Method of differences

Objective. Expose cancellation in telescoping series.

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Learn: the key idea

The method of differences evaluates a sum by expressing each term as a difference f(r) − f(r+1) (or similar), so that consecutive terms cancel when the sum is written out, leaving only a few terms at the start and end — this is called telescoping. Writing the sum explicitly for several terms exposes exactly which terms survive after cancellation. For example, Σ[1/r − 1/(r+1)] from r=1 to n telescopes to 1 − 1/(n+1), since every middle term cancels with its neighbour.

Worked example

Using the method of differences, evaluate Σ[1/(r(r+1))] for r = 1 to n, given that 1/(r(r+1)) = 1/r − 1/(r+1).

1 $Write the sum using the given identity: Σ[1/r - 1/(r+1)] for r=1 to n$
2 $List terms: (1/1-1/2) + (1/2-1/3) + ... + (1/n-1/(n+1))$
3 $All middle terms cancel in pairs, leaving only the first and last parts: 1/1 - 1/(n+1)$
4 $Simplify: 1 - 1/(n+1) = n/(n+1)$

Answer: n/(n+1)

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Common mistake. Students cancel terms incorrectly by assuming all terms vanish, forgetting that the very first and very last terms of a telescoping sum always survive. Always write out the first two and last two terms explicitly to check what remains.

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