How is the series Σ 1/(n(n+1)) at x = 3 shown to converge?

Limit comparison with Σ 1/n², whose ratio tends to 1

Interval and radius of convergence, 1/(n(n+1))

Short answer

Ratio test: |x-2| < 1, R = 1. At x=3, series ∑1/(n(n+1)) converges; at x=1, absolute convergence. Interval: 1 ≤ x ≤ 3.

Find the interval and radius of convergence of power series: (1/(1.2))(x−2) + (1/(2.3))(x−2)² + (1/(3.4))(x−2)³ + ... + (1/(n(n+1)))(x−2)ⁿ + ...

The series is Σ (x − 2)ⁿ/(n(n + 1)), n = 1, 2, 3, …, centred at x = 2.

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

Convergence for |x − 2| < 1.

Radius of convergence R = 1.

Open interval: 1 < x < 3.

x = 3: the series is Σ 1/(n(n + 1)). Compare with Σ 1/n² (limit of ratio = 1): converges.
x = 1: the series is Σ (−1)ⁿ/(n(n + 1)). Its absolute series is the same convergent series, so it converges (absolutely).

Answer: R = 1; the interval of convergence is 1 ≤ x ≤ 3.

✓ Answer checked by Claude Opus 5.5 (high effort)

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