1.3 - The method of differences

1.3 - The method of differences

Some sums become simpler when you split each term rather than combine it. You will use the method of differences in 1.3 to expose cancellation, prepare terms using partial fractions, and retain the correct boundaries in finite sums, including sums with powers.

Seeing cancellation

Can (1−12)+(12−13)+⋯+(1100−1101)\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\cdots+\left(\frac1{100}-\frac1{101}\right) be evaluated without a common denominator? Every internal fraction occurs once positively and once negatively, leaving only 1−1/1011-1/101.

Write out a short case first:

∑r=14(1r−1r+1)=(1−12)+(12−13)+(13−14)+(14−15)=1−15.\begin{aligned} \sum_{r=1}^{4}\left(\frac1r-\frac1{r+1}\right) &=\left(1-\frac12\right)+\left(\frac12-\frac13\right) +\left(\frac13-\frac14\right)+\left(\frac14-\frac15\right) \\ &=1-\frac15. \end{aligned}

The cancellation works because a negative term from one bracket matches a positive term from another. It is a property of the whole sum; you are not cancelling terms across subtraction inside a fraction.

Method of differences

The method of differences, also called telescoping, finds a sum by writing its terms as differences whose internal contributions cancel. If the summand is f(r)−f(r+1)f(r)-f(r+1), then

∑r=ab(f(r)−f(r+1))=f(a)−f(b+1).\sum_{r=a}^{b}\bigl(f(r)-f(r+1)\bigr)=f(a)-f(b+1).

The first positive contribution and the last negative contribution remain.

Writing the first few and last few terms shows which boundaries survive. The gap between the two indices determines how many terms are left at each end.

Worked example 1: a gap of one

Question. Find ∑r=1n(1r−1r+1)\displaystyle\sum_{r=1}^{n}\left(\frac1r-\frac1{r+1}\right).

Plan. Display the cancellation and keep the first positive and last negative terms.

For n≥3n\geq3,

Sn=(1−12)+(12−13)+⋯+(1n−1n+1)=1−1n+1.\begin{aligned} S_n &=\left(1-\frac12\right)+\left(\frac12-\frac13\right) +\cdots+\left(\frac1n-\frac1{n+1}\right) \\ &=1-\frac1{n+1}. \end{aligned}

Every fraction from 1/21/2 to 1/n1/n appears with both signs.

Answer. Sn=1−1n+1=nn+1\displaystyle S_n=1-\frac1{n+1}=\frac n{n+1}, also valid for n=1,2n=1,2.

Check. The answer at n=1n=1 is 1/21/2, the first term. Also Sn−Sn−1=1/n−1/(n+1)S_n-S_{n-1}=1/n-1/(n+1) for n≥2n\geq2, exactly the term added next.

Change only the second denominator from r+1r+1 to r+2r+2. A negative term now cancels two brackets later, so two contributions survive at each end.

Worked example 2: a gap of two

Question. Find ∑r=1n(1r−1r+2)\displaystyle\sum_{r=1}^{n}\left(\frac1r-\frac1{r+2}\right).

Plan. Write the first two positive terms and the last two negative terms explicitly.

For n≥5n\geq5,

Sn=(1−13)+(12−14)+(13−15)+⋯+(1n−1−1n+1)+(1n−1n+2)=1+12−1n+1−1n+2.\begin{aligned} S_n &=\left(1-\frac13\right)+\left(\frac12-\frac14\right) +\left(\frac13-\frac15\right)+\cdots \\ &\qquad+\left(\frac1{n-1}-\frac1{n+1}\right) +\left(\frac1n-\frac1{n+2}\right) \\ &=1+\frac12-\frac1{n+1}-\frac1{n+2}. \end{aligned}

The positive and negative lists overlap from 1/31/3 to 1/n1/n; those are the cancelling terms.

Answer. 32−1n+1−1n+2\displaystyle\frac32-\frac1{n+1}-\frac1{n+2}, valid for every positive integer nn.

Check. At n=2n=2, direct addition gives (1−1/3)+(1/2−1/4)=11/12(1-1/3)+(1/2-1/4)=11/12. The expression gives 3/2−1/3−1/4=11/123/2-1/3-1/4=11/12. At n=1n=1, its apparent extra boundary terms cancel, giving 2/32/3 as required.

