be.

A reliable cancellation routine

1. Decompose $u_r$ur​ into a telescoping difference (usually via partial fractions), pushing until the pieces read as shifted copies of one function $f$f.
2. Write out the first two or three terms and the last two or three terms explicitly, leaving "$\cdots$⋯" in the middle. Line up the $+$+ and $-$− parts so the cancellation is visible to the examiner.
3. Cancel the interior values, identifying every survivor — and double-check the count of survivors matches the gap.
4. Simplify the surviving boundary terms into a single expression, then read off the limit as $n\to\infty$n→∞ if asked.

The discipline of writing terms out is not busywork: it is the only reliable defence against the two classic failures — miscounting survivors (a gap-2 difference leaves two at each end, not one) and mislabelling the last survivor as $f(n)$f(n) when it is $f(n+1)$f(n+1). Even when you can see the answer immediately from $f(1) - f(n+1)$f(1)−f(n+1), a "show that" question expects the cancellation displayed, so make a habit of it.

When the gap inside the difference is bigger than one (e.g. $\tfrac1r - \tfrac{1}{r+2}$r1​−r+21​), more than one term survives at each end — this is exactly where candidates lose marks, so writing out enough terms is not optional.

Producing the difference form by partial fractions

For a rational summand the practical question is: which partial-fraction decomposition telescopes? The guiding principle is that the pieces must be shifted copies of one function, so you choose the grouping that makes that visible. Three patterns cover almost every exam case.

• Two consecutive linear factors, e.g. $\dfrac{1}{(r+a)(r+a+1)}$(r+a)(r+a+1)1​. Cover-up gives $\dfrac{1}{r+a} - \dfrac{1}{r+a+1}$r+a1​−r+a+11​, a gap-1 telescope with $f(r) = \tfrac{1}{r+a}$f(r)=r+a1​; one term survives at each end.
• Two linear factors a gap of $d$d apart, e.g. $\dfrac{1}{(r)(r+d)}$(r)(r+d)1​. Cover-up gives $\dfrac{1}{d}\big(\dfrac1r - \dfrac{1}{r+d}\big)$d1​(r1​−r+d1​), so the constant $\tfrac1d$d1​ appears and $d$d terms survive at each end. Forgetting the $\tfrac1d$d1​ is a frequent slip.
• Three (or more) consecutive factors, e.g. $\dfrac{c}{r(r+1)(r+2)}$r(r+1)(r+2)c​. Here you do not split into three single fractions; you group into a difference of two-factor pieces $\dfrac{1}{r(r+1)} - \dfrac{1}{(r+1)(r+2)}$r(r+1)1​−(r+1)(r+2)1​, which is $f(r) - f(r+1)$f(r)−f(r+1) with $f(r) = \tfrac{1}{r(r+1)}$f(r)=r(r+1)1​. Example 4 shows the technique for $(2r-1)(2r+1)(2r+3)$(2r−1)(2r+1)(2r+3).

A reliable way to find the constant in the gap-$d$d case without full partial fractions is the identity $\dfrac1r - \dfrac{1}{r+d} = \dfrac{d}{r(r+d)}$r1​−r+d1​=r(r+d)d​; rearranged, $\dfrac{1}{r(r+d)} = \dfrac1d\big(\dfrac1r - \dfrac{1}{r+d}\big)$r(r+d)1​=d1​(r1​−r+d1​). Carry that one line in your head and most rational telescopes become immediate.

A reliable cancellation routine

Example 1 — the standard telescoping sum

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

Decompose by partial fractions (cover-up rule):

$\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}. \quad (\text{M1 partial fractions})$r(r+1)1​=r1​−r+11​.(M1 partial fractions)

So this is the telescoping form with $f(r) = \tfrac1r$f(r)=r1​:

$\sum_{r=1}^{n}\left(\frac1r - \frac{1}{r+1}\right) = \left(1 - \frac12\right) + \left(\frac12 - \frac13\right) + \cdots + \left(\frac1n - \frac{1}{n+1}\right). \quad (\text{M1 write out terms})$∑r=1n​(r1​−r+11​)=(1−21​)+(21​−31​)+⋯+(n1​−n+11​).(M1 write out terms)

All interior terms cancel, leaving the first $+$+-survivor and last $-$−-survivor:

$= 1 - \frac{1}{n+1} = \frac{n}{n+1}. \quad (\text{A1})$=1−n+11​=n+1n​.(A1)

(M1 for the partial-fraction decomposition; M1 for writing out terms and identifying the cancellation; A1 for $\tfrac{n}{n+1}$n+1n​. As $n\to\infty$n→∞, $\tfrac{n}{n+1}\to 1$n+1n​→1, so the infinite series converges to 1.)

