At x = 2 the series is Σ (−1)ⁿ/n. What happens?

Diverges like the harmonic series

Diverges since terms do not tend to 0

Interval of convergence Σ(−1)ⁿxⁿ/(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.

Find the radius and interval of convergence of the power series Σ (n=1 to ∞) (−1)ⁿ xⁿ/(n·2ⁿ).

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

|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.

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.

✓ Answer checked by Claude Opus 5.5 (high effort)