At x = 7/2 the series becomes Σ 1/(n+3). What is its behaviour?

Diverges, like the harmonic series

Converges, as a p-series with p = 1

Converges absolutely because terms tend to 0

Interval of convergence Σ2ⁿ(x−3)ⁿ/(n+3)

Short answer

Radius 1/2; interval of convergence is [5/2, 7/2) (centre 3, left endpoint included, right excluded).

Find the interval and radius of convergence of the power series Σ (n=1 to ∞) 2ⁿ(x−3)ⁿ/(n+3).

The series is Σ uₙ with uₙ = 2ⁿ(x − 3)ⁿ/(n + 3), n = 1, 2, 3, … It is a power series with centre 3.

|uₙ₊₁/uₙ| = 2|x − 3| · (n + 3)/(n + 4) → 2|x − 3| as n → ∞

The series converges absolutely if 2|x − 3| < 1, i.e. |x − 3| < 1/2.

Radius of convergence R = 1/2.

The open interval is 5/2 < x < 7/2.

x = 7/2: x − 3 = 1/2, so uₙ = 1/(n + 3). Σ 1/(n + 3) behaves like the harmonic series (limit comparison with 1/n), so it diverges.
x = 5/2: x − 3 = −1/2, so uₙ = (−1)ⁿ/(n + 3). This is an alternating series with decreasing terms tending to 0, so it converges (Leibniz test).

Answer: R = 1/2; the interval of convergence is 5/2 ≤ x < 7/2 (centre 3, left end included, right end excluded).

✓ Answer checked by Claude Opus 5.5 (high effort)