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Concept 07 / 07Sequences and Series
Method of Differences and Telescoping Sums
Writing a general term as f(k) - f(k + 1) so that the sum collapses, partial fractions for reciprocal products of AP terms, the Vn method for products of consecutive terms, and telescoping with trigonometric and other functions.
Figure 1
A sum telescopes when each term can be written as a difference Tk=f(k)−f(k+1) of consecutive values of some function. Then
∑k=1nTk=[f(1)−f(2)]+[f(2)−f(3)]+⋯+[f(n)−f(n+1)]=f(1)−f(n+1).
All intermediate values cancel, like the sections of a collapsing telescope. The whole skill is finding f.
Reciprocals of products (partial fractions). For terms like k(k+1)1, partial fractions give k1−k+11, as in the figure, so ∑k=1nk(k+1)1=1−n+11=n+1n. More generally, for an AP ak with common difference d:
akak+11=d1(ak1−ak+11), so ∑k=1nakak+11=d1(a11−an+11)=a1an+1n.
For three factors: akak+1ak+21=2d1(akak+11−ak+1ak+21). The pattern: drop the last factor in one term and the first in the other, and divide by (difference of the dropped factors).
The Vn method for products. For terms that are products of consecutive AP terms, such as k(k+1)(k+2), use the forward version: k(k+1)(k+2)=41[k(k+1)(k+2)(k+3)−(k−1)k(k+1)(k+2)]. Summing telescopes to 4n(n+1)(n+2)(n+3). Rule: append the next factor, subtract the version with the previous factor prepended, and divide by (number of factors after appending) times d.
Differences of square roots: k+k+11=k+1−k after rationalising, so the sum from 1 to n is n+1−1.
Trigonometric telescoping. Use identities that produce differences. For example, sinxcoskx-type products become differences of sines; tan(k+1)x−tankx=coskxcos(k+1)xsinx, so ∑coskxcos(k+1)x1=sinxtan(n+1)x−tanx. Inverse tangents telescope via tan−11+k(k+1)1=tan−1(k+1)−tan−1k.
Products can telescope too: ∏k=2n(1−k21)=∏k⋅k(k−1)(k+1)=2nn+1, since the numerators and denominators cancel in a staggered way.
When there is no obvious f, use the method of differences on the terms of the series: compute successive differences until they become constant or geometric, deduce the form of Tk, then sum.
Key Takeaways:
- ∑k=1n[f(k)−f(k+1)]=f(1)−f(n+1).
- akak+11=d1(ak1−ak+11); the sum is a1an+1n.
- ∑k(k+1)(k+2)1=21[21−(n+1)(n+2)1].
- ∑k(k+1)(k+2)=4n(n+1)(n+2)(n+3).
- tan−11+k(k+1)1=tan−1(k+1)−tan−1k.
- Infinite telescoping sums converge when f(n+1) has a limit.
- Pitfall: in partial fractions with a gap of 2 (for example k(k+2)1), two terms survive at each end, not one.
Brain-Teaser 1: Find ∑k=1n(3k−2)(3k+1)1 and its limit as n→∞.
Solution: The factors are consecutive terms of the AP 1,4,7,… with d=3. So (3k−2)(3k+1)1=31(3k−21−3k+11), and the sum telescopes to 31(1−3n+11)=3n+1n. As n→∞, the sum tends to 31. The formula a1an+1n gives 1⋅(3n+1)n directly.
Brain-Teaser 2: Evaluate ∑k=1∞k(k+2)1.
Solution: k(k+2)1=21(k1−k+21). Now the cancellation skips one term: 11−31+21−41+31−51+⋯. The positive terms 11 and 21 never get cancelled, while everything from 31 on does. So the sum is 21(1+21)=43. With a gap of m between the factors, the first m terms survive, which is the correct form of the telescoping rule.
Brain-Teaser 3: Find ∑k=1ntan−1k2+k+11 and its limit.
Solution: Since 1+k(k+1)=k2+k+1, the term is tan−11+k(k+1)(k+1)−k=tan−1(k+1)−tan−1k, using the subtraction formula for tan−1 (valid here because both arguments are positive). The sum telescopes to tan−1(n+1)−tan−11=tan−1(n+1)−4π. As n→∞, tan−1(n+1)→2π, so the infinite sum is 4π. Recognising 1+abb−a in a denominator of the form 1+ (product) is the key step.