The opening hundred-term sum is the first pattern, so it equals 100/101100/101. The second pattern shows why a gap of two cannot be reduced to just the first and last fractions.

Watch out

For f(r)−f(r+2)f(r)-f(r+2), retaining only f(1)−f(n+2)f(1)-f(n+2) loses two uncancelled terms. Compare the full positive and negative index ranges before simplifying.

Keep the gap of two but move both denominator sequences one place forward.

Preparing a general term

Cancellation may be hidden in a single rational term, and partial fractions can reveal it.

For example, to split 1/[r(r+1)]1/[r(r+1)], write

1r(r+1)=Ar+Br+1.\frac1{r(r+1)}=\frac Ar+\frac B{r+1}.

Multiplying by r(r+1)r(r+1) gives the identity 1=A(r+1)+Br1=A(r+1)+Br. Substituting r=0r=0 gives A=1A=1, and substituting r=−1r=-1 gives B=−1B=-1. These substitutions find coefficients in a polynomial identity; they do not add zero denominators to the original series, whose indices are positive integers.

Thus 1/[r(r+1)]=1/r−1/(r+1)1/[r(r+1)]=1/r-1/(r+1). The decomposition is useful because the second denominator at one index equals the first denominator at the next index.

Worked example 3: consecutive integer factors

Question. Use partial fractions and the method of differences to find ∑r=1n1r(r+1)\displaystyle\sum_{r=1}^{n}\frac1{r(r+1)}.

Plan. Decompose the term, then write its first and last differences.

1r(r+1)=1r−1r+1,\frac1{r(r+1)}=\frac1r-\frac1{r+1},

so

∑r=1n1r(r+1)=(1−12)+(12−13)+⋯+(1n−1n+1)=1−1n+1.\begin{aligned} \sum_{r=1}^{n}\frac1{r(r+1)} &=\left(1-\frac12\right)+\left(\frac12-\frac13\right) +\cdots+\left(\frac1n-\frac1{n+1}\right) \\ &=1-\frac1{n+1}. \end{aligned}

Answer. nn+1\displaystyle\frac n{n+1}.

Check. At n=3n=3, the original fractions add to 1/2+1/6+1/12=3/41/2+1/6+1/12=3/4, which agrees with n/(n+1)n/(n+1).

Now use consecutive odd factors instead. Their spacing introduces a factor of 1/21/2, while cancellation still happens between neighbouring indices.

Worked example 4: consecutive odd factors

Question. Find ∑r=1n1(2r−1)(2r+1)\displaystyle\sum_{r=1}^{n}\frac1{(2r-1)(2r+1)}.

Plan. Find the partial fractions and follow the odd denominator sequence.

Writing 1=A(2r+1)+B(2r−1)1=A(2r+1)+B(2r-1) gives A+B=0A+B=0 and A−B=1A-B=1. Therefore

1(2r−1)(2r+1)=12(12r−1−12r+1).\frac1{(2r-1)(2r+1)} =\frac12\left(\frac1{2r-1}-\frac1{2r+1}\right).

Consequently

Sn=12[(1−13)+(13−15)+⋯+(12n−1−12n+1)]=12(1−12n+1).\begin{aligned} S_n &=\frac12\left[\left(1-\frac13\right)+\left(\frac13-\frac15\right) +\cdots+\left(\frac1{2n-1}-\frac1{2n+1}\right)\right] \\ &=\frac12\left(1-\frac1{2n+1}\right). \end{aligned}

Answer. Sn=n2n+1\displaystyle S_n=\frac n{2n+1}.

Check. Subtract successive proposed sums:

n2n+1−n−12n−1=1(2n−1)(2n+1).\frac n{2n+1}-\frac{n-1}{2n-1} =\frac1{(2n-1)(2n+1)}.

Together with S1=1/3S_1=1/3, this reproduces every term of the original sum.

Combining the fractions again is a quick way to catch a missing multiplier before summing.

Watch out

12r−1−12r+1=2(2r−1)(2r+1)\frac1{2r-1}-\frac1{2r+1}=\frac2{(2r-1)(2r+1)}. Omitting the factor 1/21/2 doubles the required sum.

The same method works when the denominator sequence increases by 33.

Finding a finite sum

A geometric factor can help terms cancel, provided you match the whole term, including its power.

