CP 4.4 - The method of differences

CP 4.4 - The method of differences

Some long sums simplify because almost all their terms cancel. You will expose that cancellation with partial fractions, keep the correct terms at both ends and use a finite partial sum to decide whether an infinite series has a sum.

Why only the boundary terms survive

How large can this sum become if you keep adding terms?

11(2)+12(3)+13(4)+⋯\frac1{1(2)}+\frac1{2(3)}+\frac1{3(4)}+\cdots

The first three terms total 3/43/4. Every later term is positive, yet even the first million terms add to less than 11: their exact total is 1 000 000/1 000 0011\,000\,000/1\,000\,001. The reason is cancellation hidden inside each fraction.

You already know

Partial fractions rewrite a rational expression as simpler fractions; multiply through by the denominator to determine their constants. A finite partial sum SnS_n adds the first nn terms. An infinite series has sum LL when SnS_n tends to the finite limit LL.

For the opening summand,

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

The right side has numerator (r+1)−r=1(r+1)-r=1, so the identity is exact whenever neither denominator is zero. Adding several such differences gives

(1−12)+(12−13)+(13−14)=1−14\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)=1-\frac14

The negative fraction in one term cancels the positive fraction in the next. This is the method of differences, also called telescoping.

Adjacent differences

For a function FF defined at all required integer arguments,

∑r=mn(F(r)−F(r+1))=F(m)−F(n+1)\sum_{r=m}^{n}\bigl(F(r)-F(r+1)\bigr)=F(m)-F(n+1)

Here m,nm,n are integers with m≤nm\le n. Cancellation leaves the first positive term and the last negative term.

Not in the formula booklet: learn it.

Cancellation acts on identical terms with opposite signs. It does not let you cancel denominators across addition. Writing the first few and last few differences makes both the repeated pattern and the surviving endpoints visible.

Key idea

Find a difference whose successive terms overlap, then retain every term that lacks a partner at the beginning or end. The number of surviving boundary terms depends on the gap in the arguments.

Check which endpoint follows from the upper index in the summand, rather than guessing it from the last number in the sigma symbol.

Partial fractions with adjacent denominators

When two linear factors differ by one, partial fractions often give adjacent differences directly; establish the identity before summing it.

For 1/[r(r+1)]1/[r(r+1)], write A/r+B/(r+1)A/r+B/(r+1). Multiplying through gives 1=A(r+1)+Br1=A(r+1)+Br, so A=1A=1, B=−1B=-1. The partial-fraction step is the algebraic reason that the cancellation is available.

Worked example 1: the opening finite sum

Question. Use the method of differences to find Sn=∑r=1n1/[r(r+1)]S_n=\sum_{r=1}^{n}1/[r(r+1)].

Plan. Decompose the summand, show the first and last differences, then collect the boundary terms.

1r(r+1)=1r−1r+1partial fractions\frac1{r(r+1)}=\frac1r-\frac1{r+1} \qquad\text{partial fractions}

For n≥4n\ge4, display the distinct first and last terms:

Sn=(1−12)+(12−13)+⋯+(1n−1−1n)+(1n−1n+1)show matching terms=1−1n+1cancel interior fractions=nn+1combine the survivors\begin{aligned} S_n &=\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\cdots\\ &\quad+\left(\frac1{n-1}-\frac1n\right)+\left(\frac1n-\frac1{n+1}\right) &&\text{show matching terms}\\ &=1-\frac1{n+1} &&\text{cancel interior fractions}\\ &=\frac n{n+1} &&\text{combine the survivors} \end{aligned}

Why this step? The last summand contains −1/(n+1)-1/(n+1), which has no positive partner. Keeping it is essential: deleting it would give the infinite limit rather than the finite sum.

Answer. Sn=n/(n+1)S_n=n/(n+1), for every positive integer nn.

Check. At n=3n=3, 1/2+1/6+1/12=3/41/2+1/6+1/12=3/4. At n=1n=1, the formula gives the single term 1/21/2; the displayed cancellation pattern is interpreted with no repeated endpoint terms.

Worked example 2 changes only the argument in the denominator, replacing rr by r+1r+1. The cancellation is identical, but both surviving endpoints shift.

Worked example 2: a shifted denominator

Question. Find ∑r=1n1/[(r+1)(r+2)]\sum_{r=1}^{n}1/[(r+1)(r+2)] using the method of differences.

Plan. Write each term as 1/(r+1)−1/(r+2)1/(r+1)-1/(r+2) and keep the new first and last fractions.

The partial-fraction identity is 1/[(r+1)(r+2)]=1/(r+1)−1/(r+2)1/[(r+1)(r+2)]=1/(r+1)-1/(r+2). For n≥4n\ge4, write

Sn=(12−13)+(13−14)+⋯+(1n−1n+1)+(1n+1−1n+2)display the boundary pattern=12−1n+2cancel the interior=n2(n+2)use a common denominator\begin{aligned} S_n&=\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)+\cdots\\ &\quad+\left(\frac1n-\frac1{n+1}\right)+\left(\frac1{n+1}-\frac1{n+2}\right) &&\text{display the boundary pattern}\\ &=\frac12-\frac1{n+2} &&\text{cancel the interior}\\ &=\frac{n}{2(n+2)} &&\text{use a common denominator} \end{aligned}

Why this step? The lower index is still 11, but the first denominator is now 22. Endpoints come from substituting into the summand, not simply copying the sigma limits.

Answer. Sn=n2(n+2)\displaystyle S_n=\frac{n}{2(n+2)}.

Check. At n=2n=2, 1/6+1/12=1/41/6+1/12=1/4, and the formula gives 2/[2(4)]=1/42/[2(4)]=1/4.