This is the prototype every other rational telescope imitates. Notice three features that recur throughout the topic: the decomposition is a gap-1 difference, so exactly one term survives at each end ($+1$+1 from $r=1$r=1, $-\tfrac{1}{n+1}$−n+11​ from $r=n$r=n); the closed form $\tfrac{n}{n+1}$n+1n​ is a single tidy fraction, which is what a "show that" will demand; and the limit is read off in one line by letting the $n$n-dependent survivor vanish. If you can reproduce this example fluently — decomposition, written-out cancellation, simplified answer, limit — you have the template for the whole lesson.

Example 2 — a gap of two, so two terms survive each end

Find $\displaystyle\sum_{r=1}^{n} \frac{1}{r(r+2)}$r=1∑n​r(r+2)1​.

By cover-up, $\dfrac{1}{r(r+2)} = \dfrac12\left(\dfrac1r - \dfrac{1}{r+2}\right)$r(r+2)1​=21​(r1​−r+21​). $(\text{M1})$(M1) Write out the terms, watching the gap of 2:

$\frac12\!\left[\left(1 - \frac13\right) + \left(\frac12 - \frac14\right) + \left(\frac13 - \frac15\right) + \cdots + \left(\frac{1}{n-1} - \frac{1}{n+1}\right) + \left(\frac1n - \frac{1}{n+2}\right)\right]. \quad (\text{M1})$21​[(1−31​)+(21​−41​)+(31​−51​)+⋯+(n−11​−n+11​)+(n1​−n+21​)].(M1)

Now cancel. The $+\tfrac13$+31​ (term $r=3$r=3) kills the $-\tfrac13$−31​ (term $r=1$r=1), and so on; the two smallest positives $1$1 and $\tfrac12$21​ have no partner, and the two largest negatives $\tfrac{1}{n+1}, \tfrac{1}{n+2}$n+11​,n+21​ have no partner:

$= \frac12\left(1 + \frac12 - \frac{1}{n+1} - \frac{1}{n+2}\right) = \frac12\left(\frac32 - \frac{1}{n+1} - \frac{1}{n+2}\right) = \frac34 - \frac12\left(\frac{1}{n+1} + \frac{1}{n+2}\right). \quad (\text{A1})$=21​(1+21​−n+11​−n+21​)=21​(23​−n+11​−n+21​)=43​−21​(n+11​+n+21​).(A1)

(M1 partial fractions; M1 writing enough terms to expose a gap-2 cancellation; A1 for the simplified result. The danger is forgetting that two terms survive at each end — a gap of $k$k in the difference leaves $k$k survivors per end. As $n\to\infty$n→∞ the sum $\to \tfrac34$→43​.)

Example 3 — a factorial difference (no fractions)

Find $\displaystyle\sum_{r=1}^{n} r\cdot r!$r=1∑n​r⋅r!.

The key identity is $r\cdot r! = (r+1)! - r!$r⋅r!=(r+1)!−r!, because $(r+1)! = (r+1)\,r! = r\cdot r! + r!$(r+1)!=(r+1)r!=r⋅r!+r!. $(\text{M1 establish identity})$(M1 establish identity) Then

$\sum_{r=1}^{n} r\cdot r! = \sum_{r=1}^{n}\big[(r+1)! - r!\big] = (2! - 1!) + (3! - 2!) + \cdots + \big((n+1)! - n!\big). \quad (\text{M1})$∑r=1n​r⋅r!=∑r=1n​[(r+1)!−r!]=(2!−1!)+(3!−2!)+⋯+((n+1)!−n!).(M1)

Telescoping (reverse form, $f(r) = r!$f(r)=r!) leaves

$= (n+1)! - 1! = (n+1)! - 1. \quad (\text{A1})$=(n+1)!−1!=(n+1)!−1.(A1)

(M1 for the algebraic identity $r\cdot r! = (r+1)! - r!$r⋅r!=(r+1)!−r!; M1 for the telescoping setup; A1 for $(n+1)! - 1$(n+1)!−1. Check at $n=2$n=2: $1\cdot1! + 2\cdot2! = 1 + 4 = 5 = 3! - 1$1⋅1!+2⋅2!=1+4=5=3!−1 ✓.)

The lesson of Example 3 is that telescoping is not about fractions — it is about expressing the summand as a difference of consecutive values of some function, and factorials work just as well as reciprocals. The hard step is purely algebraic: spotting $r\cdot r! = (r+1)! - r!$r⋅r!=(r+1)!−r!. Once you see that the awkward $r\cdot r!$r⋅r! is secretly a difference, the sum is immediate. The same mindset cracks sums like $\sum \sin r$∑sinr (via a product-to-sum identity) or $\sum \arctan\frac{1}{r^2+r+1}$∑arctanr2+r+11​ (via the $\arctan$arctan subtraction formula): in each case the entire difficulty is rewriting the term as $f(r) - f(r+1)$f(r)−f(r+1), after which telescoping does the rest. Train yourself to ask "is this a disguised difference?" before reaching for any other technique.

