Home / A-Level Further Maths 9231 / Summation of series / The method of differences
Cambridge International AS & A Level · Topic 1.3
A-Level Further Maths 9231 / Topic 1.3 / AS & A Level

Write the terms out and watch them cancel.

Some series hide a sum that collapses: each term is the difference of two values of one function, so the end of one term wipes out the start of the next. Split the general term, list the first few and last few rows, and only a handful of pieces survive.

Topic 1.3All summation of series lessons →
The key idea

If the general term can be written as \(u_r = f(r) - f(r + 1)\), then \(\displaystyle\sum_{r=1}^{n} u_r = f(1) - f(n + 1)\): every other value of \(f\) appears once with a plus sign and once with a minus sign. Write the first few and last few rows to see exactly which terms survive. A rational general term usually needs partial fractions first; a gap of two, such as \(f(r) - f(r + 2)\), leaves two terms at each end.

r 1/r −1/(r + 1) 1 1 − 1/2 2 1/2 − 1/3 3 1/3 − 1/4 ⋮ ⋮ ⋮ n − 1 1/(n − 1) − 1/n n 1/n − 1/(n + 1) cancels its partner survives sum = 1 − 1/(n + 1)        = n/(n + 1)
Fig. 1: the rows of 1/(r(r + 1)) = 1/r - 1/(r + 1) for r = 1 to n. The negative part of each row cancels the positive part of the next, so only the first and the last pieces survive
Section 01

Terms that cancel in pairs.

Suppose each term of a series is the difference of two neighbouring values of one function: \(u_r = f(r) - f(r + 1)\). Writing the sum out row by row, the \(-f(2)\) at the end of the first row meets the \(+f(2)\) at the start of the second, and so on down the list. Everything cancels except the very first piece and the very last.

\(r\)\(f(r)\)\(-f(r + 1)\)
\(1\)\(\mathbf{f(1)}\)\(-f(2)\)
\(2\)\(f(2)\)\(-f(3)\)
\(3\)\(f(3)\)\(-f(4)\)
\(\vdots\)\(\vdots\)\(\vdots\)
\(n\)\(f(n)\)\(\mathbf{-f(n + 1)}\)

Adding the rows: \(\displaystyle\sum_{r=1}^{n} \big(f(r) - f(r + 1)\big) = f(1) - f(n + 1)\). In an exam, write at least the first three rows and the last two so the examiner can see the cancellation; the answer alone does not earn the method marks.

Worked example The series in Fig. 1
Given that \(\dfrac{1}{r(r + 1)} = \dfrac{1}{r} - \dfrac{1}{r + 1}\), find \(\displaystyle\sum_{r=1}^{n} \dfrac{1}{r(r + 1)}\) in terms of \(n\).
  1. Here \(f(r) = \dfrac{1}{r}\), so each term is \(f(r) - f(r + 1)\).
  2. Rows: \(\left(1 - \tfrac{1}{2}\right) + \left(\tfrac{1}{2} - \tfrac{1}{3}\right) + \left(\tfrac{1}{3} - \tfrac{1}{4}\right) + \dots + \left(\tfrac{1}{n} - \tfrac{1}{n + 1}\right)\).
  3. Every fraction from \(\tfrac{1}{2}\) to \(\tfrac{1}{n}\) appears once with each sign and cancels.
  4. What is left: \(1 - \dfrac{1}{n + 1} = \dfrac{n + 1 - 1}{n + 1}\).
Answer: \(\dfrac{n}{n + 1}\) (check \(n = 1\): \(\dfrac{1}{2}\), and indeed \(\dfrac{1}{1 \times 2} = \dfrac{1}{2}\))
Worked example Show the identity first
Show that \(\dfrac{1}{r^{2}} - \dfrac{1}{(r + 1)^{2}} = \dfrac{2r + 1}{r^{2}(r + 1)^{2}}\), and hence find \(\displaystyle\sum_{r=1}^{n} \dfrac{2r + 1}{r^{2}(r + 1)^{2}}\).
  1. Common denominator: \(\dfrac{(r + 1)^{2} - r^{2}}{r^{2}(r + 1)^{2}} = \dfrac{r^{2} + 2r + 1 - r^{2}}{r^{2}(r + 1)^{2}} = \dfrac{2r + 1}{r^{2}(r + 1)^{2}}\).
  2. So the general term is \(f(r) - f(r + 1)\) with \(f(r) = \dfrac{1}{r^{2}}\).
  3. Rows: \(\left(1 - \tfrac{1}{4}\right) + \left(\tfrac{1}{4} - \tfrac{1}{9}\right) + \dots + \left(\tfrac{1}{n^{2}} - \tfrac{1}{(n + 1)^{2}}\right)\).
  4. The sum is \(f(1) - f(n + 1)\).
Answer: \(1 - \dfrac{1}{(n + 1)^{2}}\)
Your turn
Find \(\displaystyle\sum_{r=1}^{n} \left(\dfrac{1}{2r + 1} - \dfrac{1}{2r + 3}\right)\).
Examiner trap

