The Method of Differences

Terms that cancel in pairs collapse a sum.

A term as a difference

Some sums can be found exactly by rewriting each term as a difference of two values of one expression, one step apart. Take 1/(r(r + 1)). In partial fractions, 1/(r(r + 1)) = 1/r − 1/(r + 1). Check at r = 2: 1/(2 × 3) = 1/6, and 1/2 − 1/3 = 3/6 − 2/6 = 1/6.

The first piece is 1/r and the second is the same expression one step on, 1/(r + 1). In general, a term of the form f(r) − f(r + 1) is what the method needs. Here f(r) = 1/r.

Writing it out

Write the sum from r = 1 to n with each term split: (1/1 − 1/2) + (1/2 − 1/3) + (1/3 − 1/4) + … + (1/n − 1/(n + 1)).

The −1/2 at the end of the first bracket meets +1/2 at the start of the second, and they cancel. The −1/3 meets +1/3 in the next, and so on along the whole row: every inner piece is subtracted once and added once. Only two pieces have no partner, the first, 1/1, and the last, −1/(n + 1).

What survives

So Σ 1/(r(r + 1)) from r = 1 to n is 1 − 1/(n + 1) = n/(n + 1), exactly, for every n. For n = 3, 1/2 + 1/6 + 1/12 = 6/12 + 2/12 + 1/12 = 9/12 = 3/4, and 3/(3 + 1) = 3/4.

In general, Σ[f(r) − f(r + 1)] from r = 1 to n = f(1) − f(n + 1): the first piece of the first term and the last piece of the last term. The sum is said to telescope, because it closes up like a telescope.

Where the sum is heading

As n grows, 1/(n + 1) shrinks toward 0, so the sum n/(n + 1) climbs toward 1: 1/2, 2/3, 3/4, 4/5, and 99/100 after 99 terms. It never reaches 1, because 1/(n + 1) is never 0, and it never passes 1. The infinite series 1/(1 × 2) + 1/(2 × 3) + 1/(3 × 4) + … converges, and its sum is 1.

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The sums after one, two, three and four terms: 1/2, 2/3, 3/4, 4/5. Each hop is a term, 1/2, 1/6, 1/12, 1/20, and the sums close in on 1.

Pieces two apart

In partial fractions, 1/(r(r + 2)) = (1/2)(1/r − 1/(r + 2)). Now each piece meets its opposite two brackets later, not in the next one: (1/1 − 1/3) + (1/2 − 1/4) + (1/3 − 1/5) + (1/4 − 1/6) + …

The 1/3 subtracted in the first bracket is added in the third; the 1/4 subtracted in the second is added in the fourth. Two pieces survive at each end: 1/1 and 1/2 at the start, −1/(n + 1) and −1/(n + 2) at the end. So Σ 1/(r(r + 2)) = (1/2)(1 + 1/2 − 1/(n + 1) − 1/(n + 2)).

Check at n = 2: 1/3 + 1/8 = 11/24, and (1/2)(3/2 − 1/3 − 1/4) = (1/2)(18/12 − 4/12 − 3/12) = 11/24. As n grows the two end pieces shrink to 0, and the sum tends to (1/2)(3/2) = 3/4.

Not only fractions

Any term that is a difference of consecutive values works. Since (r + 1)! = (r + 1) × r!, the difference (r + 1)! − r! is r × r!. So Σr × r! from r = 1 to n is (2! − 1!) + (3! − 2!) + … + ((n + 1)! − n!) = (n + 1)! − 1.

For n = 3: 1 × 1 + 2 × 2 + 3 × 6 = 1 + 4 + 18 = 23, and 4! − 1 = 24 − 1 = 23. Here the larger value comes second in each bracket, so the last piece survives with a plus and the first with a minus.

The usual mistakes

Dropping the first piece. The 1/1 has nothing before it to cancel with, so the sum is 1 − 1/(n + 1), not 1/n.

Assuming one piece survives at each end. When the pieces are two apart, as in 1/(r(r + 2)), two survive at each end.

Giving the limit for a finite sum. After three terms the sum is 3/4; it is heading for 1 but is not there.

Taking the limit of the terms for the limit of the sum. The terms shrink to 0; their sum tends to 1.

A valve that seals itself

In the application below, a valve loses 240/(k(k + 1)) milliliters in hour k. The same partial fractions give the total after any number of hours, and the value it can never pass.

