4.06b - The method of differences | OCR A-Level Further Mathematics A

Add a series by exposing terms that cancel. You will use differences and partial fractions to find finite sums, keep the correct terms when the cancellation spans several rows, and take limits to find infinite sums and the error after a finite number of terms. When only part of a series cancels, you will retain the other part and use a supplied sum if one is given.

Cancellation between consecutive terms

Can infinitely many positive fractions add up to exactly one half?

\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+\cdots

The first three terms total $3/7$ , already close to $1/2$ . Each new term adds something, so no finite sum can equal one half. Yet one half can still be the limit. The route to that limit starts with a finite sum and a cancellation pattern you can see.

You already know

Partial fractions replace a rational expression by simpler fractions with the same value. A sum with limits $r = m$ to $r = n$ includes both endpoints. Its number of terms is $n-m+1$ , not $n-m$ .

Take the simpler expression $1/r-1/(r+1)$ . At $r = 1, 2, 3$ , its terms are $1-1/2$ , $1/2-1/3$ and $1/3-1/4$ . Adding them leaves $1-1/4$ . Nothing has been approximated: each middle fraction occurs once positively and once negatively.

Method of differences

The method of differences finds a sum by rewriting its terms so that intermediate contributions cancel, leaving boundary terms. This cancellation is also called telescoping. For example, $(1-1/2)+(1/2-1/3)$ telescopes to $1-1/3$ . A list of positive terms alone does not telescope until a suitable difference has been exposed.

The cancellation depends on matching expressions, including their signs. A term $-f(r+1)$ in one row is cancelled by $+f(r+1)$ in the next.

Consecutive differences

For integers $m \le n$ , provided every displayed value of $f$ is defined,

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

Not in the formula booklet: learn it.

Rather than quote this as a shortcut when asked to use the method, show enough terms to make the cancellation clear. The first two rows and the last row usually reveal both boundaries.

Worked example 1: exposing the boundaries

Question. Use the method of differences to show that

\sum_{r=1}^{n}\left(\frac1r-\frac1{r+1}\right) = 1-\frac1{n+1}

Plan. Write the first two terms and the last term, then identify the unmatched positive and negative fractions.

For $n \ge 3$ ,

\begin{aligned} S_n &= \left(1-\frac12\right)+\left(\frac12-\frac13\right)+\cdots +\left(\frac1n-\frac1{n+1}\right) && \text{expand the sum} \\ &= 1-\frac1{n+1} && \text{cancel equal opposite fractions} \end{aligned}

Why this step? The negative fraction from each row cancels the positive fraction from the next. The first positive fraction has no earlier partner, and the last negative fraction has no later partner.

Answer. $S_n = 1-1/(n+1)$ , as required. At $n = 1$ and $n = 2$ , direct addition gives the same formula, so it holds for every positive integer $n$ .

Check. At $n = 3$ , $(1-1/2)+(1/2-1/3)+(1/3-1/4) = 3/4 = 1-1/4$ .

Worked example 2 changes only the lower limit from $1$ to $3$ . The cancellation is identical, but the first surviving positive term changes.

Worked example 2: starting later

Question. Use the method of differences to find

\sum_{r=3}^{n}\left(\frac1r-\frac1{r+1}\right), \qquad n \ge 3

Plan. Start the expansion at $r = 3$ , keep the final row at $r = n$ , and cancel the fractions between them.

For $n \ge 5$ ,

\begin{aligned} S &= \left(\frac13-\frac14\right)+\left(\frac14-\frac15\right)+\cdots +\left(\frac1n-\frac1{n+1}\right) && \text{use the stated limits} \\ &= \frac13-\frac1{n+1} && \text{retain the two boundaries} \end{aligned}

Why this step? The first index is $3$ , so $1/3$ survives. The negative fraction in the last term still has denominator $n+1$ , even though the sum ends at index $n$ .

Answer. $\dfrac13-\dfrac1{n+1}$ . Directly checking $n = 3$ and $n = 4$ gives the same expression, so it is valid throughout the stated range.

Check. At $n = 4$ , $(1/3-1/4)+(1/4-1/5) = 1/3-1/5 = 2/15$ .

For short sums, write every term. Ellipses describe missing intermediate terms; they should not create extra copies of an endpoint when there are only one or two rows.

Watch out

Wrong: $\sum_{r=1}^{n}(1/r-1/(r+1)) = 1-1/n$ . That drops the final negative fraction too soon. Right: the row with $r = n$ is $1/n-1/(n+1)$ , so the surviving negative term is $-1/(n+1)$ .

Use the first and last actual rows to check a changed range before you simplify.

