At x = 4 the series is Σ (−1)ⁿ/(n+1). What happens?

Converges conditionally by the Leibniz test

Diverges, like the harmonic series

Why does the series diverge at x = 2?

(−1)ⁿ(−1)ⁿ = 1, leaving the harmonic series Σ 1/(n+1)

It is a geometric series with ratio 1

It is alternating with increasing terms

Short answer

Radius 1; interval of convergence is (2, 4] (right endpoint included, left excluded).

Find the radius and interval of convergence of the power series Σ (n=0 to ∞) (−1)ⁿ(x−3)ⁿ/(n+1).

The series is Σ (n = 0 to ∞) (−1)ⁿ(x − 3)ⁿ/(n + 1), centred at x = 3.

|uₙ₊₁/uₙ| = |x − 3| · (n + 1)/(n + 2) → |x − 3|

Absolute convergence for |x − 3| < 1.

Radius of convergence R = 1.

Open interval: 2 < x < 4.

x = 4: x − 3 = 1, series = Σ (−1)ⁿ/(n + 1). Alternating, terms decrease to 0, so it converges (conditionally).
x = 2: x − 3 = −1, so (−1)ⁿ(−1)ⁿ = 1 and the series = Σ 1/(n + 1), the harmonic series, which diverges.

Answer: R = 1; the interval of convergence is 2 < x ≤ 4.

✓ Answer checked by Claude Opus 5.5 (high effort)