---
title: "Interval and radius of convergence, 1/(n(n+1))"
subject: "Mathematics"
chapter: 6
key: "semester_2/mathematics:interval-radius-convergence-1-n"
source: https://www.bctengineer.com/q/semester_2/mathematics/interval-radius-convergence-1-n
site: BCT Engineering
---

# Interval and radius of convergence, 1/(n(n+1))

Mathematics · Chapter 6: Infinite Series

- Asked in: 2075 Baisakh, 5 marks
- Video: [Power Series & Intervals of Convergence](https://www.youtube.com/watch?v=XHoRBh4hQNU) — Dr. Trefor Bazett
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/interval-radius-convergence-1-n
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-6

## Question

Find the interval and radius of convergence of power series: (1/(1.2))(x−2) + (1/(2.3))(x−2)² + (1/(3.4))(x−2)³ + ... + (1/(n(n+1)))(x−2)ⁿ + ...

## Short answer

Ratio test: |x-2| < 1, R = 1. At x=3, series ∑1/(n(n+1)) converges; at x=1, absolute convergence. Interval: 1 ≤ x ≤ 3.

## Answer

The series is Σ (x − 2)ⁿ/(n(n + 1)), n = 1, 2, 3, …, centred at x = 2.

#### Radius (ratio test)
```text
|uₙ₊₁/uₙ| = |x − 2| · n(n+1)/((n+1)(n+2)) = |x − 2| · n/(n+2) → |x − 2|
```
Convergence for |x − 2| < 1.

**Radius of convergence R = 1.**

Open interval: 1 < x < 3.

#### End points
- x = 3: the series is Σ 1/(n(n + 1)). Compare with Σ 1/n² (limit of ratio = 1): converges.
- x = 1: the series is Σ (−1)ⁿ/(n(n + 1)). Its absolute series is the same convergent series, so it converges (absolutely).

**Answer: R = 1; the interval of convergence is 1 ≤ x ≤ 3.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For Σ (x−2)ⁿ/(n(n+1)), what are the centre and radius?
   A. Centre 2, R = 1
   B. Centre 0, R = 2
   C. Centre 2, R = 2
   D. Centre −2, R = 1
   Answer: A. The ratio tends to |x−2|, so |x−2| < 1: centre 2, R = 1, open interval 1 < x < 3. (Key point: Centre and radius)

2. How is the series Σ 1/(n(n+1)) at x = 3 shown to converge?
   A. Comparison with Σ 1/n
   B. Limit comparison with Σ 1/n², whose ratio tends to 1
   C. Leibniz test
   D. Terms tend to 0
   Answer: B. 1/(n(n+1)) divided by 1/n² tends to 1 and Σ 1/n² converges. (Key point: Endpoint x = 3)

3. At x = 1 the series is Σ (−1)ⁿ/(n(n+1)). What is the interval of convergence?
   A. 1 < x < 3
   B. 1 ≤ x < 3
   C. 1 < x ≤ 3
   D. 1 ≤ x ≤ 3
   Answer: D. At x = 1 the absolute series is the convergent one, so it converges absolutely. Both ends are included: 1 ≤ x ≤ 3. (Key point: Endpoint x = 1 and final interval)

## Close variants set in other exams

- [Interval of convergence Σ2ⁿ(x−3)ⁿ/(n+3)](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Ainterval-convergence-2-x-3) — Mathematics, chapter 6, asked 13×
- [Interval of convergence Σ(−1)ⁿ(x−3)ⁿ/(n+1)](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Ainterval-convergence-1-x-3) — Mathematics, chapter 6, asked 13×
- [Interval of convergence Σ(−1)ⁿxⁿ/(n·2ⁿ)](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Ainterval-convergence-1-x-n) — Mathematics, chapter 6, asked 13×
- [Interval of convergence Σ(−3)ⁿxⁿ/√(n+1)](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Ainterval-convergence-3-x-sqrt) — Mathematics, chapter 6, asked 13×
- [Interval, radius and centre of convergence of Σ2ⁿxⁿ/n!](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Ainterval-radius-centre-convergence-2) — Mathematics, chapter 6, asked 13×

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