---
title: "Interval of convergence Σ(−1)ⁿ(x−3)ⁿ/(n+1)"
subject: "Mathematics"
chapter: 6
key: "semester_2/mathematics:interval-convergence-1-x-3"
source: https://www.bctengineer.com/q/semester_2/mathematics/interval-convergence-1-x-3
site: BCT Engineering
---

# Interval of convergence Σ(−1)ⁿ(x−3)ⁿ/(n+1)

Mathematics · Chapter 6: Infinite Series

- Asked in: 2072 Ashwin, 5 marks; 2071 Bhadra, 5 marks; 2070 Bhadra, 5 marks; 2068 Bhadra, 5 marks (asked 4 times)
- Video: [find the radius of convergence of the given power series](https://www.youtube.com/watch?v=uPO5TW_LBKQ) — Math Eager
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/interval-convergence-1-x-3
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-6

## Question

Find the radius and interval of convergence of the power series Σ (n=0 to ∞) (−1)ⁿ(x−3)ⁿ/(n+1).

## Short answer

Radius 1; interval of convergence is (2, 4] (right endpoint included, left excluded).

## Answer

The series is Σ (n = 0 to ∞) (−1)ⁿ(x − 3)ⁿ/(n + 1), centred at x = 3.

#### Radius (ratio test)
```text
|uₙ₊₁/uₙ| = |x − 3| · (n + 1)/(n + 2) → |x − 3|
```
Absolute convergence for |x − 3| < 1.

**Radius of convergence R = 1.**

Open interval: 2 < x < 4.

#### End points
- x = 4: x − 3 = 1, series = Σ (−1)ⁿ/(n + 1). Alternating, terms decrease to 0, so it **converges** (conditionally).
- x = 2: x − 3 = −1, so (−1)ⁿ(−1)ⁿ = 1 and the series = Σ 1/(n + 1), the harmonic series, which **diverges**.

**Answer: R = 1; the interval of convergence is 2 < x ≤ 4.**

## Check yourself

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

1. For Σ (−1)ⁿ(x−3)ⁿ/(n+1), what is the radius of convergence?
   A. R = 3
   B. R = 1
   C. R = 1/2
   D. R = ∞
   Answer: B. |uₙ₊₁/uₙ| → |x−3|, so convergence needs |x−3| < 1, giving R = 1 and the open interval 2 < x < 4. (Key point: Radius via ratio test)

2. At x = 4 the series is Σ (−1)ⁿ/(n+1). What happens?
   A. Diverges, like the harmonic series
   B. Converges conditionally by the Leibniz test
   C. Converges absolutely
   D. Test is inconclusive
   Answer: B. It alternates with decreasing terms tending to 0, so it converges, but the absolute series Σ1/(n+1) diverges, so only conditionally. (Key point: Endpoint x = 4)

3. Why does the series diverge at x = 2?
   A. Terms do not tend to 0
   B. (−1)ⁿ(−1)ⁿ = 1, leaving the harmonic series Σ 1/(n+1)
   C. It is a geometric series with ratio 1
   D. It is alternating with increasing terms
   Answer: B. With x−3 = −1 the signs cancel, giving Σ 1/(n+1), which diverges. Hence the interval is 2 < x ≤ 4. (Key point: Endpoint x = 2)

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