What is the interval of convergence for ∑_{n=0}^∞ (-1)^n x^n /(n+1)?

• A. (-1, 1]
• B. (-∞, ∞)
• C. [-1, 1)
• D. [-1, 1]

Answer: (A)

This question tests how to determine where a power-series converges by combining the radius of convergence with endpoint tests. Here the series has coefficients a_n = (-1)^n/(n+1). The n-th root (or ratio) test gives |a_n|^{1/n} = (1/(n+1))^{1/n} → 1, so the radius of convergence is 1. That narrows potential convergence to -1 < x < 1, and then we check the endpoints.

At x = 1, the series becomes sum (-1)^n/(n+1), which is alternating with terms decreasing to zero, so it converges by the Alternating Series Test. It does not converge absolutely, since sum 1/(n+1) diverges, so this endpoint is a conditional convergence point.

At x = -1, the series becomes sum (-1)^n(-1)^n/(n+1) = sum 1/(n+1), which diverges (harmonic series).

So the interval of convergence is -1 < x ≤ 1.