Example 4 — when the decomposition must be grouped, not split

Find $\displaystyle\sum_{r=1}^{n} \frac{4}{(2r-1)(2r+1)(2r+3)}$r=1∑n​(2r−1)(2r+1)(2r+3)4​.

A naive three-way split into $\tfrac{A}{2r-1} + \tfrac{B}{2r+1} + \tfrac{C}{2r+3}$2r−1A​+2r+1B​+2r+3C​ does not telescope cleanly — the pieces are not shifted copies of one function. Instead, look for a two-term grouped form $f(r) - f(r+1)$f(r)−f(r+1) with $f(r) = \dfrac{1}{(2r-1)(2r+1)}$f(r)=(2r−1)(2r+1)1​. Test it:

$\frac{1}{(2r-1)(2r+1)} - \frac{1}{(2r+1)(2r+3)} = \frac{(2r+3) - (2r-1)}{(2r-1)(2r+1)(2r+3)} = \frac{4}{(2r-1)(2r+1)(2r+3)}. \quad (\text{M1 find grouped form})$(2r−1)(2r+1)1​−(2r+1)(2r+3)1​=(2r−1)(2r+1)(2r+3)(2r+3)−(2r−1)​=(2r−1)(2r+1)(2r+3)4​.(M1 find grouped form)

So the summand is exactly $f(r) - f(r+1)$f(r)−f(r+1), and telescoping gives

$\sum_{r=1}^{n} \big[f(r) - f(r+1)\big] = f(1) - f(n+1) = \frac{1}{1\cdot 3} - \frac{1}{(2n+1)(2n+3)} = \frac13 - \frac{1}{(2n+1)(2n+3)}. \quad (\text{M1 telescope; A1})$∑r=1n​[f(r)−f(r+1)]=f(1)−f(n+1)=1⋅31​−(2n+1)(2n+3)1​=31​−(2n+1)(2n+3)1​.(M1 telescope; A1)

(M1 for spotting the grouped difference $f(r) - f(r+1)$f(r)−f(r+1) with $f(r) = \tfrac{1}{(2r-1)(2r+1)}$f(r)=(2r−1)(2r+1)1​; M1 telescoping; A1 for the simplified result. As $n\to\infty$n→∞ the sum $\to \tfrac13$→31​. Check at $n=1$n=1: summand $\tfrac{4}{1\cdot3\cdot5} = \tfrac{4}{15}$1⋅3⋅54​=154​; formula $\tfrac13 - \tfrac{1}{3\cdot5} = \tfrac{5}{15} - \tfrac{1}{15} = \tfrac{4}{15}$31​−3⋅51​=155​−151​=154​ ✓. This "group, don't split" trick is the standard escalation for denominators of three or more consecutive factors.)

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Specimen-style exam question

(specimen-style — not from any past paper)

(a) Express $\dfrac{2}{r(r+1)(r+2)}$r(r+1)(r+2)2​ in partial fractions in the form $\dfrac{A}{r(r+1)} - \dfrac{A}{(r+1)(r+2)}$r(r+1)A​−(r+1)(r+2)A​, stating the value of $A$A. (b) Hence use the method of differences to show that $\displaystyle\sum_{r=1}^{n} \frac{2}{r(r+1)(r+2)} = \frac{n(n+3)}{2(n+1)(n+2)}$r=1∑n​r(r+1)(r+2)2​=2(n+1)(n+2)n(n+3)​. (c) Deduce the sum to infinity.

(a) Note $\dfrac{1}{r(r+1)} - \dfrac{1}{(r+1)(r+2)} = \dfrac{(r+2) - r}{r(r+1)(r+2)} = \dfrac{2}{r(r+1)(r+2)}$r(r+1)1​−(r+1)(r+2)1​=r(r+1)(r+2)(r+2)−r​=r(r+1)(r+2)2​, so $A = 1$A=1:

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

(b) This is telescoping with $f(r) = \dfrac{1}{r(r+1)}$f(r)=r(r+1)1​, since the second piece is $f(r+1)$f(r+1):

$\sum_{r=1}^{n}\Big[f(r) - f(r+1)\Big] = f(1) - f(n+1) = \frac{1}{1\cdot 2} - \frac{1}{(n+1)(n+2)} = \frac12 - \frac{1}{(n+1)(n+2)}.$∑r=1n​[f(r)−f(r+1)]=f(1)−f(n+1)=1⋅21​−(n+1)(n+2)1​=21​−(n+1)(n+2)1​.

Combine over the common denominator $2(n+1)(n+2)$2(n+1)(n+2):