The last surviving term is \(f(n + 1)\), not \(f(n)\): the final row is \(f(n) - f(n + 1)\), and its \(f(n)\) cancels with the row above. If the sum starts at \(r = m\) instead of \(r = 1\), the first survivor is \(f(m)\), not \(f(1)\). Writing the end rows out protects you from both slips.

Section 02

Partial fractions first.

Most exam series give the general term as a single fraction such as \(\dfrac{1}{(2r - 1)(2r + 1)}\). Split it into partial fractions; the pieces are usually values of one function at neighbouring points, and the method of differences takes over. Sometimes the gap is two steps, not one, and then two terms survive at each end.

For \(\dfrac{1}{r(r + 2)} = \dfrac{1}{2}\left(\dfrac{1}{r} - \dfrac{1}{r + 2}\right)\), each negative piece cancels the positive piece two rows further down:

\(r\)\(\tfrac{1}{r}\)\(-\tfrac{1}{r + 2}\)
\(1\)\(\mathbf{1}\)\(-\tfrac{1}{3}\)
\(2\)\(\mathbf{\tfrac{1}{2}}\)\(-\tfrac{1}{4}\)
\(3\)\(\tfrac{1}{3}\)\(-\tfrac{1}{5}\)
\(\vdots\)\(\vdots\)\(\vdots\)
\(n - 1\)\(\tfrac{1}{n - 1}\)\(\mathbf{-\tfrac{1}{n + 1}}\)
\(n\)\(\tfrac{1}{n}\)\(\mathbf{-\tfrac{1}{n + 2}}\)

The survivors (in bold) are the first two positive pieces and the last two negative pieces.