For example, 1/(r2r)1/(r2^r) and 1/((r+1)2r+1)1/((r+1)2^{r+1}) are successive members of the same sequence. Their difference is

1r2r−1(r+1)2r+1=2(r+1)−rr(r+1)2r+1=r+2r(r+1)2r+1.\begin{aligned} \frac1{r2^r}-\frac1{(r+1)2^{r+1}} &=\frac{2(r+1)-r}{r(r+1)2^{r+1}} \\ &=\frac{r+2}{r(r+1)2^{r+1}}. \end{aligned}

The numerator r+2r+2 is what makes this weighted term telescope. Splitting the rational factor without tracking the powers would hide the matching terms.

Problem-solving

After decomposing a term, substitute two or three consecutive indices. Check that the negative contribution really is the next positive contribution, including every multiplier and power.

This gives a finite sum in which only two complete boundary terms remain.

Worked example 5: matching powers of two

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

Plan. Use the difference just established and retain the first positive and last negative terms.

Sn=∑r=1n(1r2r−1(r+1)2r+1)=(12−18)+(18−124)+⋯+(1n2n−1(n+1)2n+1)=12−1(n+1)2n+1.\begin{aligned} S_n &=\sum_{r=1}^{n}\left(\frac1{r2^r}-\frac1{(r+1)2^{r+1}}\right) \\ &=\left(\frac12-\frac18\right)+\left(\frac18-\frac1{24}\right) +\cdots \\ &\qquad+\left(\frac1{n2^n}-\frac1{(n+1)2^{n+1}}\right) \\ &=\frac12-\frac1{(n+1)2^{n+1}}. \end{aligned}

The last power is 2n+12^{n+1}, because it comes from the negative contribution at r=nr=n.

Answer. Sn=12−1(n+1)2n+1\displaystyle S_n=\frac12-\frac1{(n+1)2^{n+1}}.

Check. For n=2n=2, the original terms give 3/8+1/12=11/243/8+1/12=11/24. The expression gives 1/2−1/24=11/241/2-1/24=11/24.

Change the geometric base from 22 to 33. The matching difference now produces numerator 3(r+1)−r=2r+33(r+1)-r=2r+3; the boundaries must change with the base.

Worked example 6: matching powers of three

Question. Find ∑r=1n2r+3r(r+1)3r+1\displaystyle\sum_{r=1}^{n}\frac{2r+3}{r(r+1)3^{r+1}}.

Plan. Identify consecutive terms of 1/(r3r)1/(r3^r), then display their cancellation.

2r+3r(r+1)3r+1=1r3r−1(r+1)3r+1,\frac{2r+3}{r(r+1)3^{r+1}} =\frac1{r3^r}-\frac1{(r+1)3^{r+1}},

because bringing the right side over a common denominator gives numerator 3(r+1)−r3(r+1)-r. Hence

Sn=(13−118)+(118−181)+⋯+(1n3n−1(n+1)3n+1)=13−1(n+1)3n+1.\begin{aligned} S_n &=\left(\frac13-\frac1{18}\right)+\left(\frac1{18}-\frac1{81}\right) +\cdots \\ &\qquad+\left(\frac1{n3^n}-\frac1{(n+1)3^{n+1}}\right) \\ &=\frac13-\frac1{(n+1)3^{n+1}}. \end{aligned}

Answer. Sn=13−1(n+1)3n+1\displaystyle S_n=\frac13-\frac1{(n+1)3^{n+1}}.

Check. The proposed expression has first value 5/185/18. Its successive difference is 1/(n3n)−1/((n+1)3n+1)1/(n3^n)-1/((n+1)3^{n+1}), which recombines to the original general term. Thus both the starting value and every increment agree.

These checks do more than confirm a few decimals: a correct starting sum and the correct added term determine all finite sums. For a given-answer question, show the cancellation rather than jumping from the general term to the requested result.

Exam tip

Write enough first and last terms to expose the cancellation. In a sum with powers, keep the power attached to its denominator throughout the working.

Use the same structure with one more change of geometric base.

Quick Recap

  • Express the general term as differences that actually match at successive indices.
  • Partial fractions often reveal the required difference.
  • A gap of two leaves two contributions at each boundary; a gap of one leaves one.
  • Track powers as carefully as rational factors, and display the surviving boundaries.

The opening sum is 100/101100/101, found by cancelling its internal fractions.