# Telescoping Series

> Telescoping series in JEE Mathematics: partial-fraction splits, factorial and arctangent forms, index-shift discipline and worked sums with limits.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/telescoping-series
- Exam / course: JEE · Subject: Mathematics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Telescoping Series", PrepElephant, https://prepelephant.com/topics/jee/mathematics/telescoping-series

## 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) = V_r − 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.

## Hidden telescoping and index slips

Advanced disguises the structure — arctangent terms, factorials, trig products — where no denominator factorisation advertises the split; Main hands it over through a factorable denominator. The two recurring errors are both bookkeeping. First, the off-by-one: writing Σ_(r=1)^n [f(r) − f(r+1)] as f(1) − f(n) instead of f(1) − f(n+1); checking the formula at n = 1 catches it in five seconds, a habit worth forcing. Second, taking the infinite limit carelessly: the sum converges only if f(n+1) has a finite limit, and quoting f(1) minus the wrong end gives a plausible-looking wrong answer. Before committing any telescoping result, verify it at n = 1 and, if possible, n = 2 — the two cheapest marks on the paper.

## 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.