Worked example: A Valve That Seals Itself as It Leaks: A Total Whose Middle Terms All Cancel

Question A valve is sealing itself as its sealant swells, so it leaks less every hour. In hour k it loses 240k(k+1) milliliters. (a) How much has leaked after 15 hours? (b) The maintenance log is signed as soon as the total leak first rises above 230 mL. In which hour is it signed, and what is the most that can ever leak?

  1. 1.Split the term into partial fractions. Since 1k − 1k+1 = 1k(k+1), the loss in hour k is 240k(k+1) = 240(1k − 1k+1) milliliters.

    06012018024001015202430hourstotal leaked, mLhour k loses 240/(k x (k + 1)) mL
    06012018024001015202430hourstotal leaked, mLhour k loses 240/(k x (k + 1)) mL
    The leak in hour k is 240k(k+1) mL, and the totals climb ever more slowly.

  2. 2.Write the fifteen hours out in that form: 240[(1 − 12) + (12 − 13) + … + (115 − 116)]. Every fraction after the first is subtracted once and added once, so the pairs cancel and only the two ends remain.

    06012018024001015202430hourstotal leaked, mLhour k loses 240/(k x (k + 1)) mLhour k is 240/k − 240/(k + 1)every inside fraction cancels
    06012018024001015202430hourstotal leaked, mLhour k loses 240/(k x (k + 1)) mLhour k is 240/k − 240/(k + 1)every inside fraction cancels
    Each hour is a difference: 240(1k − 1k+1), so the middle terms cancel in pairs.

  3. 3.(a) What is left is 240(1 − 116) = 240 × 1516 = 225 mL.

    06012018024001015202430hourstotal leaked, mL225 mLhour k loses 240/(k x (k + 1)) mLhour k is 240/k − 240/(k + 1)every inside fraction cancels240 x (1 − 1/16) = 225 mL
    06012018024001015202430hourstotal leaked, mL225 mLhour k loses 240/(k x (k + 1)) mLhour k is 240/k − 240/(k + 1)every inside fraction cancels240 x (1 − 1/16) = 225 mL
    (a) 240(1 − 116) = 225 mL after 15 hours.

  4. 4.The same canceling holds for any number of hours, so after n hours the total is 240(1 − 1n+1) mL. As n grows, 1n+1 tends to 0, so the total climbs to 240 mL without ever reaching it.

    06012018024001015202430hourstotal leaked, mLceiling 240 mL225 mLhour k loses 240/(k x (k + 1)) mLhour k is 240/k − 240/(k + 1)every inside fraction cancels240 x (1 − 1/16) = 225 mLafter n hours: 240 − 240/(n + 1)
    06012018024001015202430hourstotal leaked, mLceiling 240 mL225 mLhour k loses 240/(k x (k + 1)) mLhour k is 240/k − 240/(k + 1)every inside fraction cancels240 x (1 − 1/16) = 225 mLafter n hours: 240 − 240/(n + 1)
    After n hours the total is 240 − 240n+1 mL, which stays under 240 mL.

  5. 5.(b) Solve 240(1 − 1n+1) > 230, which gives 1n+1 < 10240 = 124, so n + 1 > 24 and n = 24. Check: after 23 hours the total is 240 × 2324 = 230 mL exactly, which has not passed 230; after 24 hours it is 240 × 2425 = 230.4 mL. The log is signed in hour 24, and the leak can never pass 240 mL.

    06012018024001015202430hourstotal leaked, mLceiling 240 mL225 mL230.4 mLhour k loses 240/(k x (k + 1)) mLhour k is 240/k − 240/(k + 1)every inside fraction cancels240 x (1 − 1/16) = 225 mLafter n hours: 240 − 240/(n + 1)hour 24 reaches 230.4, past 230
    06012018024001015202430hourstotal leaked, mLceiling 240 mL225 mL230.4 mLhour k loses 240/(k x (k + 1)) mLhour k is 240/k − 240/(k + 1)every inside fraction cancels240 x (1 − 1/16) = 225 mLafter n hours: 240 − 240/(n + 1)hour 24 reaches 230.4, past 230
    (b) The log is signed in hour 24, when the total reaches 230.4 mL.

Answer: (a) 225 mL; (b) in hour 24, and the leak can never pass 240 mL

Common mistakes

  • Adding the fifteen fractions on a calculator and giving the rounded decimal. The method of differences gives the exact figure, 225 mL, and it gives more besides: the same working produces the total after any number of hours, which is what part (b) needs.
  • Reading "the total climbs to 240" as "the total reaches 240". It does not. After n hours the total is 240 − 240n+1, which is below 240 for every n; 240 mL is the value the totals close on, and the leak stays under it forever.

More series and convergence problems, worked step by step →

Practice The Method of Differences in the app