- 2069 Poush · 5 marks
Interval of convergence Σ(−1)ⁿ(x−1)ⁿ/√n
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.
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]
[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)
|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.
✓ Answer checked by Claude Opus 5.5 (high effort)
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