These finite-sum identities are not in the formula booklet: derive them from the partial fractions. Pearson's method-of-differences questions require evidence of the cancellation, not just the final expression.

Watch out

Wrong: ∑r=1n1/[(r+1)(r+2)]=1−1/(n+1)\sum_{r=1}^{n}1/[(r+1)(r+2)]=1-1/(n+1), copied from the unshifted case. Right: substitute the actual first and last indices, obtaining 1/2−1/(n+2)1/2-1/(n+2).

Showing at least two terms from each end makes the cancellation unambiguous. For very small nn, check the sum directly instead of counting overlapping displayed endpoints twice.

Wider cancellation gaps and infinite sums

A gap of two in the partial-fraction arguments leaves two terms at each boundary; taking an infinite limit only comes after those finite boundary terms have been found.

For example, 1/r−1/(r+2)1/r-1/(r+2) cancels the negative term in one row with the positive term two rows later. The first two positive fractions have no earlier partners, while the last two negative fractions have no later partners.

Worked example 3: two survivors at each end

Question. Find Sn=∑r=1n2/[r(r+2)]S_n=\sum_{r=1}^{n}2/[r(r+2)]. Hence find the sum to infinity.

Plan. Use 2/[r(r+2)]=1/r−1/(r+2)2/[r(r+2)]=1/r-1/(r+2), retain two terms at each end, then take the limit.

For n≥5n\ge5, the displayed endpoint terms are distinct:

Sn=(1−13)+(12−14)+(13−15)+⋯+(1n−1−1n+1)+(1n−1n+2)show the gap of two=1+12−1n+1−1n+2retain all four boundary terms\begin{aligned} S_n&=\left(1-\frac13\right)+\left(\frac12-\frac14\right)+\left(\frac13-\frac15\right)+\cdots\\ &\quad+\left(\frac1{n-1}-\frac1{n+1}\right)+\left(\frac1n-\frac1{n+2}\right) &&\text{show the gap of two}\\ &=1+\frac12-\frac1{n+1}-\frac1{n+2} &&\text{retain all four boundary terms} \end{aligned}

Both remaining negative fractions tend to zero as nn tends to infinity.

Why this step? A wider gap changes how many boundary terms remain. Keeping only 1−1/(n+2)1-1/(n+2) would remove terms that never cancelled.

Answer. Sn=3/2−1/(n+1)−1/(n+2)S_n=3/2-1/(n+1)-1/(n+2), and the sum to infinity is 3/23/2.

Check. At n=2n=2, 2/3+1/4=11/122/3+1/4=11/12. The boundary formula gives 3/2−1/3−1/4=11/123/2-1/3-1/4=11/12. It also works directly for n=1n=1, where boundary entries overlap algebraically.

Worked example 4 shifts the denominator factors down by one. This time the series must start at r=2r=2, because r=1r=1 would make a denominator zero.

Worked example 4: respecting an excluded index

Question. For n≥2n\ge2, find ∑r=2n2/(r2−1)\sum_{r=2}^{n}2/(r^2-1). Hence find the sum to infinity.

Plan. Factor r2−1r^2-1, split into fractions separated by two indices and start the cancellation at r=2r=2.

2r2−1=1r−1−1r+1partial fractions\frac2{r^2-1}=\frac1{r-1}-\frac1{r+1} \qquad\text{partial fractions}

For n≥5n\ge5, the first and last pairs are

Sn=(1−13)+(12−14)+⋯+(1n−2−1n)+(1n−1−1n+1)use the actual lower limit=32−1n−1n+1keep two terms at each end\begin{aligned} S_n&=\left(1-\frac13\right)+\left(\frac12-\frac14\right)+\cdots\\ &\quad+\left(\frac1{n-2}-\frac1n\right)+\left(\frac1{n-1}-\frac1{n+1}\right) &&\text{use the actual lower limit}\\ &=\frac32-\frac1n-\frac1{n+1} &&\text{keep two terms at each end} \end{aligned}

Why this step? The denominator rules out r=1r=1, so a cancellation pattern beginning there would not describe a valid series. Starting at 22 gives first positive fraction 11.

Answer. The finite sum is 3/2−1/n−1/(n+1)3/2-1/n-1/(n+1), and its limit is 3/23/2.

Check. At n=2n=2, the only term is 2/32/3, and 3/2−1/2−1/3=2/33/2-1/2-1/3=2/3. For this small case use the original single term, rather than the long-form display.

The finite results and their infinite sums are not in the formula booklet: derive the finite formula and inspect its limit. A summand tending to zero is necessary for convergence but is not enough by itself; the partial sums must approach a finite value.

Watch out

Wrong: “the last summand tends to zero, so the infinite sum is zero”. Right: take the limit of the whole partial-sum expression, including its surviving initial terms. Those terms give 3/23/2 in both examples.

The opening million-term statement follows from Sn=n/(n+1)S_n=n/(n+1). Its infinite sum is 11, while each finite partial sum is smaller by exactly 1/(n+1)1/(n+1). Infinitely many positive terms can therefore have a finite total.

The next denominator has a gap of three. Write enough boundary terms to see all the unmatched fractions before taking a limit.

Quick Recap

  • Use partial fractions to expose a difference before summing.
  • Show the first and last terms with the correct indices; retain every unmatched boundary fraction.
  • A larger argument gap leaves more terms at both ends.
  • An infinite sum is the limit of the finite partial sum, provided that limit exists and is finite.

The opening sum approaches 11; after a million terms its exact value is 1 000 000/1 000 0011\,000\,000/1\,000\,001.