Differences from partial fractions

A rational summand often hides the difference that makes the whole sum collapse.

For instance, $1/[r(r+1)]$ has partial fractions $1/r-1/(r+1)$ . These fractions are larger than the original term, but their difference is exactly equal to it. That is why adding positive original terms can become a calculation with subtraction.

You already know

To find $A$ and $B$ in $P(r)/[(ar+b)(cr+d)] = A/(ar+b)+B/(cr+d)$ , multiply by the denominator, then substitute convenient values or compare coefficients. Check the identity by combining the fractions again.

Choose partial fractions first, then inspect whether a fraction at index $r$ reappears with the opposite sign at the next index. Partial fractions alone do not guarantee cancellation; the denominators must line up.

Worked example 3: consecutive integer factors

Question. Use partial fractions and the method of differences to find

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

Plan. Decompose one term, show its cancellation across consecutive rows, and simplify the boundary terms.

\begin{aligned} \frac{2}{r(r+1)} &= \frac{A}{r}+\frac{B}{r+1} && \text{choose the partial fractions} \\ 2 &= A(r+1)+Br && \text{multiply by } r(r+1) \\ A = 2, \quad B &= -2 && \text{put } r = 0,-1 \end{aligned}

For $n \ge 3$ ,

\begin{aligned} S_n &= 2\left[\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\cdots +\left(\frac1n-\frac1{n+1}\right)\right] && \text{substitute the identity} \\ &= 2\left(1-\frac1{n+1}\right) && \text{cancel the middle fractions} \\ &= \frac{2n}{n+1} && \text{combine the boundary terms} \end{aligned}

Why this step? Partial fractions turn each product in a denominator into a difference of consecutive reciprocals. Those reciprocals, rather than the original fractions, have matching neighbours.

Answer. $S_n = 2n/(n+1)$ for positive integers $n$ ; direct addition confirms the same formula for $n = 1, 2$ .

Check. At $n = 2$ , $2/(1\times2)+2/(2\times3) = 1+1/3 = 4/3$ , matching $2(2)/3$ .

Worked example 4 changes the factor progression from consecutive integers to consecutive odd integers. The gap between the factors is now $2$ , so the partial-fraction coefficients change; the denominators still match between neighbouring rows.

Worked example 4: consecutive odd factors

Question. Use the method of differences to determine

T_n = \sum_{r=1}^{n}\frac{2}{(2r-1)(2r+1)}

Plan. Find the two partial fractions, then check that $2r+1$ in one row equals $2(r+1)-1$ in the next.

\begin{aligned} \frac{2}{(2r-1)(2r+1)} &= \frac{A}{2r-1}+\frac{B}{2r+1} && \text{choose the partial fractions} \\ 2 &= A(2r+1)+B(2r-1) && \text{clear the denominator} \\ A+B = 0, \quad A-B &= 2 && \text{compare coefficients} \\ A = 1, \quad B &= -1 && \text{solve} \end{aligned}

For $n \ge 3$ ,

\begin{aligned} T_n &= \left(1-\frac13\right)+\left(\frac13-\frac15\right)+\cdots +\left(\frac1{2n-1}-\frac1{2n+1}\right) && \text{expand the differences} \\ &= 1-\frac1{2n+1} = \frac{2n}{2n+1} && \text{cancel and simplify} \end{aligned}

Why this step? The factors differ by $2$ , but their values also increase by $2$ when $r$ increases by one. Cancellation therefore still takes place between consecutive rows.

Answer. $T_n = 2n/(2n+1)$ , also valid at $n = 1, 2$ by direct addition.

Check. At $n = 2$ , $2/3+2/15 = 4/5$ , matching $2(2)/5$ .

The decompositions and the resulting special sums are not in the formula booklet: derive them when needed. A quick recombination catches a missing scale factor before it spreads through the sum.

Watch out

Wrong: $1/[(2r-1)(2r+1)] = 1/(2r-1)-1/(2r+1)$ . The right side has numerator $2$ , not $1$ . Right: include the factor $1/2$ , giving $\frac12(1/(2r-1)-1/(2r+1))$ .

In the next sum the factor progression advances by $3$ . Find the coefficients before deciding which terms cancel.

Cancellation over several rows

When the index shift is larger than one, several terms survive at each end of the sum.

Consider $1/r-1/(r+2)$ . The negative $-1/3$ from the first row cancels $+1/3$ from the third row, not the second. The first two positive fractions, $1$ and $1/2$ , survive; so do the last two negative fractions.

