- Asked 13/16
- 2073 Bhadra · 5 marks
Interval of convergence Σ(−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.
Find the interval and radius of convergence of the power series Σ (n=1 to ∞) (−3)ⁿxⁿ/√(n+1).
The series is Σ (n = 1 to ∞) (−3)ⁿ xⁿ/√(n + 1), centred at 0.
Radius (ratio test)
|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.
✓ Answer checked by Claude Opus 5.5 (high effort)
Related questions
- Interval of convergence Σ2ⁿ(x−3)ⁿ/(n+3) · Mathematics, chapter 6 · asked 13×
- Interval of convergence Σ(−1)ⁿ(x−3)ⁿ/(n+1) · Mathematics, chapter 6 · asked 13×
- Interval of convergence Σ(−1)ⁿxⁿ/(n·2ⁿ) · Mathematics, chapter 6 · asked 13×
- Interval and radius of convergence, 1/(n(n+1)) · Mathematics, chapter 6 · asked 13×
- Interval, radius and centre of convergence of Σ2ⁿxⁿ/n! · Mathematics, chapter 6 · asked 13×
- Interval of convergence Σ(−1)ⁿ(x−1)ⁿ/√n · Mathematics, chapter 6
- Convergence of series (n²−1)/(n²+1)xⁿ · Mathematics, chapter 6 · asked 2×