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

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

Mathematics · Chapter 6: Infinite Series

- Asked in: 2073 Bhadra, 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-convergence-3-x-sqrt
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-6

## Question

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

## Short answer

Ratio test: 3|x| < 1 gives R = 1/3. At x=1/3, alternating ∑(-1)ⁿ/√(n+1) converges; at x=-1/3, ∑1/√(n+1) with p=1/2 diverges. Interval: -1/3 < x ≤ 1/3.

## Answer

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

#### Radius (ratio test)
```text
|uₙ₊₁/uₙ| = 3|x| · √((n+1)/(n+2)) → 3|x|
```
Convergence for 3|x| < 1, i.e. |x| < 1/3.

**Radius of convergence R = 1/3.**

Open interval: −1/3 < x < 1/3.

#### End points
- x = 1/3: (−3)ⁿ(1/3)ⁿ = (−1)ⁿ, so the series is Σ (−1)ⁿ/√(n + 1). Alternating, terms decrease to 0, so it **converges** (conditionally).
- x = −1/3: (−3)ⁿ(−1/3)ⁿ = 1, so the series is Σ 1/√(n + 1), a p-series with p = 1/2 ≤ 1, which **diverges**.

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

## Check yourself

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

1. For Σ (−3)ⁿxⁿ/√(n+1), what is the radius of convergence?
   A. R = 3
   B. R = 1/3
   C. R = 1
   D. R = ∞
   Answer: B. The ratio tends to 3|x|, so 3|x| < 1 gives |x| < 1/3. (Key point: Radius)

2. At x = 1/3 the series is Σ (−1)ⁿ/√(n+1). What is the behaviour?
   A. Diverges, p-series with p = 1/2
   B. Converges conditionally by the Leibniz test
   C. Converges absolutely
   D. Diverges because terms do not tend to 0
   Answer: B. It alternates with terms decreasing to 0, so it converges, but not absolutely. (Key point: Endpoint x = 1/3)

3. At x = −1/3 the series is Σ 1/√(n+1). Why does it diverge, and what is the interval?
   A. p-series with p = 1/2 ≤ 1; interval −1/3 < x ≤ 1/3
   B. p-series with p = 1/2 ≤ 1; interval −1/3 ≤ x < 1/3
   C. Geometric series; interval −1/3 < x < 1/3
   D. Terms do not tend to 0; interval −1/3 ≤ x ≤ 1/3
   Answer: A. Σ n^(−1/2) diverges as p ≤ 1, so −1/3 is excluded and 1/3 included. (Key point: Endpoint x = −1/3)

## 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 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, 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×

## Related in Mathematics

- [Interval of convergence Σ(−1)ⁿ(x−1)ⁿ/√n](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Ainterval-convergence-1-x-1) — Mathematics, chapter 6
- [Convergence of series (n²−1)/(n²+1)xⁿ](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aconvergence-series-n2-1-n2) — Mathematics, chapter 6, asked 2×