At x = 1, uₙ = (1/n)(1 + 1/n)ⁿ. What is the result?

Diverges, since uₙ/(1/n) → e and Σ 1/n diverges

Converges, since (1+1/n)ⁿ is bounded

Diverges, since terms do not tend to 0

Convergence of Σ(n+1)ⁿxⁿ/n^(n+1)

Short answer

Ratio test: limit |u_{n+1}/u_n| = x; converges for 0 < x < 1. At x=1, term ~ e/n, diverges by comparison with harmonic series. Diverges for x ≥ 1.

Test the convergence of the series Σ (n+1)ⁿ xⁿ / n^(n+1).

[Read as x > 0, as in similar questions; the general term is uₙ = (n + 1)ⁿ xⁿ / n^(n+1).]

uₙ = (n + 1)ⁿ xⁿ/n^(n+1) = (1/n) · ((n + 1)/n)ⁿ · xⁿ = (1/n)(1 + 1/n)ⁿ xⁿ.

uₙ₊₁/uₙ = x · [n/(n+1)] · [(1 + 1/(n+1))ⁿ⁺¹ / (1 + 1/n)ⁿ] → x · 1 · (e/e) = x

uₙ = (1/n)(1 + 1/n)ⁿ. Compare with vₙ = 1/n: uₙ/vₙ = (1 + 1/n)ⁿ → e, finite and positive. Σ 1/n diverges, so Σ uₙ diverges.

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

✓ Answer checked by Claude Opus 5.5 (high effort)