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Convergence of 2x+3x²/8+4x³/27+...

Short answer

Converges for 0 < x ≤ 1, diverges for x > 1. Ratio test yields limit x; at x =1, limit comparison with 1/n² shows convergence.

Test the series for convergence or divergence: 2x + 3x²/8 + 4x³/27 + ... + ((n+1)xⁿ/n³) + ... (x > 0).

The series 2x + 3x²/8 + 4x³/27 + … has general term (the denominators 1, 8, 27 are n³ and the numerators 2, 3, 4 are n + 1):

uₙ = (n + 1)xⁿ/n³, n = 1, 2, 3, …

Ratio test

uₙ₊₁/uₙ = [(n+2)xⁿ⁺¹/(n+1)³] · [n³/((n+1)xⁿ)]
        = x · (n+2) n³ / (n+1)⁴
        → x as n → ∞
  • x < 1: converges.
  • x > 1: diverges.

The case x = 1

uₙ = (n + 1)/n³. Compare with vₙ = 1/n²: uₙ/vₙ = (n + 1)/n → 1, finite and non-zero. Σ 1/n² converges (p-series, p = 2 > 1), so by the limit comparison test Σ uₙ converges.

Answer: the series converges for 0 < x ≤ 1 and diverges for x > 1.

✓ Answer checked by Claude Opus 5.5 (high effort)

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