Telescoping Series

A telescoping series is a special type of infinite or finite sum in which most terms cancel with adjacent terms, leaving only a small number of terms — typically just the first and last — to determine the total. The name comes from the image of a collapsible telescope: just as the sections of a telescope slide into one another until only the outermost shell remains, the middle terms of a telescoping series collapse away, revealing a simple closed-form result. The key technique is to rewrite the general term of the series as a difference of two expressions, often using partial fraction decomposition or algebraic manipulation. Once every term is expressed in the form f(k) − f(k+1) (or equivalently f(k+1) − f(k)), the summation unfolds so that consecutive terms eliminate each other in a chain reaction of cancellation. This transforms what looks like a difficult infinite sum into a straightforward evaluation of just two values of f. Telescoping series appear throughout mathematics — in calculus, combinatorics, and number theory — and provide one of the most elegant tools for evaluating sums that would otherwise resist closed-form solutions. Mastering this technique is essential not only for competitions and exams but for building intuition about how summation and cancellation work together in analysis.

Key takeaways
  • Recognise the Structure
  • Decompose the General Term
  • Write Out the Partial Sum S_n
  • Cancel Intermediate Terms
  • Evaluate the Infinite Sum (if applicable)

Recognise the Structure

A series telescopes when its general term a_k can be written as a difference: a_k = f(k) − f(k+1) for some function f. Common signals include rational terms with factorable denominators (e.g., 1/(k(k+1))), products of consecutive integers, or differences of square roots. If you see a pattern where numerators or factors differ by exactly 1, suspect telescoping.

Decompose the General Term

Use partial fractions or algebraic identities to write a_k as f(k) − f(k+1). For example, note that 1/(k(k+1)) = 1/k − 1/(k+1). This is the most common approach and involves matching coefficients after assuming the decomposition form, then solving a small system of equations. Always verify by recombining the fractions.

Write Out the Partial Sum S_n

Express the partial sum S_n = Σ_{k=1}^{n} a_k by writing each term in its decomposed form and listing several consecutive terms explicitly. You will see that nearly every term appears once with a positive sign and once with a negative sign from the adjacent term, causing cancellation. For a_k = f(k) − f(k+1), the partial sum becomes S_n = [f(1) − f(2)] + [f(2) − f(3)] + [f(3) − f(4)] + ⋯ + [f(n) − f(n+1)].

Cancel Intermediate Terms

After writing out all terms, every f(2), f(3), …, f(n) appears with both a positive and a negative sign, so each cancels exactly. The surviving terms are only f(1) from the first group and −f(n+1) from the last group. Therefore S_n = f(1) − f(n+1). This is the power of telescoping: the partial sum collapses to a trivially computed expression.

Evaluate the Infinite Sum (if applicable)

If the series is infinite, take the limit as n → ∞. The infinite sum converges if and only if lim_{n→∞} f(n+1) exists. When that limit L exists, the sum equals f(1) − L. For 1/(k(k+1)), f(k) = 1/k, so f(1) = 1 and lim_{n→∞} 1/(n+1) = 0, giving the sum Σ_{k=1}^{∞} 1/(k(k+1)) = 1.

The Key Equation

∑k=1n[f(k)−f(k+1)]=f(1)−f(n+1)\sum_{k=1}^{n} \bigl[f(k) - f(k+1)\bigr] = f(1) - f(n+1)
  • f(k) — Telescoping function (dimensionless): Any function of the index k such that the general term a_k can be expressed as f(k) − f(k+1). Choosing f correctly is the central skill.
  • f(1) — First surviving term (dimensionless): The value of the telescoping function at the starting index k = 1; this term is not cancelled by any other term in the sum.
  • f(n+1) — Last surviving term (dimensionless): The value of the telescoping function one step beyond the last index; this term is subtracted and not cancelled by any earlier term.
  • n — Upper limit of summation (dimensionless): The index of the last term included in the partial sum. As n → ∞, the partial sum approaches the infinite series sum if f(n+1) → 0.

Worked Example

Problem. Find the exact value of the infinite series: Σ_{k=1}^{∞} 3/(k(k+3)).

116\dfrac{11}{6}

Answer. 11/6

Figure
Diagram
Telescoping Cancellation: Σ 1/(k(k+1)) = 1/k − 1/(k+1)k=1:+1/1− 1/2k=2:+1/2− 1/3k=3:+1/3− 1/4k=4:+1/4− 1/5⋮⋮⋮k=n:+1/n− 1/(n+1)cancelscancelscancelsS_n =1/1+ (cancelled terms)− 1/(n+1)S_∞ =1− lim 1/(n+1) =1 − 0 = 1
Origins · 1 minAugustin-Louis Cauchy and how this was discovered
AC
Augustin-Louis Cauchy
1789–1857 · French

Cauchy placed the theory of infinite series on a rigorous footing, introducing the concept of convergence and the notion of a partial sum as the foundation for evaluating infinite sums. His work in Cours d'analyse (1821) established the formal definitions that underpin telescoping and all other series techniques used in modern mathematics. Cauchy published over 800 research papers — so many that the French Academy of Sciences had to introduce a page limit on submissions specifically because of him.

  • Partial Fraction Decomposition — Partial fraction decomposition is the primary algebraic tool used to rewrite rational terms in the form required for telescoping, splitting complex fractions into simpler differences.
  • Geometric Series — Like telescoping series, geometric series have clean closed-form sums. Comparing the two techniques highlights when each method applies and deepens understanding of series evaluation.
  • Convergence of Infinite Series — Telescoping series are one of the few classes of series whose convergence and sum can be determined exactly; understanding convergence gives the limit argument that completes the evaluation.
  • Arithmetic Series — Arithmetic series offer another route to closed-form sums; studying them alongside telescoping series builds a toolkit of techniques for different structural types of summation problems.