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

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

Mathematics · Chapter 6: Infinite Series

- Asked in: 2069 Poush, 5 marks
- 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-1
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-6

## Question

Find the radius of convergence and the interval of convergence of the power series Σ (n=1 to ∞) (−1)ⁿ(x−1)ⁿ/√n. [Exponent of (x−1) read as n; scan unclear]

## Short answer

Ratio test: |x-1| < 1 → R=1. At x=2, ∑(-1)ⁿ/√n converges (alternating); at x=0, ∑1/√n diverges (p=1/2). Interval: 0 < x ≤2.

## Answer

[The exponent of (x − 1) was unclear in the scan; it is read as n, so the series is Σ (−1)ⁿ(x − 1)ⁿ/√n, n = 1, 2, 3, …]

The centre is x = 1.

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

**Radius of convergence R = 1.**

Open interval: 0 < x < 2.

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

**Answer: R = 1; the interval of convergence is 0 < x ≤ 2.**

## Check yourself

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

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

2. At x = 2 the series is Σ (−1)ⁿ/√n. What happens?
   A. Diverges (p = 1/2)
   B. Converges conditionally (Leibniz)
   C. Converges absolutely
   D. Diverges because terms do not tend to 0
   Answer: B. 1/√n decreases to 0 and signs alternate, so it converges conditionally. (Key point: Endpoint x = 2)

3. At x = 0 the series is Σ 1/√n. What is the interval of convergence?
   A. 0 ≤ x ≤ 2
   B. 0 ≤ x < 2
   C. 0 < x ≤ 2
   D. 0 < x < 2
   Answer: C. Σ 1/√n is a p-series with p ≤ 1, so it diverges; x = 0 is excluded and x = 2 included. (Key point: Endpoint x = 0)

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