Telescoping Series
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Direct answer
A series telescopes when its general term hides a difference of consecutive pieces, f(r) − f(r+1): in the written-out sum every interior term cancels and only the two ends survive, giving Σ_(r=1)^n [f(r) − f(r+1)] = f(1) − f(n+1). The craft is finding the split — partial fractions for rational terms, ready-made differences for factorial products, the arctangent subtraction formula for inverse tan terms. Infinite versions are limits of the closed form: the series converges exactly when f(n+1) settles to a finite limit L, the sum being f(1) − L.
What you must remember
- Core rational split: 1/(r(r+1)) = 1/r − 1/(r+1), so Σ_(r=1)^n = 1 − 1/(n+1) = n/(n+1), with limit 1.
- Three-factor split: 1/(r(r+1)(r+2)) = ½[1/(r(r+1)) − 1/((r+1)(r+2))]; the partial sum is ½[½ − 1/((n+1)(n+2))], limit 1/4.
- Factorial products: r·r! = (r+1)! − r!, so Σ_(r=1)^n r·r! = (n+1)! − 1 — the most quoted factorial telescoping.
- Arctangent form: tan⁻¹(1/(1 + r + r²)) = tan⁻¹(r+1) − tan⁻¹(r), since (r+1) − r sits over 1 + r(r+1); the sum is tan⁻¹(n+1) − tan⁻¹(1) = tan⁻¹(n/(n+2)), limit π/4.
- Squares collapse: 1/(r²(r+1)²) = 1/r² − 1/(r+1)², giving 1 − 1/(n+1)² and limit 1.
- Quadratic denominators: 1/(r² + 3r + 2) = 1/(r+1) − 1/(r+2), summing to ½ − 1/(n+2), limit 1/2.
- V-substitution fallback: when a quartic denominator resists direct partial fractions, posit 1/P(r) = Vr − V(r+1) with V_r a reciprocal quadratic and fit the numerator by comparison — the systematic last resort.
Three telescoping patterns
Pattern one, rational: Σ_(r=1)^n 1/(r(r+1)(r+2)). Split each term as ½[1/(r(r+1)) − 1/((r+1)(r+2))]; the chain cancels everything interior, leaving ½[1/(1·2) − 1/((n+1)(n+2))]. As n → ∞ the second piece dies and the limit is 1/4 — the numerical answer Main recycles.
Pattern two, arctangent: Σ_(r=1)^n tan⁻¹(1/(1 + r + r²)). Recognise (r+1) − r over 1 + (r+1)r and apply tan⁻¹A − tan⁻¹B = tan⁻¹[(A − B)/(1 + AB)], valid here since the difference lands inside the principal branch. The sum telescopes to tan⁻¹(n+1) − tan⁻¹(1) = tan⁻¹(n/(n+2)), tending to tan⁻¹(1) = π/4.
Pattern three, index discipline: Σ_(r=2)^n 1/(r² − 1) = ½ Σ [1/(r−1) − 1/(r+1)]. The negative side runs 1/1 + 1/2 + ... + 1/(n−2), the positive-cancelling side 1/3 + ... + 1/(n+1); what survives is ½[1 + ½ − 1/n − 1/(n+1)]. The shift by two (not one) is where most scripts go wrong — substituting n = 2 into the final form instantly verifies it: ½[3/2 − 1/2 − 1/6] = 1/3, matching 1/(4 − 1) = 1/3.
Frequently asked questions
What does Σ_(r=1)^n 1/(r(r+1)) equal, and what is its limit?
n/(n+1) after cancellation, tending to 1 — the anchor example of the whole method.
How do I sum tan⁻¹(1/(1 + r + r²)) from r = 1 to n?
Write it as tan⁻¹(r+1) − tan⁻¹(r); the sum collapses to tan⁻¹(n+1) − tan⁻¹(1) = tan⁻¹(n/(n+2)).
What is Σ r·r! for r = 1 to n?
(n+1)! − 1, since each term r·r! splits as (r+1)! − r!.
How do I check a telescoping answer quickly?
Substitute n = 1 into the closed form and compare with the single first term; most index-shift errors die on this test.
When does a telescoping infinite series converge?
When f(n+1) tends to a finite limit L, the partial sums f(1) − f(n+1) converge to f(1) − L.