Chapter 2 · Sequences · Concept 3
2-14 Sum of Fractional Sequences
None of the three formulas will touch 1/(1×2) + 1/(2×3) + 1/(3×4) + … . These fractions are not counting numbers, not squares, not cubes. But there is a trick, and it is the most satisfying one in the whole chapter: nearly every term destroys its neighbour.
Guiding question
What do you do when no formula fits at all?
Learning outcome. Calculate the sum of fractional sequences using fraction decomposition.
Split each fraction in two
1ab = 1b − a ( 1a − 1b )
Example
1/(4 × 10) = ⅙(1/4 − 1/10)
because b − a is 10 − 4 = 6. Check it: ⅙(1/4 − 1/10) = ⅙ × 6/40 = 1/40 ✓
Then: cancel everything that meets its opposite.
Every term leaves a + half and a − half. The minus half of one term is the plus half of a later one, so they wipe each other out. Only the very ends survive — which is why a sum of a hundred fractions can have a two-line answer.
If there is a root on the bottom, rationalise it first — multiply top and bottom by the same thing with the sign flipped. The bottom becomes a plain number and the top becomes a difference of two roots, which then cancels the same way.
Green halves are the plus parts, red halves the minus parts. Press the button and every pair that matches wipes itself out.
Each term splits into 1/(1/a − 1/b). Here are the first 6, split open:
Count the survivors before you write the answer.
When the two bottoms are one step apart in the chain, one survives at each end. When the cancelling skips a term — as with 1/(k(k+2)) — two survive at each end, and an answer with only one is wrong by a whole fraction.
(a) Find 1/(2×5) + 1/(5×8) + 1/(8×11) + … + 1/((3n−1)(3n+2)).
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Work out b − a for one term, and check it is the same every time.
5 − 2 = 3, 8 − 5 = 3, 11 − 8 = 3The 1/(b − a) out front only works if the gap is the same throughout. Check before using it.
Work with a partner. Before cancelling, both count how many terms will survive at each end. If you disagree, look at whether the chain skips.
1. Find the following sums.
2. Find the following sums.
Σ from r = 1 to 30 of 1r(r + 2)
In Exercise 1(a) the answer is n/(n+1). What happens to that as n gets enormous — and what does that tell you about adding infinitely many fractions?
Try n = 10, then 100, then 1000.
Forgetting the 1/(b − a) in front.
1/(2×5) is ⅓(1/2 − 1/5), not (1/2 − 1/5). Without the ⅓ every answer is three times too big.
Assuming one term survives at each end.
When the bottoms are two apart the chain skips, so TWO survive at each end. Write a few terms out and look.
Not rationalising before cancelling.
1/(√(k+1) + √k) cancels nothing as it stands. Multiply top and bottom by (√(k+1) − √k) first.
Cancelling the wrong way round.
A −1/5 only dies against a +1/5. Two minuses do not cancel each other.
Leaving the answer as a difference.
⅓(1/2 − 1/(3n+2)) is correct but unfinished. Put it over one denominator: n/(2(3n+2)).
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