---
title: "Interval of convergence Σ(−1)ⁿxⁿ/(n·2ⁿ)"
subject: "Mathematics"
chapter: 6
key: "semester_2/mathematics:interval-convergence-1-x-n"
source: https://www.bctengineer.com/q/semester_2/mathematics/interval-convergence-1-x-n
site: BCT Engineering
---

# Interval of convergence Σ(−1)ⁿxⁿ/(n·2ⁿ)

Mathematics · Chapter 6: Infinite Series

- Asked in: 2072 Magh, 5 marks; 2070 Magh, 5 marks (asked 2 times)
- Video: [Radius of Convergence and Interval of Convergence | L4 | TYBSc Maths | Power Series @ranjankhatu](https://www.youtube.com/watch?v=xnNiC3qwKfw) — Ranjan Khatu
- 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-n
- 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=1 to ∞) (−1)ⁿ xⁿ/(n·2ⁿ).

## Short answer

Ratio test: |x|/2 < 1 gives R = 2. At x=2, series is alternating harmonic (converges); at x=-2, harmonic (diverges). Interval: -2 < x ≤ 2.

## Answer

The series is Σ (n = 1 to ∞) (−1)ⁿ xⁿ/(n·2ⁿ), centred at 0.

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

**Radius of convergence R = 2.**

Open interval: −2 < x < 2.

#### End points
- x = 2: the series is Σ (−1)ⁿ 2ⁿ/(n 2ⁿ) = Σ (−1)ⁿ/n. Alternating with decreasing terms → 0, so it **converges**.
- x = −2: (−1)ⁿ(−2)ⁿ = 2ⁿ, series = Σ 1/n, the harmonic series, which **diverges**.

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

## Check yourself

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

1. For Σ (−1)ⁿ xⁿ/(n·2ⁿ), |uₙ₊₁/uₙ| → |x|/2. What is the radius of convergence?
   A. R = 1/2
   B. R = 2
   C. R = 1
   D. R = 4
   Answer: B. |x|/2 < 1 gives |x| < 2, so R = 2. (Key point: Radius)

2. At x = 2 the series is Σ (−1)ⁿ/n. What happens?
   A. Diverges like the harmonic series
   B. Converges by the Leibniz test
   C. Converges absolutely
   D. Diverges since terms do not tend to 0
   Answer: B. The terms alternate and decrease to 0, so it converges (conditionally). (Key point: Endpoint x = 2)

3. At x = −2 the series becomes Σ 1/n. What is the final interval?
   A. −2 ≤ x < 2
   B. −2 < x < 2
   C. −2 ≤ x ≤ 2
   D. −2 < x ≤ 2
   Answer: D. At x = −2, (−1)ⁿ(−2)ⁿ = 2ⁿ gives the divergent harmonic series, so −2 is excluded and 2 included. (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)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 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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