Worked example Odd numbers in the denominator
Find \(\displaystyle\sum_{r=1}^{n} \dfrac{1}{(2r - 1)(2r + 1)}\) in terms of \(n\).
  1. Partial fractions: \(\dfrac{1}{(2r - 1)(2r + 1)} = \dfrac{A}{2r - 1} + \dfrac{B}{2r + 1}\), so \(1 = A(2r + 1) + B(2r - 1)\).
  2. \(r = \tfrac{1}{2}\): \(1 = 2A\), so \(A = \tfrac{1}{2}\). \(r = -\tfrac{1}{2}\): \(1 = -2B\), so \(B = -\tfrac{1}{2}\).
  3. General term \(\dfrac{1}{2}\left(\dfrac{1}{2r - 1} - \dfrac{1}{2r + 1}\right)\); rows: \(\dfrac{1}{2}\left[\left(1 - \tfrac{1}{3}\right) + \left(\tfrac{1}{3} - \tfrac{1}{5}\right) + \dots + \left(\tfrac{1}{2n - 1} - \tfrac{1}{2n + 1}\right)\right]\).
  4. Survivors: \(\dfrac{1}{2}\left(1 - \dfrac{1}{2n + 1}\right) = \dfrac{1}{2} \times \dfrac{2n}{2n + 1}\).
Answer: \(\dfrac{n}{2n + 1}\)
Worked example A gap of two
Find \(\displaystyle\sum_{r=1}^{n} \dfrac{1}{r(r + 2)}\), giving your answer as a single fraction.
  1. \(1 = A(r + 2) + Br\): \(r = 0\) gives \(A = \tfrac{1}{2}\); \(r = -2\) gives \(B = -\tfrac{1}{2}\).
  2. So the sum is \(\dfrac{1}{2}\displaystyle\sum_{r=1}^{n}\left(\dfrac{1}{r} - \dfrac{1}{r + 2}\right)\), with the rows in the table above.
  3. Survivors: \(\dfrac{1}{2}\left(1 + \dfrac{1}{2} - \dfrac{1}{n + 1} - \dfrac{1}{n + 2}\right) = \dfrac{3}{4} - \dfrac{1}{2}\left(\dfrac{(n + 2) + (n + 1)}{(n + 1)(n + 2)}\right)\).
  4. \(= \dfrac{3}{4} - \dfrac{2n + 3}{2(n + 1)(n + 2)}\).
Answer: \(\dfrac{3}{4} - \dfrac{2n + 3}{2(n + 1)(n + 2)}\) (check \(n = 1\): \(\dfrac{3}{4} - \dfrac{5}{12} = \dfrac{1}{3}\), which is \(\dfrac{1}{1 \times 3}\))
Your turn
Show that \(\dfrac{1}{r(r + 1)(r + 2)} = \dfrac{1}{2r(r + 1)} - \dfrac{1}{2(r + 1)(r + 2)}\), and hence find \(\displaystyle\sum_{r=1}^{n} \dfrac{1}{r(r + 1)(r + 2)}\).
Examiner trap

Do not drop the factor \(\dfrac{1}{2}\) from the partial fractions: \(\dfrac{1}{r} - \dfrac{1}{r + 2} = \dfrac{2}{r(r + 2)}\), not \(\dfrac{1}{r(r + 2)}\). Check your split by putting \(r = 1\) into both sides. With a gap of two, list at least three rows at the start and two at the end, or you will miss the second survivor at each end.

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

What is \(\displaystyle\sum_{r=1}^{n} \big(f(r) - f(r + 1)\big)\)?

Quick check+10 XP

Which is the correct partial fraction form of \(\dfrac{1}{(2r - 1)(2r + 1)}\)?

Quick check+10 XP

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

Quick check+10 XP

In \(\displaystyle\sum_{r=1}^{n}\left(\dfrac{1}{r} - \dfrac{1}{r + 2}\right)\), which terms survive the cancellation?

Section 03

Other shapes of difference.

The difference need not be \(f(r) - f(r + 1)\). It may run the other way, \(f(r + 1) - f(r)\) or \(f(r) - f(r - 1)\), or skip a step, \(f(r + 1) - f(r - 1)\). The rule never changes: write the rows and see what is left. A question may also hand you an identity, such as \((r + 1)^{3} - r^{3} = 3r^{2} + 3r + 1\), and ask you to use differences to prove a standard result.