[DIAGRAM: asset_name: Two rows between matching fractions; asset_slug: h245_4_06_two_step_cancellation; caption: For four rows, the fractions with denominators 3 and 4 cancel, leaving the first two positive and last two negative fractions; description: Four rows labelled r = 1 to r = 4 show 1/r minus 1/(r+2), with equal denominators aligned in six columns. The rows are 1 minus 1/3, 1/2 minus 1/4, 1/3 minus 1/5 and 1/4 minus 1/6. Sage lines connect equal and opposite fractions in columns 3 and 4. Blue highlights identify surviving 1 and 1/2, and clay highlights identify surviving minus 1/5 and minus 1/6. A final line gives 1 + 1/2 - 1/5 - 1/6.]
$For four rows, the fractions with denominators 3 and 4 cancel, leaving the first two positive and last two negative fractions$

Count the shift before you cancel. For a positive integer shift $h$ , matching $-f(r+h)$ requires the row whose index is $h$ later.

Differences with an index gap

For positive integers $h$ and $n$ , with all terms defined,

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

When the boundary blocks overlap, equal terms within this expression also cancel.
Not in the formula booklet: learn it.

This formula follows by writing the second sum with index values $h+1$ to $n+h$ . The common index range disappears, leaving the leading and trailing blocks. In a worked solution, display those blocks so that the number of surviving terms is visible.

Worked example 5: a gap of two

Question. Use the method of differences to find

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

Plan. Write each term as half a difference of reciprocals, then retain two positive and two negative boundary terms.

\frac{1}{r(r+2)} = \frac12\left(\frac1r-\frac1{r+2}\right) \qquad \text{combine to check the numerator}

For $n \ge 4$ ,

\begin{aligned} S_n &= \frac12\left[\left(1-\frac13\right)+\left(\frac12-\frac14\right)+\cdots\right. \\ &\qquad\left.+\left(\frac1{n-1}-\frac1{n+1}\right)+\left(\frac1n-\frac1{n+2}\right)\right] && \text{show both trailing terms} \\ &= \frac12\left(1+\frac12-\frac1{n+1}-\frac1{n+2}\right) && \text{retain both boundaries} \\ &= \frac34-\frac1{2(n+1)}-\frac1{2(n+2)} && \text{simplify} \end{aligned}

Why this step? With a shift of two, the final negative fractions have denominators $n+1$ and $n+2$ . Neither has a matching positive term inside the sum.

Answer. $S_n = \dfrac34-\dfrac1{2(n+1)}-\dfrac1{2(n+2)}$ . Direct substitution verifies $n = 1, 2, 3$ as well.

Check. At $n = 2$ , $1/3+1/8 = 11/24$ , and $3/4-1/6-1/8 = 11/24$ .

Worked example 6 changes only the factor gap from $2$ to $3$ . Three boundary terms now survive at each end, and the partial-fraction scale becomes $1/3$ .

Worked example 6: a gap of three

Question. Use the method of differences to find

U_n = \sum_{r=1}^{n}\frac{1}{r(r+3)}

Plan. Decompose with scale $1/3$ , write enough rows to expose the three-term boundary blocks, and simplify their constant part.

\frac{1}{r(r+3)} = \frac13\left(\frac1r-\frac1{r+3}\right) \qquad \text{the difference has numerator } 3

For $n \ge 6$ ,

\begin{aligned} U_n &= \frac13\left[\left(1-\frac14\right)+\left(\frac12-\frac15\right) +\left(\frac13-\frac16\right)+\cdots\right. \\ &\qquad\left.+\left(\frac1{n-2}-\frac1{n+1}\right) +\left(\frac1{n-1}-\frac1{n+2}\right) +\left(\frac1n-\frac1{n+3}\right)\right] && \text{expose the boundary rows} \\ &= \frac13\left(1+\frac12+\frac13-\frac1{n+1}-\frac1{n+2}-\frac1{n+3}\right) && \text{cancel equal opposites} \\ &= \frac{11}{18}-\frac13\left(\frac1{n+1}+\frac1{n+2}+\frac1{n+3}\right) && 1+\frac12+\frac13 = \frac{11}{6} \end{aligned}

Why this step? The first three negative denominators are $4, 5, 6$ , so none can cancel the leading positive fractions with denominators $1, 2, 3$ . The same unmatched width appears at the other end.

Answer. $U_n = \dfrac{11}{18}-\dfrac13\left(\dfrac1{n+1}+\dfrac1{n+2}+\dfrac1{n+3}\right)$ . The boundary-block formula also applies for $n < 6$ ; overlapping fractions cancel within it.

Check. At $n = 1$ , the expression is $11/18-(1/3)(1/2+1/3+1/4) = 1/4$ , the first term.

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4.06b - The method of differences

Cancellation between consecutive terms

Differences from partial fractions

Cancellation over several rows

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