Telescoping sums and products: fractions that cancel

A sum of fractions may look long and laborious. Sometimes, instead of calculating every term separately, we need only change its representation: intermediate parts disappear and just the endpoints survive. The same can happen in a product. This is the principle of telescoping expressions.

1. The fundamental idea

A sum telescopes when, after a suitable decomposition, almost all its terms cancel in pairs. Look for consecutive terms with opposite signs. The most important decomposition is 1/[k(k+1)] = 1/k − 1/(k+1), since 1/k − 1/(k+1) = [(k+1)−k]/[k(k+1)] = 1/[k(k+1)].

Example 1 — The first chain

Calculate 1/2 + 1/6 + 1/12 + 1/20. The denominators are 1·2, 2·3, 3·4, 4·5. Thus the sum becomes (1−1/2) + (1/2−1/3) + (1/3−1/4) + (1/4−1/5). The −1/2 cancels the +1/2, then −1/3 cancels +1/3, and −1/4 cancels +1/4. We are left with 1−1/5 = 4/5.

For n terms, Σ(k=1…n) 1/[k(k+1)] = 1 − 1/(n+1) = n/(n+1).

2. A step of two: denominators k(k+2)

When the second factor is k+2, the same idea works with coefficient 1/2: 1/[k(k+2)] = (1/2)·[1/k − 1/(k+2)].

Example 2

1/(1·3) + 1/(3·5) + 1/(5·7) + 1/(7·9) becomes (1/2)[(1−1/3) + (1/3−1/5) + (1/5−1/7) + (1/7−1/9)]. The intermediate terms disappear and leave (1/2)(1−1/9) = (1/2)(8/9) = 4/9.

The general step-a formula

For a ≠ 0, with nonzero denominators, 1/[k(k+a)] = (1/a)·[1/k − 1/(k+a)]. This identity is a machine for building telescoping sums. We must still check that the chosen indices link each term to the next.

Example 3 — A step of three

1/(1·4) + 1/(4·7) + 1/(7·10) becomes (1/3)[(1−1/4) + (1/4−1/7) + (1/7−1/10)] = (1/3)(1−1/10) = 3/10.

3. Telescoping expressions with square roots

Sums involving roots often telescope after rationalisation. The key identity is 1/[√(k+1)+√k] = √(k+1)−√k. Indeed, multiply numerator and denominator by √(k+1)−√k: the denominator becomes (k+1)−k=1.

Example 4

1/(1+√2) + 1/(√2+√3) + 1/(√3+2) can be rewritten (√2−1) + (√3−√2) + (2−√3). The intermediate roots cancel, leaving 2−1=1.

Example 5 — General form

Σ(k=1…n) 1/[√k+√(k+1)] becomes (√2−1)+(√3−√2)+⋯+(√(n+1)−√n). Therefore Σ(k=1…n) 1/[√k+√(k+1)] = √(n+1)−1.

4. Three consecutive factors

Three consecutive factors also permit a useful decomposition: 1/[k(k+1)(k+2)] = (1/2)·{1/[k(k+1)] − 1/[(k+1)(k+2)]}.

Example 6

Calculate 1/(1·2·3) + 1/(2·3·4) + 1/(3·4·5). Substitution gives (1/2)[1/(1·2)−1/(2·3)] + (1/2)[1/(2·3)−1/(3·4)] + (1/2)[1/(3·4)−1/(4·5)]. The middle terms cancel, leaving (1/2)[1/2−1/20] = (1/2)(9/20) = 9/40.

In general, Σ(k=1…n) 1/[k(k+1)(k+2)] = (1/2)[1/2 − 1/((n+1)(n+2))].

5. Differences of squares

When n²−1 appears, factor it: n²−1=(n−1)(n+1). Then 1/(n²−1) = (1/2)[1/(n−1)−1/(n+1)].

Example 7

For 1/3 + 1/8 + 1/15 + 1/24, write 1/(2²−1)+1/(3²−1)+1/(4²−1)+1/(5²−1). The decomposition creates two staggered chains: (1/2)[(1−1/3)+(1/2−1/4)+(1/3−1/5)+(1/4−1/6)]. The pairs −1/3 and +1/3, then −1/4 and +1/4, cancel. The result is (1/2)(1+1/2−1/5−1/6) = 17/30.

Verification note: the source PDF gives 7/10 on the final line of this example. The exact result, also checked by adding the four fractions directly, is 17/30.

6. Products can telescope too

The principle applies to products as well as sums. In a product, consecutive factors may cancel between numerator and denominator.

Example 8 — The simplest product

(1−1/2)(1−1/3)(1−1/4)⋯(1−1/n). Each factor is (k−1)/k. Hence (1/2)(2/3)(3/4)⋯((n−1)/n) = 1/n.

Example 9 — A product with squares

Π(k=2…n)(1−1/k²). Factor 1−1/k² = (k²−1)/k² = [(k−1)/k]·[(k+1)/k]. The product splits into two chains: [(1/2)(2/3)⋯((n−1)/n)] · [(3/2)(4/3)⋯((n+1)/n)]. The first equals 1/n, the second (n+1)/2. Thus Π(k=2…n)(1−1/k²) = (n+1)/(2n).

7. How to recognise a telescoping expression

1. Factor the denominators: look for k(k+1), k(k+a) or (k−1)(k+1).
2. Decompose into differences: try expressing each fraction as two simpler fractions.
3. Rationalise roots: use the conjugate of √(k+1)+√k.
4. Write out the first terms: spot identical parts with opposite signs.
5. Inspect the endpoints: often only the first and last terms survive.

If the cancellation is not immediately visible, explicitly write three or four consecutive terms. The pattern often emerges.

8. Exercises

Try them before opening their solutions.

1. A. 1/2 + 1/6 + 1/12 + 1/20 + 1/30
   Show the solution

   The denominators are k(k+1) for k=1,…,5. The sum is 1−1/6 = 5/6.

2. B. 1/(1·4) + 1/(4·7) + 1/(7·10) + 1/(10·13)
   Show the solution

   Use 1/[k(k+3)] = (1/3)(1/k−1/(k+3)). What remains is (1/3)(1−1/13) = 4/13.

3. C. 1/(1+√2) + 1/(√2+√3) + 1/(√3+2) + 1/(2+√5)
   Show the solution

   Each rationalised term becomes a difference of consecutive roots. The interior parts disappear, leaving √5−1.

4. D. 1/(1·2·3) + 1/(2·3·4) + 1/(3·4·5) + 1/(4·5·6)
   Show the solution

   The three-factor identity gives (1/2)[1/(1·2)−1/(5·6)] = (1/2)(1/2−1/30) = 7/30.

5. E. (1−1/2)(1−1/3)(1−1/4)(1−1/5)(1−1/6)
   Show the solution

   This becomes (1/2)(2/3)(3/4)(4/5)(5/6) = 1/6.

6. F. Π(k=2…5)(1−1/k²)
   Show the solution

   Use (n+1)/(2n) with n=5 to get 6/10 = 3/5.

Closing idea

The real trick is not to calculate faster but to change representation. A complicated fraction can become a difference; a radical can be rationalised; a product can be factored. Once opposite consecutive terms appear, the expression shortens itself.

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