Worked example Proving the sum of squares
Use the identity \((r + 1)^{3} - r^{3} = 3r^{2} + 3r + 1\) and the result \(\displaystyle\sum_{r=1}^{n} r = \tfrac{1}{2}n(n + 1)\) to show that \(\displaystyle\sum_{r=1}^{n} r^{2} = \tfrac{1}{6}n(n + 1)(2n + 1)\).
  1. Sum the left side from \(r = 1\) to \(n\): rows \((2^{3} - 1^{3}) + (3^{3} - 2^{3}) + \dots + ((n + 1)^{3} - n^{3}) = (n + 1)^{3} - 1\).
  2. Sum the right side: \(3\displaystyle\sum r^{2} + 3\sum r + \sum 1 = 3S + \tfrac{3}{2}n(n + 1) + n\), where \(S = \displaystyle\sum_{r=1}^{n} r^{2}\) and \(\displaystyle\sum_{r=1}^{n} 1 = n\).
  3. So \(3S = (n + 1)^{3} - 1 - n - \tfrac{3}{2}n(n + 1) = (n + 1)^{3} - (n + 1) - \tfrac{3}{2}n(n + 1)\).
  4. Take out \((n + 1)\): \(3S = (n + 1)\left[(n + 1)^{2} - 1 - \tfrac{3}{2}n\right] = (n + 1)\left(n^{2} + \tfrac{1}{2}n\right) = \tfrac{1}{2}n(n + 1)(2n + 1)\).
Answer: \(S = \tfrac{1}{6}n(n + 1)(2n + 1)\), as required
Worked example A difference that skips a step
Show that \(\dfrac{1}{(r - 1)^{2}} - \dfrac{1}{(r + 1)^{2}} = \dfrac{4r}{(r^{2} - 1)^{2}}\), and hence find \(\displaystyle\sum_{r=2}^{n} \dfrac{r}{(r^{2} - 1)^{2}}\).
  1. \(\dfrac{(r + 1)^{2} - (r - 1)^{2}}{(r - 1)^{2}(r + 1)^{2}} = \dfrac{4r}{\big((r - 1)(r + 1)\big)^{2}} = \dfrac{4r}{(r^{2} - 1)^{2}}\).
  2. So \(\dfrac{r}{(r^{2} - 1)^{2}} = \dfrac{1}{4}\big(f(r - 1) - f(r + 1)\big)\) with \(f(r) = \dfrac{1}{r^{2}}\).
  3. Rows from \(r = 2\): \(\left(1 - \tfrac{1}{9}\right) + \left(\tfrac{1}{4} - \tfrac{1}{16}\right) + \left(\tfrac{1}{9} - \tfrac{1}{25}\right) + \dots + \left(\tfrac{1}{(n - 2)^{2}} - \tfrac{1}{n^{2}}\right) + \left(\tfrac{1}{(n - 1)^{2}} - \tfrac{1}{(n + 1)^{2}}\right)\).
  4. Survivors: \(f(1) + f(2) - f(n) - f(n + 1) = 1 + \dfrac{1}{4} - \dfrac{1}{n^{2}} - \dfrac{1}{(n + 1)^{2}}\); multiply by \(\dfrac{1}{4}\).
Answer: \(\dfrac{5}{16} - \dfrac{1}{4}\left(\dfrac{1}{n^{2}} + \dfrac{1}{(n + 1)^{2}}\right)\) (check \(n = 2\): \(\dfrac{5}{16} - \dfrac{1}{4}\left(\dfrac{1}{4} + \dfrac{1}{9}\right) = \dfrac{2}{9} = \dfrac{2}{3^{2}}\))
Your turn
Let \(f(r) = \tfrac{1}{3}r(r + 1)(r + 2)\). Show that \(f(r) - f(r - 1) = r(r + 1)\), and hence find \(\displaystyle\sum_{r=1}^{n} r(r + 1)\).
Examiner trap

When you sum an identity over \(r = 1\) to \(n\), a constant term sums to \(n\), not to itself: \(\displaystyle\sum_{r=1}^{n} 1 = n\). Forgetting this in the \(3r^{2} + 3r + 1\) proof leaves you with a wrong 'standard result'. Test your final formula with \(n = 1\) and \(n = 2\).

Stage 2 · Exam

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Find \(\displaystyle\sum_{r=1}^{n} \dfrac{1}{(r + 1)(r + 2)}\) in terms of \(n\).

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

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Given that \(\dfrac{2r + 1}{r^{2}(r + 1)^{2}} = \dfrac{1}{r^{2}} - \dfrac{1}{(r + 1)^{2}}\), find \(\displaystyle\sum_{r=n+1}^{2n} \dfrac{2r + 1}{r^{2}(r + 1)^{2}}\).

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Find \(\displaystyle\sum_{r=1}^{n} \big(\sqrt{r + 1} - \sqrt{r - 1}\big)\).

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Series split, rows cancelled, survivors collected.

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Six original Cambridge-style questions
Telescoping sums from a given split, partial fractions with gaps of one and two, a three-factor denominator, a sum between two variable limits, and the sum of cubes from a given identity.